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https://oeis.org/A333985/internal | 1,653,633,968,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00273.warc.gz | 508,475,245 | 2,920 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A333985 a(0) = 0; a(n) = 6^(n-1) + (1/n) * Sum_{k=1..n-1} binomial(n,k)^2 * 6^(k-1) * (n-k) * a(n-k). 4
%I
%S 0,1,8,102,2448,99576,6070032,517803840,58901955840,8614609282944,
%T 1574889814326528,351896788824053760,94354291010501932032,
%U 29899137879209196380160,11053567519385396409446400,4715135497874174650128617472,2298676381054790419739595571200,1270045124912998373344157769891840
%N a(0) = 0; a(n) = 6^(n-1) + (1/n) * Sum_{k=1..n-1} binomial(n,k)^2 * 6^(k-1) * (n-k) * a(n-k).
%F Sum_{n>=0} a(n) * x^n / (n!)^2 = -log((7 - BesselI(0,2*sqrt(6*x))) / 6).
%t a[0] = 0; a[n_] := a[n] = 6^(n - 1) + (1/n) Sum[Binomial[n, k]^2 6^(k - 1) (n - k) a[n - k], {k, 1, n - 1}]; Table[a[n],{n, 0, 17}]
%t nmax = 17; CoefficientList[Series[-Log[(7 - BesselI[0, 2 Sqrt[6 x]])/6], {x, 0, nmax}], x] Range[0, nmax]!^2
%Y Cf. A102223, A333981, A333982, A333983, A333984, A337597.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Sep 04 2020
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Last modified May 27 02:41 EDT 2022. Contains 354093 sequences. (Running on oeis4.) | 608 | 1,405 | {"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-2022-21 | latest | en | 0.438728 |
https://www.stat.math.ethz.ch/pipermail/bioconductor/2010-August/034866.html | 1,643,068,456,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00474.warc.gz | 1,056,768,819 | 3,557 | # [BioC] how does limma calculate the coefficients with using different design matrix?
Gordon K Smyth smyth at wehi.EDU.AU
Tue Aug 17 01:27:40 CEST 2010
```Dear Xiaokuan,
Your example simply shows that the two design matrices are different, as
they naturally are. You haven't actually used contrasts.fit() at all.
Do you understand how it is intended to be used?
You need something like this:
fit<-lmFit(dat,designcontr)
fit <- contrasts.fit(fit,cont.matrix_contr)
topTable(eBayes(fit))
and
fit<-lmFit(dat,design)
fit <- contrasts.fit(fit,cont.matrix)
topTable(eBayes(fit))
When I do run the above code, I get exactly the same results from the two
approaches, just as it supposed to happen.
Best wishes
Gordon
On Mon, 16 Aug 2010, Xiaokuan Wei wrote:
Dear Gordon,
Thank you for your detailed explaination.
In fact, my data don't have missing values and in function of lmFit, I didn't
use weights argument.
I dig deeper for this issue trying to see what the coefficients are after lmFit.
When use designcontr:
fit<-lmFit(dat,designcontr)
tp1<-topTable(eBayes(fit))
tp1<-topTable(eBayes(fit),n=Inf)
tp1[tp1\$ID=='feature1',]
ID g0 g1 g2 g3 g4 g5
g6 g7 g8 g9 AveExpr F
1 feature1 11.32317 -0.07976296 2.900216 3.06401 3.210565 0.2295532 1.545972
2.433364 2.62326 2.719911 13.18788 111975.0
1 5.284493e-27 5.296354e-25
When use design:
fit<-lmFit(dat,design)
tp2<-topTable(eBayes(fit),n=Inf)
tp2[tp2\$ID=='feature1',]
ID g0 g1 g2 g3 g4 g5
g6 g7 g8 g9 AveExpr F
1 feature1 11.32317 11.24341 11.55272 12.86914 13.75653 13.94643 14.04308
14.22338 14.38718 14.53373 13.18788 111975.0
1 5.284493e-27 5.296354e-25
So, as you said, if I use contrasts.fit with designcontr and cont.matrix_contr,
I get exact the coefficents of g1 to g9 for coef1 to coef9.
but when I used design and contr.matrix, only the coef1 is
g1-g0=11.24341-11.32317=-0.07976 which is the same vaule as using designcontr
and cont.matrix_contr. however, the other coefficents from coef2 to
coef9 ARE the difference between each g[i] with g0. but the results are
DIFFERENT from using designcontr and cont.matrix_contr; as you can see that I
didn't use weights, and my data have no missing values.
Do you have any comments on this? Thank you very much.
Xiaokuan
________________________________
From: Gordon K Smyth <smyth at wehi.EDU.AU>
To: Xiaokuan Wei <weixiaokuan at yahoo.com>
Cc: Bioconductor mailing list <bioconductor at stat.math.ethz.ch>
Sent: Sun, August 15, 2010 10:40:58 PM
Subject: [BioC] how does limma calculate the coefficients with using different
design matrix?
Dear Xiaokuan,
You haven't told us about your data, but it must contain missing values or
weights, because otherwise the two topTable results you give would have
been identical. In the absence of missing values or weights, the results
would have been as you expected.
With missing values or weights, the first topTable result you give (the
one with designcontr) is exact whereas the second result is approximate.
The fact that contrast.fit() gives approximate results in the presence of
missing values or weights is mentioned on the documentation page for
contrast.fit, and has been discussed a few times on this list, see
https://www.stat.math.ethz.ch/pipermail/bioconductor/2009-December/030947.html
Professor Albyn Jones is trying to pursuade me to implement something
similar to contrasts.fit() that would be exact in all cases. The only way
I could do this would be to add an argument 'contrasts' to lmFit. I may
yet do that!
Best wishes
Gordon
> Date: Sat, 14 Aug 2010 12:30:57 -0700 (PDT)
> From: Xiaokuan Wei <weixiaokuan at yahoo.com>
> To: bioconductor <bioconductor at stat.math.ethz.ch>
> Subject: [BioC] how does limma calculate the coefficients with using
> different design matrix?
>
> Dear list,
>
> I have two design matrixes:
>
> design<-model.matrix(~-1+factor(grp))
> designcontr<-model.matrix(~factor(grp))
>
> then make contrast matrix:
> cont.matrix<-rbind(rep(-1,9),diag(9))
> cont.matrix_contr <- rbind(rep(0,9),diag(9))
>
> after using topTable to get the specific probe values, I found:
> e.g.
> using designcontr and cont.matrix_contr:
> ID Coef1 Coef2 Coef3 Coef4 Coef5 Coef6 Coef7 Coef8 Coef9 AveExpr F P.Value
>
> feature1 -0.07976 2.900216 3.06401 3.210565 0.229553 1.545972 2.433364 2.62326
> 2.719911 13.18788 1133.135 1.60E-15 7.20E-11
>
>
> using design and cont.matrix:
> ID Coef1 Coef2 Coef3 Coef4 Coef5 Coef6 Coef7 Coef8 Coef9 AveExpr F P.Value
>
> feature1 -0.07976 0.229553 1.545972 2.433364 2.62326 2.719911 2.900216 3.06401
> 3.210565 13.18788 1133.135 1.60E-15 7.20E-11
>
>
>
> I try to understand why there is difference between cont.matrix_contr and
> cont.matrx results, since both of them is try to get the difference between
>each
> grp(grp2 to grp10) and grp1. i.e. grp[i](i=2-10) - grp[1]
>
> even I used the different contrasts, the coef should be the same. I am
> confused.
>
>
> Thank you for your help.
>
> Regards,
> Xiaokuan
``` | 1,615 | 5,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.622294 |
https://thisrunsforyou.com/qa/quick-answer-what-is-percentage-sign-called.html | 1,607,011,465,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00469.warc.gz | 456,684,616 | 8,011 | # Quick Answer: What Is Percentage Sign Called?
## How do you write percentages in English?
Percent, Per Cent and PercentagesNumbers up to ten are usually written as words, while larger numbers are written as numerals.When a percentage is written as a word, it should be followed by “percent”When a percentage is written as a numeral, it should be followed by the “%” sign..
## What is a friendly percent?
Friendly percents are percents, such as 10%, 50%, and 25%. The chart shows a summary of how to use each friendly percent to find the percent of a number. Percent. Fraction.
## What does this sign mean in math?
This symbol > means greater than, for example 4 > 2. ≤ ≥ These symbols mean ‘less than or equal to’ and ‘greater than or equal to’ and are commonly used in algebra. In computer applications <= and >= are used. ≪ ≫ These symbols are less common and mean much less than, or much greater than.
## What does the word most mean?
adjective, superlative of much or many, with more as comparative. in the greatest quantity, amount, measure, degree, or number: to win the most votes. in the majority of instances: Most operations are successful.
## Can percent be more than 100?
“It’s impossible to give more than 100%,” they’ll say. “That’s what ‘percent’ means.” But of course percentages greater than 100 are possible. … It all depends on what your baseline is — x percent of what.
## What is God’s number in the Bible?
“God” or “Lord” appear in the King James Bible a total of 10,875 times. 1 plus 8 plus 7 plus 5 equals 21, or 7 times 3. There are 22 chapters in Revelation (see # 9), and the last chapter contains no violence of any kind.
## What is the percentage symbol called?
%The percent (per cent) sign % is the symbol used to indicate a percentage, a number or ratio as a fraction of 100.
## What does percent mean?
Definition: A percent is a ratio whose second term is 100. Percent means parts per hundred. The word comes from the Latin phrase per centum, which means per hundred. … This time the table includes percents.
## How do you write a percent sign?
The percent sign ( % ) is made on most American keyboards by holding the SHIFT key and pressing the number 5 on the top number line. The percent sign can also be made by choosing INSERT SYMBOL and selecting the % sign.
## What is the meaning of 100 %?
: completely, entirely I agree with her 100 percent.
## Can most mean all?
most – this is in the direction of ‘all’ and is comparative, meaning more than any other. The general idea is that it is ‘almost all’ but because it is so vague, it can be used for anything that is more than anything else.
## What percent is most of the time?
FrequencyAdverbFrequency it representsalways100 percent of the time (or almost 100%).usuallyabout 80 percent of the time.oftenmore than half of the time.sometimesless than half of the time.2 more rows
## What percentage is most?
While “most” literally just means more than half, it is best used to mean something like 60% to 90%. Below 60% you should consider terms like “more than half” or “a small majority”. Above 90% you get in the territory of “almost all” or “a large majority”. Of course, this depends on context.
## What is meant by 10%?
So it tells you a ratio “out of 100” Example: 10% means 10 out of every 100. So if 10% of 500 people have ice cream, then 50 people have ice cream. Introduction to Percentages.
## What kind of number is 100?
100 or one hundred (Roman numeral: C) is the natural number following 99 and preceding 101.
## How do you write less than 1?
L Method. This method is pretty simple—”less than” starts with a letter L, so the symbol that looks most like an L is the one that means “less than.” < looks more like an L than >, so < means “less than.”
## Can you end a sentence with a percentage?
This is not the case with the percent sign. You still need to add a full stop after, if it is written at the end of a sentence. If the percent sign is used in the middle of a sentence then we do not use a full stop here. I received a 5% increase in my salary.
## How do you describe percentages in words?
The noun percentage requires an adjective to describe its size (eg, “a large percentage”) when it does not refer to specific numbers in the sentence. Some style guides recommend a numeral and the symbol % rather than the word percent, even if the number is less than 10 (eg, “A total of 5%…”). | 1,068 | 4,429 | {"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} | 3.40625 | 3 | CC-MAIN-2020-50 | latest | en | 0.94239 |
https://learn.careers360.com/ncert/questions/cbse_11_class-maths-conic_section-miscellaneous_exercise/ | 1,585,600,830,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00170.warc.gz | 565,878,450 | 94,138 | ## Filters
Clear All
P Pankaj Sanodiya
As we know that if a point moves in a plane in such a way that its distance from two-point remain constant then the path is an ellipse. Now, According to the question, the distance between the point from where the sum of the distance from a point is constant = 10 Now, the distance between the foci=8 Now, As we know the relation, Hence the equation of the ellipse is, Hence the path of...
P Pankaj Sanodiya
Given, an equilateral triangle inscribed in parabola with the equation. The one coordinate of the triangle is A(0,0). Now, let the other two coordinates of the triangle are and Now, Since the triangle is equilateral, The coordinates of the points of the equilateral triangle are, So, the side of the triangle is
P Pankaj Sanodiya
Given the parabola, Comparing this equation with , we get Now, As we know the coordinates of ends of latus rectum are: So, the coordinates of latus rectum are, Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3) Widht of the triangle = 2*6=12 Height of the triangle = 3 So The area = Hence the required area is 18 unit square.
P Pankaj Sanodiya
Let be the angle that rod makes with the ground, Now, at a point 3 cm from the end, At the point touching the ground Now, As we know the trigonometric identity, Hence the equation is,
P Pankaj Sanodiya
The equation of the semi-ellipse will be of the form Now, According to the question, the length of major axis = 2a = 8 The length of the semimajor axis =2 Hence the equation will be, Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5 So, the height at this point is Hence the height of the required point is 1.56 m. | 458 | 1,689 | {"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": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-16 | latest | en | 0.836556 |
https://www.ajdesigner.com/phpgutter/curb_opening_inlet_orifice_equation_gravity.php | 1,713,582,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817474.31/warc/CC-MAIN-20240420025340-20240420055340-00025.warc.gz | 589,166,606 | 7,030 | # Curb Opening Inlet Orifice Equations Formulas Design Calculator
Solving for acceleration of gravity.
Note: equation applies to flow depths greater than 1.4 the opening height
## Inputs:
capture flow rate (Qi)
area of opening (A)
flow depth (d)
height of opening (h)
## Conversions:
capture flow rate (Qi)
= 0
= 0
foot^3/second
area of opening (A)
= 0
= 0
foot^2
flow depth (d)
= 0
= 0
foot
height of opening (h)
= 0
= 0
foot
## Solution:
acceleration of gravity (g)
= NOT CALCULATED
## Other Units:
Change Equation
Select to solve for a different unknown
curb type gutter
gutter flow rate Manning's coefficient roadway cross slope longitudinal slope flow width
curb opening inlet with weir operation
capture flow rate or interception capacity curb opening length curb opening width flow depth at curb
curb opening inlet with weir operation
no depression
capture flow rate or interception capacity curb opening length depth of flow
curb opening inlet with orifice operation
capture flow rate or interception capacity curb opening area acceleration of gravity flow depth at curb opening height
gutter opening inlet
capture flow rate or interception capacity grate perimeter flow depth
slotted pipe drain inlet
pipe length for complete interception flow rate longitudinal slope Manning's coefficient roadway cross slope
gutter capture efficiency
gutter capture efficiency capture flow rate total flow rate
gutter carryover or bypass
gutter carry over or bypass total flow rate capture flow rate
Where
Q = gutter flow rate n = Manning's coefficient Sx = roadway cross slope S = longitudinal slope T = flow width d = flow depth at curb a = depressing depth
References - Books:
Martin Wanielista, Robert Kersten and Ron Eaglin. 1997. Hydrology Water Quantity and Quality Control. John Wiley & Sons. 2nd ed.
Online Web Apps, Rich Internet Application, Technical Tools, Specifications, How to Guides, Training, Applications, Examples, Tutorials, Reviews, Answers, Test Review Resources, Analysis, Homework Solutions, Worksheets, Help, Data and Information for Engineers, Technicians, Teachers, Tutors, Researchers, K-12 Education, College and High School Students, Science Fair Projects and Scientists
By Jimmy Raymond
Contact: aj@ajdesigner.com | 502 | 2,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-18 | latest | en | 0.76684 |
http://www.caraudio.com/forums/enclosure-design-construction-help/81483-calculating-cu-ft-fiberglass.html | 1,529,318,024,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860168.62/warc/CC-MAIN-20180618090026-20180618110026-00617.warc.gz | 390,902,547 | 14,283 | 1. ## Calculating Cu.ft. In Fiberglass
Here Is A ?. How Would You Calculate The Area Of A Sealed Fiberlass Enclosure (for Subs) With An Ambiguous Shape, Its Not Like You Can Break It Down Into Sections And Calculate It, Its All Convex And Concave Curves!!!! I See All These Crazy Ameoba Shaped Enclosures And I Wonder How They Calculated The Area To Match The Subs Requirements
2. ## Re: Calculating Cu.ft. In Fiberglass
I always say sand or water... more accurate the packing peanuts.
3. ## Re: Calculating Cu.ft. In Fiberglass
Originally Posted by NjccBflo
I always say sand or water... more accurate the packing peanuts.
GOOD POINT, BUT THATS ONLY GOOD IF THE MAGICAL SHAPE YOU CHOOSE JUST BY LUCK HAPPEN TO MEASURE OUT OK, MY REAL FOCUS IS IN THE PREDESIGN, ITS NOT GOOD IF YOU BUILD IT FILL IT FULL OF WATER AND THEN FIND OUT ITS A RIDCULOUSLY USELESS DESIGN, ALSO WATER DOESNT WORK IF ITS ALSO PARTIALLY MDF!!!
4. ## Re: Calculating Cu.ft. In Fiberglass
Originally Posted by chevy6998
{CAPS LOCK}
Best to build too big and back fill it. You get the hang of "guestimating" after the first couple of projects.
-and-
Thats why SAND is always an option.
5. ## Re: Calculating Cu.ft. In Fiberglass
second you should have a rough design build it a tad big then you can add small junks of wood to make it the right size. if it happens to be too small add a little polyfill.
third good luck and its not an exact science sort of guess and checkl
Danny
6. ## Re: Calculating Cu.ft. In Fiberglass
Originally Posted by Nitropyro666
second you should have a rough design build it a tad big then you can add small junks of wood to make it the right size. if it happens to be too small add a little polyfill.
third good luck and its not an exact science sort of guess and checkl
Danny
sorry not being mean, im just short and to the point, didn't realize how it sounded till i read it the second time, i apologize
7. ## Re: Calculating Cu.ft. In Fiberglass
i just thought there may be more o f a science to it
8. ## Re: Calculating Cu.ft. In Fiberglass
Originally Posted by chevy6998
i just thought there may be more o f a science to it
there can be. Whoa there can be.
But... You (and alot of others on this forum) should realize that, unles you are very seriously competing then:
-If your tuning is 1.03hz off, it's not the end of the world.
-If your vB is off 0.0000786^cu ft, it's not the end of the world.
-If your SSF isn't dialed in exactly right, your woofer will not launch it's cone.
-Nick
9. ## Re: Calculating Cu.ft. In Fiberglass
Originally Posted by NjccBflo
there can be. Whoa there can be.
But... You (and alot of others on this forum) should realize that, unles you are very seriously competing then:
-If your tuning is 1.03hz off, it's not the end of the world.
-If your vB is off 0.0000786^cu ft, it's not the end of the world.
-If your SSF isn't dialed in exactly right, your woofer will not launch it's cone.
-Nick
Theres and exact science to just about any thing, but most of the answers I got were guestimation solutions. if you know any sites or books on the subject let me know
10. ## Re: Calculating Cu.ft. In Fiberglass
You dont want to get yourself into it...its much more complex than just plugging numbers in.
11. ## Re: Calculating Cu.ft. In Fiberglass
Originally Posted by chevy6998
Theres and exact science to just about any thing, but most of the answers I got were guestimation solutions. if you know any sites or books on the subject let me know
Well yeah there is an exact science to this too, just simply build it and plan on it being large, figure how much air space there is when you are done, now look at your numbers, you needed 1ft^3 you are at 1.1, 1.5, or 1.9549349, it doesnt matter which, now just add the correct amount of filling and your done, it doesnt take long at all and it isnt too hard. Sorry if i come off rude im not trying to be at all im just laying it out for you.
12. ## Re: Calculating Cu.ft. In Fiberglass
there is another way---if you have the room. you simply build the enclosure out of mdf to the right specs and then use fiberglass purely as cosmetics to fit the contour of the space you want it to go into.
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• You may edit your posts | 1,129 | 4,348 | {"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-2018-26 | latest | en | 0.805374 |
https://stackoverflow.com/questions/7610631/glsl-mod-vs-hlsl-fmod | 1,702,139,221,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00169.warc.gz | 569,386,243 | 43,533 | # Glsl mod vs Hlsl fmod
I've implemented the spiral GLSL shader described in this question in HLSL, but the results are not the same. I think it's because of the `mod` function in GLSL that I've translated to `fmod` in HLSL. I suspect that this problem only happens when we have negative numbers in the input of the `fmod` function.
I've tried replacing the call to `mod` by a call to a function that I've made which does what is described in the GLSL documentation and it works:
`mod` returns the value of `x` modulo `y`. This is computed as `x - y * floor(x/y)`.
The working code I use instead of `fmod` is:
``````float mod(float x, float y)
{
return x - y * floor(x/y)
}
``````
By contrast to GLSL `mod`, MSDN says the HLSL `fmod` function does this:
The floating-point remainder is calculated such that `x = i * y + f`, where `i` is an integer, `f` has the same sign as `x`, and the absolute value of `f` is less than the absolute value of `y`.
I've used an HLSL to GLSL converter, and the `fmod` function is translated as `mod`. However, I don't know if I can assume that `mod` translates to `fmod`.
### Questions
1. What are the differences between GLSL `mod` and HLSL`fmod`?
2. How can I translate MSDN's cryptic description of `fmod` to a pseudo-code implementation?
``````uniform float time;
uniform vec2 resolution;
uniform vec2 aspect;
void main( void ) {
vec2 position = -aspect.xy + 2.0 * gl_FragCoord.xy / resolution.xy * aspect.xy;
float angle = 0.0 ;
if (position.x != 0.0 && position.y != 0.0){
angle = degrees(atan(position.y,position.x)) ;
}
float amod = mod(angle+30.0*time-120.0*log(radius), 30.0) ;
if (amod<15.0){
gl_FragColor = vec4( 0.0, 0.0, 0.0, 1.0 );
} else{
gl_FragColor = vec4( 1.0, 1.0, 1.0, 1.0 );
}
}
``````
``````struct Psl_VertexShaderInput
{
float3 pos : POSITION;
};
{
float4 pos : POSITION;
};
{
float4 Output0 : COLOR0;
};
float3 psl_positionOffset;
float2 psl_dimension;
{
psl_output.pos = float4(psl_input.pos + psl_positionOffset, 1);
return psl_output;
}
float time : TIME;
float2 resolution : DIMENSION;
{
float2 aspect = float2(resolution.x / resolution.y, 1.0);
float2 position = -aspect.xy + 2.0 * pos.xy / resolution.xy * aspect.xy;
float angle = 0.0;
if (position.x != 0.0 && position.y != 0.0)
{
angle = degrees(atan2(position.y, position.x));
}
float amod = fmod((angle + 30.0 * time - 120.0 * log(radius)), 30.0);
if (amod < 15.0)
{
psl_output.Output0 = float4(0.0, 0.0, 0.0, 1.0);
return psl_output;
}
else
{
psl_output.Output0 = float4(1.0, 1.0, 1.0, 1.0);
return psl_output;
}
}
technique Default
{
pass P0
{
As you've noted, they're different. The GLSL `mod` will always have the same sign as `y` rather than `x`. Otherwise it's the same -- a value `f` such that `x = i*y + f` where `i` is an integer and `|f| < |y|`. If you're trying to make a repeating pattern of some kind, the GLSL `mod` is generally what you want.
For comparison, the HLSL `fmod` is equivalent to `x - y * trunc(x/y)`. They're the same when `x/y` is positive, different when `x/y` is negative. | 967 | 3,048 | {"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} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.742115 |
https://pythonexamples.org/python-numpy-reshape-array/ | 1,716,957,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00208.warc.gz | 379,603,290 | 16,586 | # Reshape Array in NumPy
Contents
## NumPy – Reshape Array
To reshape a given array to specific shape using NumPy library, we can use numpy.reshape() function.
Pass the given array, and required shape (as tuple) as arguments to the numpy.reshape() function.
## Examples
### 1. Reshape numpy array from (3,4) to (2,6)
In the following program, we reshape a numpy array of shape (3, 4) to (2, 6).
Python Program
``````import numpy as np
# Reshape (3, 4) array to (6, 2)
arr = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
shape = (6, 2)
output = np.reshape(arr, shape)
print(output)``````
Run Code Copy
Output
``````[[ 1 2]
[ 3 4]
[ 5 6]
[ 7 8]
[ 9 10]
[11 12]]``````
### 2. Reshape numpy array from (3,4) to (-1,2)
If any of the dimension in the input shape is given as -1, then this dimension is adjusted based on the length in other dimensions.
For example, in the following program, we reshape a numpy array of shape (3, 4) to (-1, 2). Since we have -1 for the first dimension in the output shape, that dimension’s length is computed from (3*4)/(2) which is 6.
Python Program
``````import numpy as np
# Reshape (3, 4) array to (6, 2)
arr = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
shape = (-1, 2)
output = np.reshape(arr, shape)
print(output)``````
Run Code Copy
Output
``````[[ 1 2]
[ 3 4]
[ 5 6]
[ 7 8]
[ 9 10]
[11 12]]``````
## Summary
In this NumPy Tutorial, we learned how to reshape a given numpy array in Python using numpy.reshape() function.
## Related Tutorials
Code copied to clipboard successfully 👍 | 540 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.804866 |
https://www.debate.org/polls/does-mathematics-prove-objectivisim-or-absolute-truths | 1,611,367,539,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531702.36/warc/CC-MAIN-20210123001629-20210123031629-00487.warc.gz | 744,537,248 | 9,940 | # Does Mathematics prove Objectivisim or Absolute Truths?
Posted by: Dishoungh
• Yes, duh
• No, it doesn't
(Maximum 900 words)
ArcTImes says2014-06-01T01:10:30.7702329-05:00
Define absolute truth and objectivism.
discomfiting says2014-06-01T01:31:54.3727418-05:00
Any particular reason why Lana Del Rey is "pro" side?
missrepresented says2014-06-01T14:56:08.8777254-05:00
In general, absolute truth is whatever is always valid, regardless of parameters or context. The absolute in the term connotes one or more of: a quality of truth that cannot be exceeded; complete truth; unvarying and permanent truth. It can be contrasted to relative truth or truth in a more ordinary sense in which a degree of relativity is implied.
missrepresented says2014-06-01T14:57:39.4825254-05:00
In pure mathematics , however, there is said to be a proof for the existence of absolute truth. A common tactic in mathematical proofs is the use of reductio ad absurdum , in which the statement to be proved is denied as a premise, and then that premise is shown to lead to a contradiction. When it can be demonstrated that the negation of a statement leads to a contradiction, then the original statement is proved true. The logical proof of the statement, "There exists an absolute truth," is almost trivial in its simplicity. Suppose we assert the negation of the statement, that is, that there is no such thing as absolute truth. By making that assertion, we claim that the sentence "There exists no absolute truth" is absolutely true. The statement is self-contradictory, so its negation, "There exists an absolute truth," is true. This proof applies only to logic. It does not tell us whether any particular statement other than itself is true. It does not prove the existence (or non-existence) of God, the devil, heaven, hell, or little green people from another galaxy. Neither does it assert that we can always ascertain the truth or falsity of any arbitrary statement. The Incompleteness Theorem , proved by Kurt Gödel and published in 1931, actually showed that there exist logical statements whose truth value is undecidable, that is, they cannot be proved either true or false. November 2010 Posted by: Margaret Rouse | 539 | 2,213 | {"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} | 3.15625 | 3 | CC-MAIN-2021-04 | latest | en | 0.913509 |
https://www.esaral.com/q/a-12-pf-capacitor-is-connected-to-a-50v-battery-64953 | 1,726,826,143,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00305.warc.gz | 695,773,783 | 11,530 | # A 12 pF capacitor is connected to a 50V battery.
Question:
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Solution:
Capacitor of the capacitance, $C=12 \mathrm{pF}=12 \times 10^{-12} \mathrm{~F}$
Potential difference, = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
$E=\frac{1}{2} C V^{2}$
$=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^{2}$
$=1.5 \times 10^{-8} \mathrm{~J}$
Therefore, the electrostatic energy stored in the capacitor is $1.5 \times 10^{-8} \mathrm{~J}$. | 194 | 578 | {"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} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.729367 |
https://www.mdpi.com/2075-1680/7/2/22/htm | 1,542,528,112,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744320.70/warc/CC-MAIN-20181118073231-20181118095231-00125.warc.gz | 936,361,473 | 44,555 | Metrics 0
## Export Article
Axioms 2018, 7(2), 22; doi:10.3390/axioms7020022
Article
Special Numbers and Polynomials Including Their Generating Functions in Umbral Analysis Methods
Department of Mathematics, Faculty of Science University of Akdeniz, TR-07070 Antalya, Turkey
Received: 20 January 2018 / Accepted: 30 March 2018 / Published: 1 April 2018
## Abstract
:
In this paper, by applying umbral calculus methods to generating functions for the combinatorial numbers and the Apostol type polynomials and numbers of order k, we derive some identities and relations including the combinatorial numbers, the Apostol-Bernoulli polynomials and numbers of order k and the Apostol-Euler polynomials and numbers of order k. Moreover, by using p-adic integral technique, we also derive some combinatorial sums including the Bernoulli numbers, the Euler numbers, the Apostol-Euler numbers and the numbers $y 1 n , k ; λ$. Finally, we make some remarks and observations regarding these identities and relations.
Keywords:
Apostol-Bernoulli polynomials and numbers; Apostol-Euler polynomials and numbers; Sheffer sequences; Appell sequences; Fibonacci numbers; umbral algebra; p-adic integral
MSC:
05A40; 11B68; 11B73; 11B83; 11S80; 26C05; 30B10
## 1. Introduction
In order to give the results presented in this paper, we use two techniques which are p-adic integral technique and the umbral calculus technique. In [1,2,3,4,5], we constructed generating functions for families of combinatorial numbers which are used in counting techniques and problems and also computing negative order of the first and the second kind Euler numbers and other combinatorial sums. In this paper, by applying umbral algebra and umbral analysis methods and their operators to generating functions of the combinatorial numbers and the Apostol type polynomials and numbers, we give many identities and relations including the Fibonacci numbers, the combinatorial numbers, the Apostol-Bernoulli polynomials and numbers of higher order and the Apostol-Euler polynomials and numbers of higher order.
Throughout this paper, we use the following notations, definitions and relations.
Here and in the following, let $C$, $R$, $Z$, and $N$ be the sets of complex numbers, real numbers, integers, and positive integers, respectively, and let $N 0 : = N ∪ 0$. We assume $0 0 = 1$.
Moreover, throughout this paper, $log z$ is tacitly assumed to denote the principal branch of the many-valued function $log z$ with the imaginary part $log z$ constrained by
$− π < Im ( log z ) ≤ π$
(cf. [6,7,8,9]).
The Apostol-Bernoulli polynomials $B n k ( x , λ )$ of order k are defined by
$F B t , x ; λ , k = t λ e t − 1 k e t x = ∑ n = 0 ∞ B n k ( x , λ ) t n n ! ,$
where $λ$ is an arbitrary (real or complex) parameter and $x ∈ R$, and $| t | < 2 π$ when $λ = 1$ and $| t | < 2 π | log λ |$ when $λ ≠ 1$. Moreover, $B n k ( λ ) : = B n k ( 0 , λ )$ denote the Apostol-Bernoulli numbers of order. $B n k : = B n k ( 1 )$ denote the Bernoulli numbers of order k and also $B n : = B n 1$ denote the Bernoulli numbers (cf. see, for details, [6,8,9,10,11,12,13,14], and the references cited therein).
The Apostol-Euler polynomials $E n k ( x , λ )$ of order k are defined by
$F E t , x ; λ , k = 2 λ e t + 1 k e t x = ∑ n = 0 ∞ E n k ( x , λ ) t n n ! ,$
where $λ$ is an arbitrary (real or complex) parameter and $x ∈ R$, and $| t | < π$ when $λ = 1$ and $| t | < π | log λ |$ when $λ ≠ 1$. Moreover, $E n k ( λ ) : = E n k ( 0 , λ )$ denote the Apostol-Euler numbers of order k. $E n k : = E n k ( 1 )$ denote the Euler numbers of order k and also $E n : = E n 1$ denote the Euler numbers (cf. see, for details, [6,8,9,10,11,12,13,14,15], and the references cited therein).
The $λ$-array polynomials $S v n x ; λ$ are defined by
$F S ( t , x , v ; λ ) = λ e t − 1 v v ! e x t = ∑ n = 0 ∞ S v n x ; λ t n n !$
where $v ∈ N 0$ and $λ ∈ C$ (cf. [16]). Furthermore,
$S 2 n , v ; λ : = S v n 0 ; λ$
where, as usual, $S 2 n , v ; λ$ denote the $λ$-Stirling numbers (cf. [8,12]). Substituting $λ = 1$ into (3), we have the array polynomials:
$S v n x : = S v n x ; 1$
(cf. [16,17,18] and (Theorem 2 [19])).
In (cf. Equation (8) [1]), we defined the combinatorial numbers $y 1 n , k ; λ$ by means of the following generating function:
$F y 1 t , k ; λ = 1 k ! λ e t + 1 k = ∑ n = 0 ∞ y 1 n , k ; λ t n n !$
where $k ∈ N 0$ and $λ ∈ C$.
By using (4), we have
$y 1 n , k ; λ = 1 k ! ∑ j = 0 k k j j n λ j$
where $n ∈ N 0$ (cf. Equation (9) [1]).
Relationships between the $λ$-array polynomials $S v n x ; λ$ and the numbers $y 1 n , k ; λ$ and the Stirling numbers of the second kind $S 2 ( n , k )$ are given below, respectively:
$S k n 0 ; λ = S 2 n , v ; λ = − 1 k y 1 n , k ; − λ$
and
$S 2 ( n , k ) = − 1 k y 1 n , k ; − 1$
(cf. [1,17,20,21,22,23,24,25]).
The Fibonacci numbers $F j$ are defined by the following generating function
$t 1 − t − t 2 = ∑ n = 0 ∞ F n t n$
(cf. (p. 229. [26])). We need the following well-known formulas for the Fibonacci numbers. Let $λ = 1 + 5 2$ and $β = 1 − 5 2$. Let $j ∈ N$, we have
$λ j = λ F j + F j − 1$
and
$β j = β F j + F j − 1$
(cf. (p. 78, Lemma 5.1. [26])). Using the above identities, one easily derives the following Binet’s formula:
$F j = λ j − β j λ − β .$
Substituting $− n$ with $n ∈ N$, into the above equation, we easily have
$F − n = ( − 1 ) n + 1 F n$
(cf. (p. 84 [26])).
In the last section, we will give some combinatorial sums with p-adic integrals technique. Hence, let us give definitions of these integrals and a few properties of them.
Let $f ( x ) ∈ C 1 ( Z p → K )$, a set of continuous derivative functions, and $K$ is a field with a complete valuation.
The Volkenborn integral (the bosonic p-adic integral) is defined by
$∫ Z p f ( x ) d μ 1 ( x ) = lim N → ∞ 1 p N ∑ x = 0 p N − 1 f ( x ) ,$
where $μ 1 ( x ) = μ 1 x + p N Z p$ is the Haar distribution on $Z p$:
$μ 1 x + p N Z p = 1 p N ,$
(cf. [27,28]). On the other hand, the p-adic fermionic integral is defined by
$∫ Z p f x d μ − 1 x = lim N → ∞ ∑ x = 0 p N − 1 − 1 x f x$
where
$μ − 1 x = μ − 1 x + p N Z p = ( − 1 ) x p N$
(cf. [29]).
The Bernoulli numbers and the Euler numbers are related to the following p-adic integrals representations, respectively,
$B n = ∫ Z p x n d μ 1 x ,$
(cf. [27,28]) and
$E n = ∫ Z p x n d μ − 1 x ,$
(cf. [27]).
#### 1.2. Umbral Algebra and Calculus
Throughout this section, we use the notations and definitions of the Roman’s book (cf. [13]). Let $P = C x$ be the algebra of polynomials in the single variable x over the field of complex numbers. Let $P *$ be the vector space of all linear functionals on $P$. Let $L ∣ p ( x )$ be the action of a linear functional L on a polynomial $p ( x )$. Let $F$ denote the algebra of formal power series
$f t = ∑ k = 0 ∞ a k t k k ! ,$
(cf. [13]). Furthermore, for all $n ∈ N 0$, one has
$f t ∣ x n = a n$
and also
$f ( t ) g ( t ) ∣ p x = f ( t ) ∣ g ( t ) p x ,$
where $f ( t )$, $g ( t )$ are in $F$ (cf. [13]).
For $p x ∈ P$, as a linear functional, we have
$e y t ∣ p x = p y .$
and as a linear operator, we have
$e y t p x = p x + y$
(cf. [13]). The Sheffer polynomials for pair $( g ( t ) , f ( t ) )$, where $g ( t )$ must be invertible and $f ( t )$ must be delta series. The Sheffer polynomials for pair $( g ( t ) , t )$ is the Appell polynomials or Appell sequences for $g ( t )$. The Appell polynomials are defined by means of the following generating function
$∑ k = 0 ∞ s k x k ! t k = 1 g ( t ) e x t ,$
(cf. [13]). Some properties of the Appell polynomials are given as follows.
$s n x = g ( t ) − 1 x n ,$
(p. 86, Theorem 2.5.5 [13]), derivative formula
$t s n x = n s n − 1 x$
We summarize the results presented in this paper as follows:
In Section 2, by applying the umbral algebra and umbral calculus methods to generating functions of the special numbers and polynomials, we derive some identities and relations including the numbers $y 1 n , k ; λ$, combinatorial sums, the Fibonacci numbers, Apostol-Bernoulli type numbers and polynomials, and the Apostol-Euler type numbers and polynomials. Finally, we give some remarks and observations.
In Section 3, by using the p-adic integrals, we give many combinatorial sums related to the Bernoulli numbers, the Euler numbers, the Apostol-Euler numbers and the numbers $y 1 n , k ; λ$.
## 2. Identities Including the Numbers $y 1 n , k ; λ$, Combinatorial Sums, and Apostol-Euler Type Numbers and Polynomials
In this section, by using the umbral algebra and umbral calculus methods, we derive many identities and relations containing the numbers $y 1 n , k ; λ$, combinatorial sums, the Fibonacci numbers, Apostol-Bernoulli type numbers and polynomials, and the Apostol-Euler type numbers and polynomials.
Theorem 1.
$1 k ! ∑ j = 0 k k j λ j ∑ v = 0 m m v x m − v j v = 1 k ! ∑ j = 0 k k j λ j x + j m$
or
$∑ v = 0 m m v v ! x m − v y 1 v , k ; λ = 1 k ! ∑ j = 0 k k j λ j x + j m$
Proof.
By applying linear operators in (15) and (18) to (4), respectively, we obtain
$1 k ! λ e t + 1 k x m = 1 k ! ∑ j = 0 k k j λ j x + j m$
and
$1 k ! λ e t + 1 k x m = ∑ n = 0 ∞ y 1 n , k ; λ 1 n ! t n x m = 0 , n > m y 1 n , k ; λ , n = m ∑ v = 0 m m v v ! x m − v y 1 v , k ; λ n < m .$
Combining (20) with (21), we get the desired results. ☐
Remark 1.
Substituting $x = 0$ into (19), we arrive at (5).
By applying the action of a linear operator $λ e t + 1 k$ to the Apostol-Euler polynomial $E n a x , λ$, we obtain the following result.
Theorem 2.
$∑ j = 0 k k j λ j E n a x + j , λ = ∑ j = 0 k k j ∑ v = 0 j ( − 1 ) j − v j v 2 v E n a − v x , λ .$
Proof.
By applying the action of a linear operator $λ e t + 1 k$ to the Apostol-Euler polynomial $E n a x , λ$, we obtain
$λ e t + 1 k E n a x , λ = ∑ j = 0 k k j λ j e t j E n a x , λ .$
Applying linear operators in (15) to the above equation, we have
$λ e t + 1 k E n a x , λ = ∑ j = 0 k k j λ j E n a x + j , λ .$
Combining the following relation with (23)
$E n a x , λ = 2 λ e t + 1 a x n$
(cf. p. 101 [13]), we have
$λ e t + 1 k E n a x , λ = ∑ j = 0 k k j ∑ v = 0 j ( − 1 ) j − v j v 2 v 2 λ e t + 1 a − v x n .$
After some elementary calculation in the above equation, we have
$λ e t + 1 k E n a x , λ = ∑ j = 0 k k j ∑ v = 0 j ( − 1 ) j − v j v 2 v E n a − v x , λ .$
Combining (24) and (25), we arrive at the desired result. ☐
Substituting $λ = k = 1$ into (22), we arrive at the following corollary, which was proved by Roman (p. 103, Equation (4.2.11) [13]), see also (cf. [32]).
Corollary 1.
$2 E n ( a − 1 ) ( x ) = E n ( a ) ( x + 1 ) + E n ( a ) ( x )$
We assume that, $λ ≠ 1$ and $a ∈ N$, we have the following well-known relationships between the polynomials $B n a x , λ$ and $E n a x , − λ$:
$∑ n = 0 ∞ E n a x , − λ t n n ! = 2 − λ e t + 1 a e t x = − 2 t a ∑ n = 0 ∞ B n a x , λ t n n ! .$
Therefore
$n a E n − a a x , − λ = − 2 a B n a x , λ$
or
$E n a x , λ = − 2 a n + a a B n + a a x , − λ .$
Substituting the above relation into (22), we get the following result.
Theorem 3.
$∑ j = 0 k k j λ j − 2 a n + a a B n + a a x + j , − λ = ∑ j = 0 k k j ∑ v = 0 j ( − 1 ) j + a j v 2 a n + a − v a − v B n + a − v a − v x , − λ .$
Setting $k = 1$ in (26), we get the following corollary.
Corollary 2.
$B n + a a x , − λ + λ B n + a a x + 1 , − λ = − n + a B n + a − 1 a − 1 x , − λ .$
Remark 2.
Another proof of the Equation (27) is given by Dere et al. [6] and see also (cf. [32]).
Remark 3.
Substituting $n + a = m$ and $λ = − 1$ into (27), we get
$B m ( a ) ( x + 1 ) = B m ( a ) ( x ) + m B m − 1 ( a − 1 ) ( x )$
(cf. p. 95, Equation (4.2.6) [13]).
The following theorem was proved in (cf. [1]). Here, we give a proof different from that in (cf. [1]).
Theorem 4.
Let n and k be nonnegative integers. Then we have
$y 1 n , k ; λ = 2 k k ! E n − k 0 , λ .$
Proof.
Using (12), we obtain
$y 1 n , k ; λ = 1 k ! λ e t + 1 k ∣ x n .$
From the above equation, we have
$y 1 n , k ; λ = 2 k k ! λ e t + 1 2 k ∣ x n = 2 k k ! t 0 ∣ λ e t + 1 2 − k x n = 2 k k ! E n − k 0 , λ .$
Therefore, we arrive at the desired result. ☐
Theorem 5.
$2 k k ! ∑ j = 0 n n j E j a λ E n − j − k λ = ∑ j = 0 n n j E j a λ y 1 n − j , k ; λ$
Proof.
We set
$1 k ! λ e t + 1 k ∣ E n a λ = 1 k ! λ e t + 1 k ∣ ∑ j = 0 n n j E j a λ x n − j = 1 k ! ∑ j = 0 n n j E j a λ λ e t + 1 k ∣ x n − j$
Combining the above equation with (28), we get
$1 k ! λ e t + 1 k ∣ E n a λ = 1 k ! ∑ j = 0 n n j E j a λ y 1 n − j , k ; λ .$
On the other hand
$1 k ! ∑ j = 0 n n j E j a λ λ e t + 1 k ∣ x n − j = 2 k k ! ∑ j = 0 n n j E j a λ t 0 ∣ λ e t + 1 2 − k x n − j .$
Therefore, combining (29) with (30), we arrive at the desired result. ☐
Theorem 6.
$y 1 m , k ; λ = 1 2 k ! ∑ j = 0 k k j λ j 2 E m j + 1 + E m j .$
Proof.
We set the following functional equation
$F y 1 t , k ; λ = 1 2 k ! ∑ j = 0 k k j λ j F E t , j + 1 ; 1 , k + F E t , j ; 1 , 1 .$
By combining the above equation with (4) and (2), we get
$∑ n = 0 ∞ y 1 n , k ; λ t n n ! = ∑ n = 0 ∞ 1 2 k ! ∑ j = 0 k k j λ j E n j + 1 + E n j t n n ! .$
Comparing the coefficients of $t n n !$ on both sides of the above equation yields the desired result. ☐
Theorem 7.
Let $m ∈ N$. Then we have
$S 2 m − 1 , k ; λ = 1 m k ! ∑ j = 0 k ( − 1 ) k − j k j λ j B m j + 1 , λ − B m j , λ .$
Proof.
We also set the following functional equation
$F S ( t , 0 , k ; λ ) = 1 t k ! ∑ j = 0 k ( − 1 ) k − j k j λ j F B t , j + 1 ; λ , 1 − F B t , j ; λ , 1 .$
By combining the above equation with (1) and (3), we get
$∑ m = 0 ∞ S 2 m , k ; λ t m m ! = ∑ m = 0 ∞ 1 t k ! ∑ j = 0 k ( − 1 ) k − j k j λ j B m j + 1 , λ − B m j , λ t m m ! .$
Comparing the coefficients of $t m m !$ on both sides of the above equation yields the desired result. ☐
Theorem 8.
Let $2 λ = 1 + 5$. Then we have
$∑ j = 0 n n j E j a λ y 1 n − j , k ; λ = 1 k ! ∑ j = 0 k k j λ F j + F j − 1 E n a j , λ .$
Proof.
We define the following functional equation:
$F y 1 t , k ; λ F E t , 0 ; λ , a = 1 k ! ∑ j = 0 k k j λ j F E t , j ; λ , a .$
By combining the above equation with (4), (2), and (7), we get
$∑ n = 0 ∞ ∑ j = 0 n n j E j a λ y 1 n − j , k ; λ t n n ! = ∑ n = 0 ∞ 1 k ! ∑ j = 0 k k j λ F j + F j − 1 E n a j , λ t n n ! ,$
where $λ = 1 + 5 2$. Comparing the coefficients of $t n n !$ on both sides of the above equation yields the desired result. ☐
## 3. Combinatorial Sums via $p$-Adic Integral
In this section, by using the p-adic integrals, we derive some combinatorial sums containing the Bernoulli numbers, the Euler numbers, the Apostol-Euler numbers and the numbers $y 1 n , k ; λ$.
Theorem 9.
$B n = k ! 2 k ∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v E v k λ B j − v .$
Proof.
Combining (2) with (4), we set the following functional equation:
$F y 1 t , k ; λ F E t , x ; λ , k = 2 k k ! e t x .$
By using the above equation, we get
$∑ n = 0 ∞ E n k ( x , λ ) t n n ! ∑ n = 0 ∞ y 1 n , k ; λ t n n ! = 2 k k ! ∑ n = 0 ∞ x n t n n ! .$
Therefore
$∑ n = 0 ∞ ∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v x j − v E v k λ t n n ! = 2 k k ! ∑ n = 0 ∞ x n t n n ! .$
Comparing the coefficients of $t n n !$ on both sides of the above equation yields the following relation:
$∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v x j − v E v k λ = 2 k k ! x n .$
By applying the Volkenborn integral to (31), we get
$∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v E v k λ ∫ Z p x j − v d μ 1 x = 2 k k ! ∫ Z p x n d μ 1 x .$
Combining the above equation with (10), we arrive at the desired result. ☐
Remark 4.
Replacing x by k and λ by $λ 2$, the Equation (31) is reduced to the following relation:
$∑ j = 0 n n j y 1 n − j , k ; λ 2 E j k k , λ 2 = 2 k k ! k n .$
Since
$λ k 2 n E n k ( k , λ 2 ) = ∑ m = 0 n n m k m E n − m * k λ$
where $E n * k λ$ denote the Apostol-type Euler numbers of the second kind of order k (cf. [25,33]), the Equation (32) yields
$∑ j = 0 n n j y 1 n − j , k ; λ 2 2 j ∑ m = 0 j j m k m E j − m * k λ = 2 λ k k ! k n .$
By combining the above equation with the following identity
$∑ m = 0 n n m 2 m y 1 m , k ; λ 2 E n − m * k λ = 2 λ k k ! k n$
(cf. [33]), we get the following combinatorial sums
$∑ j = 0 n n j y 1 n − j , k ; λ 2 2 j ∑ m = 0 j j m k m E j − m * k λ = ∑ m = 0 n n m 2 m y 1 m , k ; λ 2 E n − m * k λ .$
Theorem 10.
$E n = k ! 2 k ∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v E v k λ E j − v .$
Proof.
By applying the fermionic p-adic integral to (31), we have
$∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v E v k λ ∫ Z p x j − v d μ − 1 x = 2 k k ! ∫ Z p x n d μ − 1 x .$
Combining the above equation with (11), we arrive at the desired result. ☐
Theorem 11.
$∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v E v k λ j + 1 − v = 2 k ( n + 1 ) k ! .$
Proof.
Integrate Equation (31) with respect to x from 0 to 1, we obtain
$∑ j = 0 n n j y 1 n − j , k ; λ ∑ v = 0 j j v E v k λ ∫ 0 1 x j − v d x = 2 k k ! ∫ 0 1 x n d x .$
After some calculations, we get the desired result. ☐
Theorem 12.
$∑ j = 0 n n j y 1 j , k ; λ B n − j − E n − j = 1 k ! ∑ j = 0 k k j λ j B n ( j ) − E n ( j ) .$
Proof.
Setting
$e t x F y 1 t , k ; λ = 1 k ! ∑ j = 0 k k j λ j e ( x + j ) t .$
Combining (4), we have
$∑ n = 0 ∞ ∑ j = 0 n n j x n − j y 1 j , k ; λ t n n ! = ∑ n = 0 ∞ 1 k ! ∑ j = 0 k k j λ j ( x + j ) n t n n ! .$
Comparing the coefficients of $t n n !$ on both sides of the above equation yields the following relation:
$∑ j = 0 n n j x n − j y 1 j , k ; λ = 1 k ! ∑ j = 0 k k j λ j ( x + j ) n .$
By applying the bosonic p-adic integral to (33) and combining with (10), we have
$∑ j = 0 n n j B n − j y 1 j , k ; λ = 1 k ! ∑ j = 0 k k j λ j B n ( j ) .$
By applying the fermionic p-adic integral to (33) and combining with (11), we obtain
$∑ j = 0 n n j E n − j y 1 j , k ; λ = 1 k ! ∑ j = 0 k k j λ j E n ( j ) .$
Subtracting both sides of Equations (34) and (35), after some elementary calculations, we arrive at the desired result. ☐
Theorem 13.
$∑ j = 0 n n j y 1 j , k ; λ n + 1 − j = 1 k ! ∑ j = 0 k k j λ j ( j + 1 ) n + 1 − j n + 1 n + 1 .$
Proof.
Integrate Equation (33) with respect to x from 0 to 1, we obtain
$∑ j = 0 n n j y 1 j , k ; λ ∫ 0 1 x n − j d x = 1 k ! ∑ j = 0 k k j λ j ∫ 0 1 ( x + j ) n d x .$
After some calculations, we get the desired result. ☐
## Acknowledgments
The present investigation was supported by the Scientific Research Project Administration of Akdeniz University. The author would like to thank to all referees for their valuable comments.
## Conflicts of Interest
The authors declare no conflict of interest.
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33. Kucukoglu, I.; Simsek, Y. Identities and derivative formulas for the combinatorial and Apostol-Euler type numbers by their generating functions. preprint. | 8,900 | 24,276 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 185, "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-2018-47 | longest | en | 0.86317 |
https://www.coep.in/mcq-mechanical-engineering/fluid-mechanics-13/ | 1,660,686,231,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00318.warc.gz | 634,064,293 | 14,925 | # Fluid Mechanics 13
Mechanical Engineering MCQ Question Papers: DRDO, ISRO, Interview
Subject: Fluid Mechanics 13
Part 1: List for questions and answers of Fluid Mechanics
Q1. One litre of a certain fluid weighs 8N. What is its specific volume?
a) 2.03 x 10^– 3 m^3/kg
b) 20.3 x 10^– 3 m^3/kg
c) 12.3 x 10^– 3 m^3/kg
d) 1.23 x 10^– 3 m^3/kg
Q2. According to Archimede’s principle, if a body is immersed partially or fully in a fluid then
the buoyancy
a) Force is _______ the weight of fluid displaced by the body
b) Equal to
c) Less than
d) More than
Q3. What is the correct formula for absolute pressure?
a) Pabs = Patm – Pgauge
b) Pabs = Pvacuum – Patm
c) Pabs = Pvacuum + Patm
d) Pabs = Patm+ Pgauge
Q4. When is a liquid said to be not in a boiling or vaporized state?
a) If the pressure on liquid is equal to its vapour pressure
b) If the pressure on liquid is less than its vapour pressure
c) If the pressure on liquid is more than its vapour pressure
d) Unpredictable
Q5. Bulk modulus is the ratio of
a) Shear stress to volumetric strain
b) Volumetric strain to shear stress
c) Compressive stress to volumetric strain
d) Volumetric strain to compressive stress
Q6. When the angle between surface tension with the liquid (θ) Is greater than 90 deg. The liquid becomes
a) Flat
b) Concave upward
c) Convex upward
d) Unpredictable
Q7. Which property of the fluid offers resistance to deformation under the action of shear force?
a) Density
b) Viscosity
c) Permeability
d) Specific gravity
Q8. The specific weight of the fluid depends upon
a) Gravitational acceleration
b) Mass density of the fluid
c) Both a and b
d) None of the above
Q9. Inter molecular cohesive force in the fluids is
a) Less than that of the solids
b) More than that of the solids
c) Equal to that of the solids
d) Unpredictable
Q10. Which branch of fluid mechanics deals with translation, rotation and deformation of the fluid element? Without considering the force and energy causing such motion is called as
a) Fluid dynamics
b) Fluid kinematics
c) Fluid kinetics
d) Hydraulics
Q11. Shear stress in static fluid is
a) Always zero
b) Always maximum
c) Between zero to maximum
d) Unpredictable
Q12. When there are no external forces, the shape of a liquid drop is determined by
a) Surface tension of the liquid
b) Density of liquid
c) Viscosity of liquid
d) Temperature of air only
Q13. Rain drops are spherical in shape because of
a) Surface tension
b) Capillary
c) Downward motion
d) Acceleration due to gravity
Q14. Rain drops are spherical because of
a) Gravitational force
b) Surface tension
c) Air resistance
d) Low viscosity of water
Q15. What is the main result of adding surfactants into a liquid composed of two immiscible phases such as oil and water?
a) Reduction in the interfacial tension between the phases
b) Increase in the interfacial tension between the phases
c) Catalysation of a chemical reaction between the phases
d) Nothing happens
Q16. A surfactant with a very large Hydrophile-Lipophile Balance (HLB) value (e.g. 40) is expected to function as a
a) Anti-foaming agent
b) Water in oil (w/o) emulsifier
c) Oil in water (o/w) emulsifier
d) Solubility enhancer
Q17. Which of the following drug delivery systems comprises a hydrophobic core and a hydrophilic surface?
a) Liposomes
b) Micelles
c) Reverse micelles
d) None of the above
Q18. Measuring zeta potential is useful in determining which property of a liquid
formulation?
a) Viscosity
b) Stability
c) Solubility
d) Particle size
Q19. Plants get water through the roots because of
a) Capillarity
b) Viscosity
c) Gravity
d) Elasticity
Q20. Kerosene in the wick of lantern rises up because
a) Of negligible viscosity
b) The diffusion of the oil through the wick
c) Of the surface tension of the oil
d) Wick attracts the kerosene
Part 1: List for questions and answers of Fluid Mechanics | 1,092 | 3,860 | {"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-2022-33 | latest | en | 0.836619 |
keinze.wordpress.com | 1,548,326,775,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584520525.90/warc/CC-MAIN-20190124100934-20190124122934-00593.warc.gz | 549,856,888 | 20,490 | Effort Estimation Techniques
My starting point for agile estimation was the book “Aufwandschätzung bei Softwareprojekten”, Steve McConnell, 2006, Microsoft Press, However, at the end I found the classification by Boehm (Barry W. Boehm, Software Engineering Economics, Englewood Cliffs, NJ : Prentice-Hall, 1981).
In my blog I’ll follow the classification of estimation methods from Boehm 1981 (Barry W. Boehm, Software Engineering Economics, Englewood Cliffs, NJ : Prentice-Hall, 1981):
• Algorithmic cost modeling
• Parametric Models (e.g. COCOMO)
• Function Points / Lines of Code
• Proxy Based
• Process Simulation
• Estimation by analogy
• StoryPoints
• T-Shirt Sizes
• Expert judgment
• Single experts
• Group of experts (Wideband Delphi, Planning Poker)
• Variances and subsidiary techniques
• Top-down / Bottom-up estimation
• Combinations of techniques
• PERT / Fuzzy estimation
• Parkinson’s Law
An estimation is an approximation based on input data that may be incomplete or uncertain (Dictionary, Wikipedia).
If the estimation is given as a single value (“one-point-estimation”), it’s assumed that there is a 50% probability that the real value is higher, 50% that it is lower than estimated.
It is more accurate to provide an estimation range, with a min, max, and confidence intervals. However this requires more mathematics.
Note: An estimate is different from a project plan: The project plan is designed to hit a target that is a statement of a desirable business objective. A commitment is a promise to hit the target.
Analysis is the act of breaking something into parts to get a better understanding of it.
Law of Large Numbers (LLN) The average of the results obtained from a large number of trials is close to the expected value. Therefore you gain more accuracy, if you involve a group of experts rather than one expert.
Economy of scale Refers to the cost advantages that an enterprise obtains due to expansion.
Diseconomy of Scale Estimation of smaller projects / demands do not scale to bigger projects / demands, due to e.g. communication/management costs, duplication of effort
Closer to the end of a project, uncertainty becomes smaller. This can be visualized as a “cone of uncertainty” – e.g. look at http://construx.com/Page.aspx?cid=1648
Now I’ll present an overview of the estimation methods:
One big class of estimation techniques are algorithmic methods: Algorithmic methods use mathematical relations/formulas for the estimation. The formulas are based on research and historical data and use inputs such as Lines of Code (LOC), number of functions to perform, Defects, … you get from analysis. The advantage of these methods is that they are very precise and easy to apply. The limiting factor is the availability of the input data. Also they are unable to handle exceptional conditions, input data might not be available or of poor quality.
Also (dynamic) process simulations are algorithmic methods: They use a dynamic model with assumptions about the project and organization, e.g. their velocities and error rates (need to be calibrated with real data).
Another class of estimation techniques is estimation by analogy: They compare new tasks / project with other tasks / projects already known to derive the estimated effort from historical data. One needs to find areas that can be counted (e.g. number of tables, screens, use cases etc.). Best known techniques areStoryPoints and T-Shirt Sizes:
StoryPoints
• Assign numbers to the categories that are related to the complexity
• Typical categories are
• Powers of 2: 1, 2, 4, 8, 16,…
• Fibonacci: 1, 2, 3, 5, 8, 13, …
• Story points are relative to a defined anchor, to compare to
T-Shirt Size (S, M, L, XL, …)
• Generalize the story point categories (maybe “8” story points do not exactly relate to twice the effort of “4”, e.g. due to diseconomy of scale
• The average size of a category is determined by historical data
The comparison is usually done by expert judgment: The judgment can relay on individual experts: developers, architects, etc. who are asked about the expected effort (one-point estimation) or ranges (min / max) or clusters. It can also rely on a group of experts – e.g. done in the planning Poker (Scrum poker) or poker party.
Last but not least there is the not so seriously meant estimation techniques, the Parkinson’s Law: “Work expands so as to fill the time available for its completion” (Cyril Northcote Parkinson, published in The Economist in 1955). Therefore, the cost is determined by available resources rather than by objective assessment. The estimated effort depends on the customer’s budget and not on the software functionality – e.g. If the software has to be delivered in 12 months and 5 people are available, the effort required is estimated to be 60 person-months.
For more variants of the Parkinson’s law see Wikipedia http://de.wikipedia.org/wiki/Parkinsonsche_Gesetze
Last but not least I’ll provide some variances and subsidiary techniques
Top-down approach
Split Requirements (Epics) into smaller elements (Stories) and assign some relative measure like story points, percentages. Split some (at least one) element further until you can achieve some good estimation
Bottom-up Judgment:
Best results, if tasks < 2d (otherwise details will be overseen)
Sum up to get the total effort. This yields to highly accurate estimated due to the Law of Large Numbers:
Fuzzy Effort Estimation
Fuzzy Numbers represent the physical world more realistically than single valued numbers (“Optimization Criteria for Effort Estimation using Fuzzy Technique”, Harish Mittal/ Pradeep Bhatia, 2007, CLEI ELECTRONIC JOURNAL, http://www.clei.cl/cleiej/papers/v10i1p2.pdf).
Some further reading I suggest: “The Comparison of the Software Cost Estimating Methods” by Liming Wu (http://www.compapp.dcu.ie/~renaat/ca421/LWu1.html).
4 thoughts on “Effort Estimation Techniques”
1. StoryPoints
“Assign numbers to the categories that directly relate to the expected effort”
/// effort /// or complexity ?
effort = f( complexity ) ?
cu
@Boeffi
1. Thank you for the comment – I’ve updated the section.
2. Good compilation, Mirko! One question: Regardless of the estimation technique, how do you use the numbers that were estimated? Do you compare them against real numbers or do you use them as categories and measure how long something in that category really takes, on average? I wouldn’t recommend the former but rather the latter.
1. We use the estimation techniques
a) during the demand process to give our customers cost input for their business case calculation
b) to prepare planning sessions (effectively limit the input for the Scrum teams embedded in an outer waterfall)
c) in combination with real data used to measure performance (e.g. velocities, CFD) | 1,530 | 6,815 | {"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} | 3.21875 | 3 | CC-MAIN-2019-04 | latest | en | 0.825904 |
https://forums.space.com/threads/earths-weight-capacity.11529/ | 1,720,965,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00374.warc.gz | 247,570,880 | 23,266 | # Earth's weight capacity
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R
#### red67
##### Guest
How much weight (manmade) can the Earth withstand?<br /><br />
G
#### grooble
##### Guest
infinite, you can't exactly add more mass than is already here without bringing it in from other planets or Roids.
M
#### meteo
##### Guest
I don't really understand the question either, be more specific. Are you talking about a city causing land to sink?
A
##### Guest
Or do you mean how heavy (tall) a building the Earth could support????
J
#### jcdenton
##### Guest
<font color="yellow">How much weight (manmade) can the Earth withstand? </font><br /><br />That would depend on the soil you're building on. I believe this field of study is called geotechnic engineering. <div class="Discussion_UserSignature"> </div>
N
#### nexium
##### Guest
We expect the ocean level to rise when and if the ice caps melt. At enourmous cost we could dredge the shallow parts of the ocean and pile the the dirt on saltwater swamps until the new surface was 200 meters above sealevel. Dreging would reduce the amount of ocean level rise very slightly. As the coastal plains (including many of our cities) became the new salt marsh, we could cover them with with almost 200 meters of dreged material. At some point this new land height would unbalance the plate tetonics, and some of the artificially raised land would sink toward the center of Earth a few meters. We would then have exceeded the "Earth's weight capacity" Neil
P
#### paleo
##### Guest
The Earth's 'weight' capacity doesn't change. <br /><br /> Anything man could do has no more effect than a feather on a mountain. Man can't shift ten trillionth of the Earth's mass.<br /><br /> Puting weight on a particular geographic point (a city block)...and having the ground sink a few meters only has an impact from the immediacy of a human perspective but zilcho at a geologic level.
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1K | 511 | 2,064 | {"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-2024-30 | latest | en | 0.929687 |
http://www.mathworks.com/matlabcentral/fileexchange/41232-flauta/content/DoRe.m | 1,427,859,283,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131302478.63/warc/CC-MAIN-20150323172142-00203-ip-10-168-14-71.ec2.internal.warc.gz | 657,766,572 | 6,726 | Code covered by the BSD License
# Flauta
by
### Edgar (view profile)
flauta en matlab
DoRe.m
```%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Dr. Miguel Aurelio Duarte Villaseor
% Docente en el ITS de Poza Rica
% Poza rica, Veracruz, Mxico. Marzo de 2013
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%Descripcin: saca varios tonos por la bocina,
%se puede definir: amplitud y duracion
amplitud=20;
duracion=0.5; %en segundos
Fs=0.00001;
t=[0:Fs:duracion];
Do=261.626;
Do=amplitud*sin(2*pi*Do*t);
Do1=277.183;
Do1=amplitud*sin(2*pi*Do1*t);
Re=293.665;
Re=amplitud*sin(2*pi*Re*t);
Re1=311.127;
Re1=amplitud*sin(2*pi*Re1*t);
Mi=329.628;
Mi=amplitud*sin(2*pi*Mi*t);
Fa=349.228;
Fa=amplitud*sin(2*pi*Fa*t);
Fa1=369.994;
Fa1=amplitud*sin(2*pi*Fa1*t);
Sol=391.995;
Sol=amplitud*sin(2*pi*Sol*t);
Sol1=415.305;
Sol1=amplitud*sin(2*pi*Sol1*t);
La=440;
La=amplitud*sin(2*pi*La*t);
La1=466.164;
La1=amplitud*sin(2*pi*La1*t);
Si=493.883;
Si=amplitud*sin(2*pi*Si*t);
% sound(Do,1/Fs);
% pause(0.2)
% sound(Do1,1/Fs);
% pause(0.2)
% sound(Re,1/Fs);
% pause(0.2)
% sound(Re1,1/Fs);
% pause(0.2)
% sound(Mi,1/Fs);
% pause(0.2)
% sound(Fa,1/Fs);
% pause(0.2)
% sound(Fa1,1/Fs);
% pause(0.2)
% sound(Sol,1/Fs);
% pause(0.2)
% sound(Sol1,1/Fs);
% pause(0.2)
% sound(La,1/Fs);
% pause(0.2)
% sound(La1,1/Fs);
% pause(0.2)
% sound(Si,1/Fs);
% pause(0.2)
%ejemplo de como unidos forman una tonada (flauta de pan)
%primero para las notas sostenidas le doy ms tiempo
t2=[0:Fs:duracion*2];
Sol2=391.995;
Sol2=amplitud*sin(2*pi*Sol2*t2);
Fa12=369.994;
Fa12=amplitud*sin(2*pi*Fa12*t);
Do2=261.626;
Do2=amplitud*sin(2*pi*Do2*t);
sound(Sol,1/Fs);
sound(Sol,1/Fs);
sound(Sol,1/Fs);
sound(La,1/Fs);
sound(Si,1/Fs);
sound(Si,1/Fs);
sound(Sol2,1/Fs);
pause(0.3)
sound(Sol,1/Fs);
sound(Sol,1/Fs);
sound(Sol,1/Fs);
sound(La,1/Fs);
sound(Si,1/Fs);
sound(Si,1/Fs);
sound(Fa12,1/Fs);
pause(0.3)
sound(Sol,1/Fs);
sound(Sol,1/Fs);
sound(Sol,1/Fs);
sound(Fa1,1/Fs);
sound(Re,1/Fs);
sound(Do2,1/Fs);
sound(Re,1/Fs);
```
Contact us | 889 | 2,096 | {"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-2015-14 | longest | en | 0.223006 |
https://eudml.org/doc/215935 | 1,496,094,982,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612553.95/warc/CC-MAIN-20170529203855-20170529223855-00082.warc.gz | 914,068,306 | 12,318 | # Automorphisms and derivations of a Fréchet algebra of locally integrable functions
Studia Mathematica (1992)
• Volume: 103, Issue: 1, page 51-69
• ISSN: 0039-3223
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## Abstract
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We find representations for the automorphisms, derivations and multipliers of the Fréchet algebra $L{¹}_{loc}$ of locally integrable functions on the half-line ${ℝ}^{+}$. We show, among other things, that every automorphism θ of $L{¹}_{loc}$ is of the form $\theta ={\phi }_{a}{e}^{\lambda X}{e}^{D}$, where D is a derivation, X is the operator of multiplication by coordinate, λ is a complex number, a > 0, and ${\phi }_{a}$ is the dilation operator $\left({\phi }_{a}f\right)\left(x\right)=af\left(ax\right)$ ($f\in L{¹}_{loc}$, $x\in {ℝ}^{+}$). It is also shown that the automorphism group is a topological group with the topology of uniform convergence on bounded sets and is the semidirect product of a connected subgroup and a discrete group which is isomorphic to the discrete group of real numbers.
## How to cite
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Ghahramani, F., and McClure, J.. "Automorphisms and derivations of a Fréchet algebra of locally integrable functions." Studia Mathematica 103.1 (1992): 51-69. <http://eudml.org/doc/215935>.
@article{Ghahramani1992,
abstract = {We find representations for the automorphisms, derivations and multipliers of the Fréchet algebra $L¹_\{loc\}$ of locally integrable functions on the half-line $ℝ^+$. We show, among other things, that every automorphism θ of $L¹_\{loc\}$ is of the form $θ = φ _ae^\{λX\}e^D$, where D is a derivation, X is the operator of multiplication by coordinate, λ is a complex number, a > 0, and $φ_a$ is the dilation operator $(φ_af)(x) = af(ax)$ ($f ∈ L¹_\{loc\}$, $x ∈ ℝ^+$). It is also shown that the automorphism group is a topological group with the topology of uniform convergence on bounded sets and is the semidirect product of a connected subgroup and a discrete group which is isomorphic to the discrete group of real numbers.},
author = {Ghahramani, F., McClure, J.},
journal = {Studia Mathematica},
keywords = {representations; automorphisms; derivations; multipliers; Fréchet algebra; locally integrable functions on the half-line; dilation operator; automorphism group; topology of uniform convergence on bounded sets; semidirect product of a connected subgroup and a discrete group},
language = {eng},
number = {1},
pages = {51-69},
title = {Automorphisms and derivations of a Fréchet algebra of locally integrable functions},
url = {http://eudml.org/doc/215935},
volume = {103},
year = {1992},
}
TY - JOUR
AU - Ghahramani, F.
AU - McClure, J.
TI - Automorphisms and derivations of a Fréchet algebra of locally integrable functions
JO - Studia Mathematica
PY - 1992
VL - 103
IS - 1
SP - 51
EP - 69
AB - We find representations for the automorphisms, derivations and multipliers of the Fréchet algebra $L¹_{loc}$ of locally integrable functions on the half-line $ℝ^+$. We show, among other things, that every automorphism θ of $L¹_{loc}$ is of the form $θ = φ _ae^{λX}e^D$, where D is a derivation, X is the operator of multiplication by coordinate, λ is a complex number, a > 0, and $φ_a$ is the dilation operator $(φ_af)(x) = af(ax)$ ($f ∈ L¹_{loc}$, $x ∈ ℝ^+$). It is also shown that the automorphism group is a topological group with the topology of uniform convergence on bounded sets and is the semidirect product of a connected subgroup and a discrete group which is isomorphic to the discrete group of real numbers.
LA - eng
KW - representations; automorphisms; derivations; multipliers; Fréchet algebra; locally integrable functions on the half-line; dilation operator; automorphism group; topology of uniform convergence on bounded sets; semidirect product of a connected subgroup and a discrete group
UR - http://eudml.org/doc/215935
ER -
## References
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1. [1] W. G. Bade and H. G. Dales, Norms and ideals in radical convolution algebras, J. Funct. Anal. 41 (1) (1981), 77-109. Zbl0463.46046
2. [2] H. G. Dales, Convolution algebras on the real line, in: Radical Banach Algebras and Automatic Continuity, J. M. Bachar et al. (eds.), Lecture Notes in Math. 975, Springer, Berlin 1983, 180-209.
3. [3] H. G. Dales, Positive Results in Automatic Continuity, forthcoming book.
4. [4] H. G. Diamond, Characterization of derivations on an algebra of measures, Math. Z. 100 (1967), 135-140. Zbl0145.38801
5. [5] H. G. Diamond, Characterization of derivations on an algebra of measures II, ibid. 105 (1968), 301-306. Zbl0167.14802
6. [6] W. F. Donoghue, Jr., The lattice of invariant subspaces of a completely continuous quasi-nilpotent transformation, Pacific J. Math. 7 (1957), 1031-1035. Zbl0078.29504
7. [7] I. Gelfand, D. Raikov and G. Shilov, Commutative Normed Rings, Chelsea, 1964.
8. [8] F. Ghahramani, Homomophisms and derivations on weighted convolution algebras, J. London Math. Soc. (2) 21 (1980), 149-161. Zbl0435.43005
9. [9] F. Ghahramani, Isomorphisms between radical weighted convolution algebras, Proc. Edinburgh Math. Soc. 26 (1983), 343-351. Zbl0518.43002
10. [10] F. Ghahramani, The connectedness of the group of automorphisms of L¹(0,1), Trans. Amer. Math. Soc. 302 (2) (1987), 647-659. Zbl0634.46041
11. [11] F. Ghahramani, The group of automorphisms of L¹(0,1) is connected, ibid. 314 (2) (1989), 851-859. Zbl0682.46037
12. [12] F. Ghahramani and J. P. McClure, Automorphisms of radical weighted convolution algebras, J. London Math. Soc. (2) 41 (1990), 122-132. Zbl0725.46035
13. [13] S. Grabiner, Derivations and automorphisms of Banach algebras of power series, Mem. Amer. Math. Soc. 146 (1974). Zbl0295.46074
14. [14] N. P. Jewell and A. M. Sinclair, Epimorphisms and derivations on L¹(0,1) are continuous, Bull. London Math. Soc. 8 (1976), 135-139. Zbl0324.46048
15. [15] H. Kamowitz and S. Scheinberg, Derivations and automorphisms of L¹(0,1), Trans. Amer. Math. Soc. 135 (1969), 415-427. Zbl0172.41703
16. [16] E. A. Michael, Locally multiplicatively-convex topological algebras, Mem. Amer. Math. Soc. 11 (1952). Zbl0047.35502
17. [17] A. P. Robertson and W. J. Robertson, Topological Vector Spaces, Cambridge Univ. Press, London 1964. Zbl0123.30202
18. [18] M. P. Thomas, Approximation in the radical algebra ℓ¹(w) when ${w}_{n}$ is star-shaped, in: Radical Banach Algebras and Automatic Continuity, J. M. Bachar et al. (eds.), Lecture Notes in Math. 975, Springer, Berlin 1983, 258-272.
19. [19] F. Treves, Topological Vector Spaces, Distributions and Kernels, Academic Press, 1967. Zbl0171.10402
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Note: Best practice suggests putting the JavaScript code just before the closing </body> tag. | 2,162 | 7,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-22 | longest | en | 0.837281 |
http://www.sosmath.com/algebra/logs/log4/log42/log42206/log42206.html | 1,542,882,458,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746205.96/warc/CC-MAIN-20181122101520-20181122123520-00338.warc.gz | 510,697,511 | 3,580 | ### GRAPHS OF LOGARITHMIC FUNCTIONS
In this section we will illustrate, interpret, and discuss the graphs of logarithmic functions. Recall that whenever there is a minus sign in front of the logarithmic term, it means that there is a reflection across the x-axis. Recall also that whenever there is a minus sign in front of the x in the argument, it means that there is a reflection across the y-axis.
Horizontal and Vertical Shifts and Reflection Across the x-axis: The following examples discuss the difference between the graph of f(x) and the graph of -f(x + A) + B
Example 11: Graph the function and the function on the same rectangular coordinate system. and answer the following questions about each graph:
1. In what quadrants in the graph of the function located? In what quadrants is the graph of the function located?
2. What is the x-intercept and the y-intercept on the graph of the function ? What is the x-intercept and the y-intercept on the graph of the function ?
3. Find the point (2, f(2)) on the graph of and find (4, g(4)) on the graph of . What do these two points have in common?
4. Describe the relationship between the two graphs.
5. How would you moved the graph of so that it would be superimposed on the graph of ? When you moved the graph, where would the point (1, 0) on be after the move?
1. You can see that the both graphs are located in quadrants I and IV.
2. You can see that neither of the graphs crosses the y-axis; therefore, neither of the graphs has a y-intercept. Notice that the graph of f(x) crosses the x-axis at 1 because . The graph of g(x) crosses the x-axis at 22.085537 because when x = 22,085537.
3. The point is located on the graph of . The point is located on the graph of .
4. Both graphs have the same shape. It appears that the graph of is the result of reflecting the graph over the x-axis, then shifting the graph to the right and upward.
5. After we reflect the graph across the x-axis, the point (1,0) stays the same. After the graph is then moved to the right 2 units and up 3 units, it would be superimposed on the graph of . The point (1, 0) on the graph of would first be shifted to the right 2 units and up 3 units to (1 + 2, 0 + 3) or (3, 3).
Example 12: Graph the function and the function on the same rectangular coordinate system. Answer the following questions about each graph:
1. In what quadrants in the graph of the function located? In what quadrants is the graph of the function located?
2. State the domain of both functions.
3. What is the x-intercept and the y-intercept on the graph of the function ? What is the x-intercept and the y-intercept on the graph of the function ?
4. Find the point (2, f(2)) on the graph of and find (6, g(6)) on the graph of
5. What do these two points have in common?
6. Describe the relationship between the two graphs.
7. Describe how you would move the graph of moved so that it would be superimposed on the graph of . Where would the point (1, 0) on the graph of wind up on after the move?
1. The graph of is located in quadrants I and IV. The graph of is located in quadrants II, III and IV.
2. The domain of is the set of all positive real numbers. The graph of is the set of real numbers less than 8.
3. The graph of does not cross the y-axis. It crosses the x-axis at 1.
4. The graph of crosses the x-axis at -12.085537 because when x = -12.085537. It crosses the y-axis at -0.920558 because .
5. The point is located on the graph of . The point is located on the graph of .
6. The point (6, g(6)) is located to the right and down from the point (2, f(2)).
7. Both graphs have the same shape. The graph of opens to the left whereas the graph of opens to the right. There are horizontal and vertical shifts.
8. Whenever there is a minus in front of the x, it means there is a reflection across the y-axis. Whenever something is added or subtracted from the x in the argument, it means there is a horizontal shift. Whenever a constant is added to the logarithmic term, it means there is a vertical shift.
9. Rewrite the equation by factoring out -1 from the argument to read .
From the equation, you can see that there is a reflection across the y-axis (minus sign in front of the x), a horizontal shift will be to the right 8 units, and a vertical shift down 3 units.
Therefore, reflect the graph of over the y-axis and then shift (move) the reflected graph to the right 8 units and down 3 units. The point (1, 0) would be located (-1, 0) after the reflection and (1 + 8, 0 - 3) or (9, - 3) after the horizontal and vertical shifts..
If you would like to review another example, click on Example.
[Exponential Rules] [Logarithms]
[Algebra] [Trigonometry ] [Complex Variables] | 1,185 | 4,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2018-47 | latest | en | 0.946755 |
https://www.exceldemy.com/learn-excel/finance/calculate-interest-rate/ | 1,721,087,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00388.warc.gz | 666,576,463 | 70,656 | # How to Calculate Interest Rate in Excel (3 Ways)
Consider the following dataset that we’ll use to calculate interest rates. We’ll calculate the Monthly and Yearly Interest Rates, as well as the Effective and Nominal Interest Rates, using formulas and functions.
### Method 1 – Use a Formula to Calculate the Periodic Interest Rate in Excel
We will compute the interest rate for months and then for years in the two subsections below.
We will apply the RATE function.
=RATE(nper, pmt, pv, [fv], [type], [guess])
Here,
• Nper (required) – total payment periods number (years, months)
• Pmt (required) – the pre-set payment amounts each period that cannot be varied over the annuity’s lifetime. It generally involves principal and interest but excludes taxes.
• Pv (required) – the present worth of the loan.
• Fv (optional) – the future worth, or the cash balance you want after the last installment It defaults to 0 if not specified.
• Type (optional) – specifies the date on which the payments are made:
• Payment is due at the end of the period if 0 or absent (default).
• 1 – the first payment is required at the start of the period.
• Guess (optional) – your best guess as to what the rate may be If you leave it blank, it defaults to 10%.
#### Case 1.1 – Monthly Interest Rate
Steps:
• Select cell F4.
• Insert the following formula.
`=RATE(C4,-C5, C6)`
• Press Enter.
#### Case 1.2 – Annual Interest Rate
We’ll compute the monthly interest rate first, then multiply by 12 to get the annual interest rate.
Steps:
• Insert the following formula in cell F4:
`=RATE(C4,-C5, C6)`
• Multiply the previous calculation by the value of C7 or use the following formula to get the annual interest rate:
`=RATE(C4, -C5,C6)*C7`
• The Annual Interest Rate will be shown in cell F6.
Pay attention to the C5 cell. It is in negative form since the payment will be reduced after each transaction. It will show a #NUM error if you don’t put the value in negative.
### Method 2 – Apply a Formula to Calculate the Effective Interest Rate in Excel
Here:
• i= the rate of interest.
• n= number of the compounding periods.
#### Case 2.1 Use the Formula of Compound Interest
Steps:
• Use this formula.
• The formula becomes:
`=(1+C4/C5)^C5 - 1`
• Press Enter.
#### Case 2.2 – Use the EFFECT Function
Steps:
`=EFFECT(C4, C5)`
• Press Enter to get the Effective Interest Rate.
### Method 3 – Convert Effective Interest to Nominal Interest
You can get back to the Nominal/Simple Interest Rate from the Effective Interest Rate.
Steps:
• Enter the following formula of the NOMINAL function.
`=NOMINAL(C7, C5)`
• Press Enter to get the Nominal Interest Rate.
Download the Practice Workbook
## How to Calculate Interest Rate in Excel: Knowledge Hub
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Advanced Excel Exercises with Solutions PDF | 825 | 3,471 | {"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} | 3.875 | 4 | CC-MAIN-2024-30 | latest | en | 0.85786 |
https://forum.pjrc.com/index.php?threads/teensyduino-1-54-beta-2.64023/page-2#post-257111 | 1,713,731,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00519.warc.gz | 223,173,501 | 24,366 | # Teensyduino 1.54 Beta #2
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I should have mentioned, another property some people are expecting is identical result for "reverse" mapping.
In other words, ideally these should give the same result for all inputs:
map(x, 0, 5, 0, 1023);
map(x, 5, 0, 1023, 0);
And these 2 should give the same result as the 2 above, but "reversed"
map(x, 0, 5, 1023, 0);
map(x, 5, 0, 0, 1023);
Of course Arduino's map() doesn't do any of this, and I do not believe any of the proposed map() functions on their issue trackers do either, especially when considering extrapolating beyond the mapped range.
Thanks Paul - will check that next. Out of curiosity I was looking for a mathematical approach to map and found something on rosetta code for a map function: https://rosettacode.org/wiki/Map_range. I added this to the code just to see what it would give me:
Code:
``````#include <vector>
template<typename tVal>
tVal map_value(std::pair<tVal,tVal> a, std::pair<tVal, tVal> b, tVal inVal)
{
tVal inValNorm = inVal - a.first;
tVal aUpperNorm = a.second - a.first;
tVal normPosition = inValNorm / aUpperNorm;
tVal bUpperNorm = b.second - b.first;
tVal bValNorm = normPosition * bUpperNorm;
tVal outVal = b.first + bValNorm;
return outVal;
}
// This is the original map() function in Arduino Core
long map1(long x, long in_min, long in_max, long out_min, long out_max)
{
return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min;
}
// This is the same but with the +1 "range extrension" as suggested by st42
long mapPlus1(long x, long in_min, long in_max, long out_min, long out_max)
{
return (x - in_min) * (out_max - out_min + 1) / (in_max - in_min + 1) + out_min;
}
// This is another version of map with rounding done only with integer calculations
// as suggested by M.Kooijman
long mapRound(long x, long in_min, long in_max, long out_min, long out_max)
{
return ((x - in_min) * (out_max - out_min) + (in_max - in_min)/2) / (in_max - in_min) + out_min;
}
void setup(void) {
Serial.begin(115200);
delay(2000);
long x;
Serial.printf("Range 0-20 to 0-4\n");
//rosetta version
//https://rosettacode.org/wiki/Map_range#C.2B.2B
Serial.printf(" x map map1 map(+1) map(round) Rosetta\n");
std::pair<float,float> a(0,20), b(0,4);
for (x=-10; x<=30; x++) {
Serial.printf("%6ld %6ld %6ld %6ld %6ld %f\n", x,
map(x, 0, 20, 0, 4),
map1(x, 0, 20, 0, 4),
mapPlus1(x, 0, 20, 0, 4),
mapRound(x, 0, 20, 0, 4),
map_value(a, b, (float)x));
}
Serial.printf("\n\n");
Serial.printf("Range 0-5 to 0-1023\n");
std::pair<float,float> c(0,5), d(0,1023);
Serial.printf(" x map map1 map(+1) map(round) Rosetta\n");
for (x=-5; x<=10; x++) {
Serial.printf("%6ld %6ld %6ld %6ld %6ld %f\n", x,
map(x, 0, 5, 0, 1023),
map1(x, 0, 5, 0, 1023),
mapPlus1(x, 0, 5, 0, 1023),
mapRound(x, 0, 5, 0, 1023),
map_value(c, d, (float)x));
}
}
void loop() {}``````
Output:
Code:
``````Range 0-20 to 0-4
x map map1 map(+1) map(round) Rosetta
-10 -2 -2 -2 -1 -2.000000
-9 -2 -1 -2 -1 -1.800000
-8 -2 -1 -1 -1 -1.600000
-7 -1 -1 -1 0 -1.400000
-6 -1 -1 -1 0 -1.200000
-5 -1 -1 -1 0 -1.000000
-4 -1 0 0 0 -0.800000
-3 -1 0 0 0 -0.600000
-2 0 0 0 0 -0.400000
-1 0 0 0 0 -0.200000
0 0 0 0 0 0.000000
1 0 0 0 0 0.200000
2 0 0 0 0 0.400000
3 1 0 0 1 0.600000
4 1 0 0 1 0.800000
5 1 1 1 1 1.000000
6 1 1 1 1 1.200000
7 1 1 1 1 1.400000
8 2 1 1 2 1.600000
9 2 1 2 2 1.800000
10 2 2 2 2 2.000000
11 2 2 2 2 2.200000
12 2 2 2 2 2.400000
13 3 2 3 3 2.600000
14 3 2 3 3 2.800000
15 3 3 3 3 3.000000
16 3 3 3 3 3.200000
17 3 3 4 3 3.400000
18 4 3 4 4 3.600000
19 4 3 4 4 3.800000
20 4 4 4 4 4.000000
21 4 4 5 4 4.200000
22 4 4 5 4 4.400000
23 5 4 5 5 4.600000
24 5 4 5 5 4.800000
25 5 5 5 5 5.000000
26 5 5 6 5 5.200000
27 5 5 6 5 5.400000
28 6 5 6 6 5.600000
29 6 5 6 6 5.800000
30 6 6 7 6 6.000000
Range 0-5 to 0-1023
x map map1 map(+1) map(round) Rosetta
-5 -1023 -1023 -853 -1022 -1023.000000
-4 -819 -818 -682 -818 -818.400024
-3 -614 -613 -512 -613 -613.800049
-2 -409 -409 -341 -408 -409.200012
-1 -205 -204 -170 -204 -204.600006
0 0 0 0 0 0.000000
1 205 204 170 205 204.600006
2 409 409 341 409 409.200012
3 614 613 512 614 613.800049
4 818 818 682 818 818.400024
5 1023 1023 853 1023 1023.000000
6 1228 1227 1024 1228 1227.600098
7 1432 1432 1194 1432 1432.199951
8 1637 1636 1365 1637 1636.800049
9 1841 1841 1536 1841 1841.399902
10 2046 2046 1706 2046 2046.000000``````
and it looks like it is working - unless you used the same method so would get identical results
Next up - reverse range.
Ok - here is a side by side comparison. looks like its working both in forward and reversed ! ....
Code:
`````` Range 0-20 to 4-0 Range 0-20 to 0-4
x map map1 map(+1) map(round) Rosetta x map map1 map(+1) map(round) Rosetta
-10 6 6 5 6 6 -10 -2 -2 -2 -1 -2
-9 6 5 5 6 5.8 -9 -2 -1 -2 -1 -1.8
-8 6 5 5 6 5.6 -8 -2 -1 -1 -1 -1.6
-7 5 5 5 5 5.4 -7 -1 -1 -1 0 -1.4
-6 5 5 4 5 5.2 -6 -1 -1 -1 0 -1.2
-5 5 5 4 5 5 -5 -1 -1 -1 0 -1
-4 5 4 4 5 4.8 -4 -1 0 0 0 -0.8
-3 5 4 4 5 4.6 -3 -1 0 0 0 -0.6
-2 4 4 4 4 4.4 -2 0 0 0 0 -0.4
-1 4 4 4 4 4.2 -1 0 0 0 0 -0.2
0 4 4 4 4 4 0 0 0 0 0 0
1 4 4 4 4 3.8 1 0 0 0 0 0.2
2 4 4 4 4 3.6 2 0 0 0 0 0.4
3 3 4 4 4 3.4 3 1 0 0 1 0.6
4 3 4 4 4 3.2 4 1 0 0 1 0.8
5 3 3 4 4 3 5 1 1 1 1 1
6 3 3 4 4 2.8 6 1 1 1 1 1.2
7 3 3 3 4 2.6 7 1 1 1 1 1.4
8 2 3 3 3 2.4 8 2 1 1 2 1.6
9 2 3 3 3 2.2 9 2 1 2 2 1.8
10 2 2 3 3 2 10 2 2 2 2 2
11 2 2 3 3 1.8 11 2 2 2 2 2.2
12 2 2 3 3 1.6 12 2 2 2 2 2.4
13 1 2 3 2 1.4 13 3 2 3 3 2.6
14 1 2 2 2 1.2 14 3 2 3 3 2.8
15 1 1 2 2 1 15 3 3 3 3 3
16 1 1 2 2 0.8 16 3 3 3 3 3.2
17 1 1 2 2 0.6 17 3 3 4 3 3.4
18 0 1 2 1 0.4 18 4 3 4 4 3.6
19 0 1 2 1 0.2 19 4 3 4 4 3.8
20 0 0 2 1 0 20 4 4 4 4 4
21 0 0 1 1 -0.2 21 4 4 5 4 4.2
22 0 0 1 1 -0.4 22 4 4 5 4 4.4
23 -1 0 1 0 -0.6 23 5 4 5 5 4.6
24 -1 0 1 0 -0.8 24 5 4 5 5 4.8
25 -1 -1 1 0 -1 25 5 5 5 5 5
26 -1 -1 1 0 -1.2 26 5 5 6 5 5.2
27 -1 -1 1 0 -1.4 27 5 5 6 5 5.4
28 -2 -1 0 -1 -1.6 28 6 5 6 6 5.6
29 -2 -1 0 -1 -1.8 29 6 5 6 6 5.8
30 -2 -2 0 -1 -2 30 6 6 7 6 6``````
Code:
`````` Range 0-5 to 1023-0 Range 0-5 to 0 -1023
x map map1 map(+1) map(round) Rosetta x map map1 map(+1) map(round) Rosetta
-5 2046 2046 1874 2046 2046 -5 -1023 -1023 -853 -1022 -1023
-4 1842 1841 1704 1841 1841.400024 -4 -819 -818 -682 -818 -818.400024
-3 1637 1636 1534 1637 1636.800049 -3 -614 -613 -512 -613 -613.800049
-2 1432 1432 1363 1432 1432.199951 -2 -409 -409 -341 -408 -409.200012
-1 1228 1227 1193 1228 1227.599976 -1 -205 -204 -170 -204 -204.600006
0 1023 1023 1023 1023 1023 0 0 0 0 0 0
1 818 819 853 819 818.400024 1 205 204 170 205 204.600006
2 614 614 683 615 613.799988 2 409 409 341 409 409.200012
3 409 410 512 410 409.199982 3 614 613 512 614 613.800049
4 205 205 342 205 204.599991 4 818 818 682 818 818.400024
5 0 0 172 1 0 5 1023 1023 853 1023 1023
6 -205 -204 1 -204 -204.600052 6 1228 1227 1024 1228 1227.600098
7 -409 -409 -169 -408 -409.199982 7 1432 1432 1194 1432 1432.199951
8 -614 -613 -339 -613 -613.800049 8 1637 1636 1365 1637 1636.800049
9 -818 -818 -510 -818 -818.399963 9 1841 1841 1536 1841 1841.399902
10 -1023 -1023 -680 -1022 -1023 10 2046 2046 1706 2046 2046``````
I'm not eager to extend the SPI library public API this way, so as written, no, probably not.
But I would be happy to merge this if it didn't change the public API. Maybe some trickery with inline methods and possibly __builtin_const_p could let these be private functions that get automatically used when the inputs are conducive?
Paul I commented in the PR, that I also wanted to add these (16 bit versions) earlier and we removed it from the PR, back then. And as I commented on this PR, if we do add the 16 bit extensions we should add them to 3.x and LC as well.
I believe inttypes.h were added in C99, and I think C++11. I left the C standards committee between the C90 and C99 standards, so I don't recall exactly when it got added. It relies on string concatenation that was in the original C standard (and in fact was one of the changes added by the committee that weren't in the original K&R C and later pcc, GNU compilers).
String concatenation is when the lexer sees two or more adjacent string literals together, and pastes them into one string. I tend to use it for printf formats to break long lines, but it is also useful for substitution of the formats:
Code:
`````` printf ("This is the first line\n"
"This is the second line\n"
"This is the third line\n");``````
IMHO, most of the code written here is tightly coupled to the used processor anyway. So, using all this complicated syntax just to get platform independent printfs for platform dependent code seems a bit, say, overengineered to me.
I appreciate having the ability in the times when the size of the item needs to be a given size (since I have used many different platforms over the years). While I haven't programmed an AVR 32u4 (such as the Leonardo) or AVR 328p (such as the Uno) in a few years, I have moved some of my code between the platforms. There int was 16-bits, instead of 32-bits And who knows, maybe the Teensy 5 systems will have different defaults.
MichaelMeissner said:
I believe inttypes.h were added in C99, and I think C++11. I left the C standards committee between the C90 and C99 standards, so I don't recall exactly when it got added. It relies on string concatenation that was in the original C standard (and in fact was one of the changes added by the committee that weren't in the original K&R C and later pcc, GNU compilers).
You are correct Michael. When I was looking at the c++ man pages it showed C99 and a tab for C++11.
Ok - here is a side by side comparison. looks like its working both in forward and reversed ! ....
Thanks for testing!
Hopefully we will finally have a map() function that meets all expectations.
Thanks for testing!
Hopefully we will finally have a map() function that meets all expectations.
Not problem glad to help. This is deja vu. Seem to remember you working on the map() function a couple of years ago as well.
If there is anything else you need checked let me know.
I tried making use of the SPI 16 and 32 bit block transfer functions. The Ethernet library fails to run if they're used.
Can anyone see a mistake?
https://github.com/PaulStoffregen/SPI/commit/aedb543cb374217c1ee4c731ffcb8f06f7533afa
I'm about to turn on my oscilloscope and start checking if these functions actually work...
I am guessing byte ordering.
For example earlier when I had transfer16 with buffers, I could use it do an easy faster display driver for 16 bit colors, which worked great and did not work correctly to use the transfer(.,..) buffer version as the LSB and MSB bytes were swapped....
Note: On T3.x code we have stuff that when you call transfer with buffers, we try packing the data into words and unpack as we push the stuff onto the FIFO and pull it off ... Don't think we ever did that on T4 code yet.
Looking at the SPI.h looks like you still have some 1052 defines (line 1139 or so not sure still needed unless for posterity );
Don't really use ethernet much but looking at the lib looks like the only place that uses SPI is in W5100.cpp especially in the read write function (line 298) (not sure I even have a wiznet board around but might). The 32 bit write looks like this (not sure what chip == 52 or 51 or 55 means though):
Code:
`````` } else if (chip == 52) {
setSS();
cmd[2] = ((len >> 8) & 0x7F) | 0x80;
cmd[3] = len & 0xFF;
SPI.transfer(cmd, 4);
#ifdef SPI_HAS_TRANSFER_BUF
SPI.transfer(buf, NULL, len);
#else
// TODO: copy 8 bytes at a time to cmd[] and block transfer
for (uint16_t i=0; i < len; i++) {
SPI.transfer(buf[i]);
}
#endif``````
Don't remember what we did on the transfer16 in the display libs so will defer to Kurt
Turns out this optimization makes only a small improvement.
Here's a little test program which transmits 12 bytes at 33 Mbit/sec.
Code:
``````#include <SPI.h>
void setup() {
SPI.begin();
pinMode(10, OUTPUT);
digitalWrite(10, HIGH);
}
void loop() {
uint32_t data[] = {0x12345678, 0xDEADC0DE, 0x55AA964C};
SPI.beginTransaction(SPISettings(33000000, MSBFIRST, SPI_MODE0));
digitalWriteFast(10, LOW);
SPI.transfer(data, sizeof(data));
digitalWriteFast(10, HIGH);
SPI.endTransaction();
delay(10);
}``````
With the original byte-oriented block transfer, total time taken is 3.67 us.
Using the 32 bit word-oriented block transfer, total time taken is 3.54 us.
You can see the bytes are in the wrong order, but otherwise it seems to be working. Not sure if this modest speedup is worth the complexity? Any thoughts?
I will try to take a look through some of the code for the fun of it.
Some of the more recent changes were sort of a hack to get around some interesting features of the TCR register.
That is if you do something like:
spi_registers->TCR = <Set to 16 bit mode >
Code:
`````` spi_registers->TDR = my_word;
spi_registers->TCR = <Set to 8 bit mode>
....
Serial.print(spi_registers->TCR, HEX);``````
The print will show you are in 16 bit mode. This was the issue we had awhile ago if someone did: transfer(x);transfer16; transfer16(z); transfer(a);
The a would output 16 bits... Why, because the transfer 16, Read in the current TCR, set it to 16 bit mode, output its data, set it back to the saved TCR... Well when you call the transfer16 the
second time it reads that it is in 16 bit mode and restores it to 16 bit mode....
And yes, the KinetisK code originally always reversed the bytes when it packed them to the PUSHR register, but that screwed things up when code was LSBFIRST
@KurtE is the expert here but I think why the speed up for 1 32-bit word transfer or 16-bit (my guess) is small if you are doing alot of transfers like for a display it may add up to a greater savings - not sure I said that correctly.
Turns out this optimization makes only a small improvement.
You can see the bytes are in the wrong order, but otherwise it seems to be working. Not sure if this modest speedup is worth the complexity? Any thoughts?
Yep that was sort of what I saw earlier on, which is why at the time I punted on it at the time...
But I did like the transfer16(buf, retbuf, cnt) like functions.
as one could write pretty quickly functions like we use in display libraries pretty quickly...
Like:
Code:
``````void ILI9341_t3::writeRect(int16_t x, int16_t y, int16_t w, int16_t h, const uint16_t *pcolors)
{
beginSPITransaction(_clock);
writecommand_cont(ILI9341_RAMWR);
for(y=h; y>0; y--) {
for(x=w; x>1; x--) {
writedata16_cont(*pcolors++);
}
writedata16_last(*pcolors++);
}
endSPITransaction();
}``````
Code:
``````void ILI9341_t3::writeRect(int16_t x, int16_t y, int16_t w, int16_t h, const uint16_t *pcolors)
{
beginSPITransaction(_clock);
writecommand_cont(ILI9341_RAMWR);
_pspi->transfer16(pcolors, nullptr, w*h);
endSPITransaction();
}``````
Note: I did not do the full conversion here showing what the other functions would be without touching hardware registers. Again on T3.x it would still be reasonably slower than the current code in that with T3.x the PUSHR can handle the automatic switching of DC pin, where here the transfers have to complete and then change. But still would not be bad...
Side note: the above is why the ESP32's SPI has the methods:
Code:
`````` void writePixels(const void * data, uint32_t size);//ili9341 compatible
void writePattern(uint8_t * data, uint8_t size, uint32_t repeat);``````
For my use case I was using transfer32 to write 2 pixels at a time to 6 displays which offered a significant increase in performance. Of course for the pixel data the byte order didn’t have to be changed so I could just write all the buffers as is and the displays showed up correctly. I know Paul doesn’t like making changes to public API, but it would be nice to access these functions in some type of way for the special cases that can make use of them.
From : File-abstraction-and-SdFat-integration
@KurtE / @mjs513:: ... @Paul
the Println() is ugly - it prints like:
Code:
``````size_t Print::println(void)
{
uint8_t buf[2]={'\r', '\n'};
return write(buf, 2);
}``````
I suppose the fix is to have the Windows IDE SerMon act like it does under Mac/Linux and turn "\r\n" into a single "\n" for GUI display?
Or maybe the '\r' and '\n' are not needed for Mac/Linux either and they just happen to eat the Return [ like a good line printer ] then do the New Line without showing a double line feed like on Windows.
the Println() is ugly
You need to be much more specific about how to reproduce this problem!
I tried just now with Arduino 1.8.13 and 1.54-beta2 on Windows 10. I believe you can see in this screenshot it looks perfectly fine.
I'm sure you're doing something else that does bring out some sort of problem. But how am I supposed to guess what you're actually doing that's giving ugly printing?
You need to be much more specific about how to reproduce this problem!
I tried just now with Arduino 1.8.13 and 1.54-beta2 on Windows 10. I believe you can see in this screenshot it looks perfectly fine.
I'm sure you're doing something else that does bring out some sort of problem. But how am I supposed to guess what you're actually doing that's giving ugly printing?
Oops - guess I forgot the other piece of the puzzle.
When you dump to the SerMon everything looks fine, as you have shown, but now try and copy the contents of the window and paste it into your editor of choice including the forum. You will see extra blank lines. Using your example:
Code:
``````Hello World 1
Hello World 2
Hello World 3
Hello World 4
Hello World 5
Hello World 6``````
@Paul and all - yep that is the issue I mentioned earlier.
As for stuff for the next beta - Wonder about the Hardware Serial Half duplex support I did earlier in the Pull Request: https://github.com/PaulStoffregen/cores/pull/489
It was sort of fun to do an T3.x as no real additional overhead to the Serial Writes and the ISR as in this case simply changed the bitband register value we were saving to instead of setting or clearing the IO pins it instead changed the bit in the IO register that set the TX pin to transmit or receive.
Ah, yes, I see it now. Double space with copy & paste. This problem is on the Arduino side. To see this, select the non-Teensy COM port from the Tools > Ports menu. That will use Arduino's normal (slow) serial monitor. If you copy & paste from that one, you'l get single space lines.
I've put this on my bug list. But I'm going to wait until at least beta4. I want to get beta3 wrapped up for testing just as soon as the new File and SD / SdFat stuff is stable enough for widespread testing.
Installed on macOS Catalina 10.15.7
No new issues so far. Haven't looked at any of the new features but all old stuff seems to work.
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Not open for further replies. | 7,212 | 20,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.556082 |
https://www.mymathgenius.com/Library/Math133_Unit_2_Discussion_Question | 1,721,765,279,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00081.warc.gz | 772,599,032 | 14,091 | # Math133 Unit 2 Discussion Question
Assignment Name: Unit 2 Discussion Board
Deliverable Length: 2–3 paragraphs
Details:
The Discussion Board (DB) is part of the core of online learning. Classroom discussion in an online environment requires the active participation of students and the instructor to create robust interaction and dialogue. Every student is expected to create an original response to the open-ended DB question as well as engage in dialogue by responding to posts created by others throughout the week. At the end of each unit, DB participation will be assessed based on both level of engagement and the quality of the contribution to the discussion.
At a minimum, each student will be expected to post an original and thoughtful response to the DB question and contribute to the weekly dialogue by responding to at least two other posts from students. The first contribution must be posted before midnight (Central Time) on Wednesday of each week. Two additional responses are required after Wednesday of each week. Students are highly encouraged to engage on the Discussion Board early and often, as that is the primary way the university tracks class attendance and participation.
The purpose of the Discussion Board is to allow students to learn through sharing ideas and experiences as they relate to course content and the DB question. Because it is not possible to engage in two-way dialogue after a conversation has ended, no posts to the DB will be accepted after the end of each unit.
Modeling Costs for the General Store
Remember the form of a quadratic equation: ax2+bx+c
You will use: W=-.01x2+100x+c where (-.01x2+100x) represents the store's variable costs and c is the store's fixed costs.
So, W is the stores total monthly costs based on the number of items sold, x.
Think about what the variable and fixed costs might be for your fictitious storefront business - and be creative. Start by choosing a fixed cost, c, between \$5,000 and \$10,000, according to the following class chart:
If your last name starts with the letter Choose a fixed cost between
A-E \$5000-\$5700
F-I \$5800-\$6400
J-L \$6500-\$7100
M-O \$7200-\$7800
P-R \$7800-\$8500
S-T \$8600-\$9200
U-Z \$9300-\$10,000
Post your chosen c value in your subject line, so your classmates can easily scan the discussion thread and try to avoid duplicating your c value. (Different c values make for more discussion.)
Your monthly cost is then, W = -.01x2+100x+c.
Substitute the c value chosen in the previous step to complete your unique equation predicting your monthly costs.
Next, choose two values of x (number of items sold) between 100 and 200. Again, try to choose different values from classmates.
Plug these values into your model for W and evaluate the monthly business costs given that sales volume.
Discuss results and alternate business decisions.
In your own words, please post a response to the Discussion Board and comment on other postings. You will be graded on the quality of your postings.
Course Materials
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Level: College/University
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### Algebra,Math133,Unit 2, Discussion Question, Store
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https://blog.innerdrive.co.uk/5-ways-to-use-dual-coding | 1,660,714,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00511.warc.gz | 160,479,993 | 15,735 | <img alt="" src="https://secure.coat0tire.com/222470.png" style="display:none;">
# 5 easy ways to use dual coding for studying
Have you heard of dual coding? It had been gaining popularity as a study and revision technique, so you may well know all about what it is, and perhaps even why it is such an important learning strategy. But do your students actually know how to use it?
In this blog, we’re exploring 5 different ways to use dual coding in revision. But first, let’s have a quick recap on what dual coding is…
## What is dual coding?
Dual coding is the process of blending both words and pictures while learning. Viewing those two formats gives us two different representations of the same piece of information. This works because visual and verbal information are processed differently, which when both viewed simultaneously improves learning - hence the term “dual coding”.
This helps form a stronger connection and gives the information a better chance of cementing into our long-term memory, as well as helping us to recall this information more easily when needed.
Dual coding expert, Oliver Caviglioli (who has written a great book on the topic which we’ll be reviewing soon) suggests 4 key principles for using dual coding. These are:
• Cut - Reduce the amount of content, be selective and only use the most important information. It’s all too easy to just copy down everything about a topic, with a couple of added pictures, in an attempt to make an exhaustive resource. However, this just causes an overload of information which is difficult to read and remember.
• Chunk - Divide the content into groups of related information;
• Align - Make sure that words and pictures are neatly ordered, making them easier to read;
• Restrain - Avoid “overdoing” it. In other words, don’t go crazy with different colours and fonts.
So, what are some different ways that you and your students can use dual coding?
## 5 Ways to use Dual Coding
Drawings
This does what it says on the tin: we’re talking about simply drawing pictures. Research suggests that creating a drawing from a piece of information requires you to elaborate on its meaning. You have to really consider the information to decipher how best to represent it, create the necessary motor movements with a pencil, and use pictorial processing to inspect your drawing once it’s finished. This deeper processing helps to cement the information into long-term memory.
Many students will be quick to dismiss this strategy because they “can’t draw”. The fact is, there’s no need to be an artist - the most important thing is that your drawings have meaning. As long as they represent the information you need to remember, even if this is only meaningful to you, drawings are a great way to boost student learning.
Graphic organisers
A graphic organiser is a way to organise information (both words and pictures) according to the relationships between different concepts.
There are loads of different types of graphic organisers, all corresponding to different ways of thinking: mind maps and tree diagrams organise information by “chunking” it into related groups; Venn diagrams organise information through comparing its similarities and differences; flowcharts organise information as it occurs within a sequence; other graphic organisers show information with cause-and-effect relationships.
If you need some inspiration, you can actually see some great examples of graphic organiser examples on Oliver Caviglioli’s website.
Diagrams
Diagrams are perfect for learning and revising complex concepts or processes that need breaking down to understand. They can represent a lot of information within a relatively small space, compared to writing it all out in one long passage, making difficult concepts easier to read and get to grips with. Here are several ways you can make your diagram dual coding more effective:
• Make use of signalling. This means drawing attention to the most important aspects of the diagram, such as key words. You could do so by highlighting, underlining, putting writing in bold, using arrows, circling… To maximise the impact of these techniques, use them sparingly.
• Place text within diagrams rather than around the outside. For example, if your diagram is simply an annotated picture, have the labels within the picture rather than around it or in a key. It’s much easier to process the words and pictures when they’re right next to each other, than if you have to keep switching your attention back and forth. This is called the spatial contiguity principle.
• One creative option is to leave diagram labels blank. You can then re-visit your diagram at a later date to test yourself on filling in the labels. This combines dual coding with two other extremely effective revision strategies that you should know about, retrieval practice and spacing.
Posters
Posters are great for combining writing, pictures and diagrams all within one page of information. They can be used to present all the most important information you know on a specific topic, for example. To ensure you can easily learn and revise from your poster, here are some key tips:
• Use hierarchy - Be sure to include titles and subtitles, and to emphasise these with larger writing or writing in bold. This not only make the poster easier to read, but also gives the information an element of order to help you understand it better.
• Be consistent - Consistency of organisation, colour coding, fonts, and other styling aspects will all help to make your poster more readable.
• Use contrast - Writing large chunks of text in one colour or highlighting everything on your poster doesn’t make anything stand out. Instead, use one or two key colours to create contrast and allow key information to stand out. The more sparsely you use colour, the more effective it will be in helping you to remember the key bits of information. It’s all about that 4th principle: restraint.
Timelines
For any information that happens in a particular order or sequence, timelines should be a go-to. And the same rules apply: collect the information you need to know, cut irrelevant information, chunk information into different sections of time, and put these into an organised sequence, all the time using restraint with colours and text and in general not trying to be too “arty”. As always, less is more.
## Final thoughts
We hope we’ve given you some practical knowledge so that you can go off and start using dual coding to learn and remember information. We should add that, while we’ve aimed the tips in this blog towards students, these different ways of using dual coding all apply to teachers wishing to use dual coding in their lessons, too. | 1,335 | 6,728 | {"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} | 2.703125 | 3 | CC-MAIN-2022-33 | longest | en | 0.924984 |
https://gmatclub.com/reviews/comments/manhattan-gmat-9-session-in-person-prep-course-345355375 | 1,621,144,528,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989690.55/warc/CC-MAIN-20210516044552-20210516074552-00392.warc.gz | 302,159,933 | 25,409 | GMAT Club
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February 02, 2018
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Verified score:
720 Q46 V42
## 720 from 590 - Money well spent
REVIEWER IDENTITY VERIFIED by score report [?]
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Got my target 720 score after approx 4 months that comprised self study, Manhattan Prep's 9-week course on Wall street Manhattan (David Mahler was our instructor), and private tutoring (also with Dave Mahler). I started with a 590 Manhattan Prep GMAT practice test. Quite frankly, I could not have managed the 720 without Manhattan Prep.
1. The 9-week course: This was a great start for me as my math skills were very rusty (haven't touched math since high school and my day-to-day work doesn't require the use of any meaningful quant skill - I'm a lawyer). The math foundations and GMAT test strategies explored in class were concise and immediately helpful. Dave Mahler was our instructor - I highly recommend him for the clarity with which he explains tough concepts (particularly Data Sufficiency and Critical Reasoning). FYI the 9-week course comes with the full suite of Manhattan Prep Books, which were all great - concise, clear, and effective - no verbiage which is good for busy professionals.
2. Private tutoring: After the 9 week course, I spent a couple of weeks wrapping up the GMAT OG and then I took a real GMAT test. Bombed it. I decided I needed more targetted tutoring, so I splurged on the private tutoring (8 lessons in total). This was worth every penny. Dave diagnosed my flawed test-taking strategies, and his private lessons sharpened my quant and verbal reasoning skills. These lessons were challenging but crucial to my success on the test - he would often ask me to explain my thought processes WHILE I solved tough problems (DS, PS, SC, CR), and this invariably forces you to think critically about every single step in your own problem-solving process. Every step in my problem-solving process was critiqued and improved upon. As important, Dave was extremely optimistic and encouraging throughout this whole process. He was a firm supporter and you feel that he's invested in your success.
TL DR: Manhattan Prep's 9 week course and private tutoring are both very strongly recommended. Manhattan Prep's books are very strongly recommended. Dave Mahler is very strongly recommended.
This reviewer has not participated on GMAT Club but it is a REAL person and a REAL review. GMAT Club has verified this test-taker's identity through GMAC/Pearson Vue Score Reporting system and confirmed that this reviewer indeed took the GMAT, is unique, and has not submitted multiple reviews. | 586 | 2,659 | {"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 | 0.965012 |
https://tradersinsight.news/ibkr-quant-news/stochastic-oscillator-part-i/ | 1,680,003,560,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00545.warc.gz | 656,555,443 | 14,941 | # Stochastic Oscillator – Part I
Contributor:
QuantInsti
Visit: QuantInsti
This blog tells us how to calculate and plot the Stochastic Oscillator. We also understand the difference between the Slow, Fast and Full Stochastic Oscillator along with the advantages, as well as disadvantages of the Stochastic Oscillator.
The Stochastic Oscillator is not as popular as let’s say RSI, Bollinger Bands or Moving Averages, but it is useful if we use it with other indicators.
If you have read a few of our blogs about indicators in the past, you would realise that we usually pick one indicator and give an in-depth brief about it along with a few applications, as well as the strategies associated with it. Today we will be trying to tackle two indicators, i.e., the Stochastic Oscillator and Average True Range indicator. We will also see how they can be used to strengthen your trading skills and help you create a formidable strategy.
Stochastic Oscillator definition
The Stochastic Oscillator is a momentum indicator which compares the recent closing price of an asset to a range of its prices over a specific period of time. While the Stochastic Oscillator is supposed to be similar to RSI, another technical indicator, we will see later on in the article how the two indicators are different.
Why is Stochastic Oscillator used?
In the words of Dr George Lane, who helped create the Stochastic Oscillator, it is an indicator which does not follow price or volume but signifies the speed, or momentum of a price. In this manner, it helps us anticipate a change in the direction of the price.
Thus, the Stochastic Oscillator has become a well-known indicator among traders to identify a bullish or bearish trend in the market.
We will now understand how to plot the Stochastic Oscillator with the help of an example.
Example to understand the Stochastic Oscillator
The Stochastic Oscillator consists of two lines, %K and %D.
The story goes that in the 1950s, while George Lane and his colleagues were trying to plot different oscillators by hand, they would run out of chart paper due to the range of values. Thus they tried to express them as percentages.
After trial and error, with each formula expressed as an alphabet, they finally managed to create a stable indicator with %K.
We will look into %K in more detail now.
Formula for calculating %K
The %K can be expressed as,
``[(Current closing price - lowest low price in selected range) / (highest high price in selected range - lowest low price in selected range)] * 100``
Stay Tuned for Part II in which Rekhit will discuss Calculating Stochastic Oscillator Live Market Example.
##### Disclosure: Interactive Brokers
Information posted on IBKR Traders’ Insight that is provided by third-parties and not by Interactive Brokers does NOT constitute a recommendation by Interactive Brokers that you should contract for the services of that third party. Third-party participants who contribute to IBKR Traders’ Insight are independent of Interactive Brokers and Interactive Brokers does not make any representations or warranties concerning the services offered, their past or future performance, or the accuracy of the information provided by the third party. Past performance is no guarantee of future results.
This material is from QuantInsti and is being posted with permission from QuantInsti. The views expressed in this material are solely those of the author and/or QuantInsti and IBKR is not endorsing or recommending any investment or trading discussed in the material. This material is not and should not be construed as an offer to sell or the solicitation of an offer to buy any security. To the extent that this material discusses general market activity, industry or sector trends or other broad based economic or political conditions, it should not be construed as research or investment advice. To the extent that it includes references to specific securities, commodities, currencies, or other instruments, those references do not constitute a recommendation to buy, sell or hold such security. This material does not and is not intended to take into account the particular financial conditions, investment objectives or requirements of individual customers. Before acting on this material, you should consider whether it is suitable for your particular circumstances and, as necessary, seek professional advice.
In accordance with EU regulation: The statements in this document shall not be considered as an objective or independent explanation of the matters. Please note that this document (a) has not been prepared in accordance with legal requirements designed to promote the independence of investment research, and (b) is not subject to any prohibition on dealing ahead of the dissemination or publication of investment research.
Any trading symbols displayed are for illustrative purposes only and are not intended to portray recommendations. | 966 | 4,941 | {"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-2023-14 | latest | en | 0.950318 |
http://www.listserv.uga.edu/cgi-bin/wa?A2=ind0609c&L=sas-l&D=1&O=A&F=&S=&P=23488 | 1,371,717,116,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711005723/warc/CC-MAIN-20130516133005-00061-ip-10-60-113-184.ec2.internal.warc.gz | 545,385,991 | 4,091 | ```Date: Mon, 18 Sep 2006 10:52:03 -0400 Reply-To: wing wah Sender: "SAS(r) Discussion" From: wing wah Subject: hash Dear folks, I am trying to reconstruct the real time orderbook in financial market. If the quotes below are bid price that is posted by traders, i am interested to rank the posted quote as the bestbid secondbid thirdbid fourthbid and so on. The quantity of each rank (bestbid, secondbid,....) is sum of accumulated quantity of those rank before it. i.e. quantity of bestbid=quantity of bestbid, accumulated quantity of secondbid=quantity of secondbid+quantity of bestbid, and so on... I hope the output of the code will provide a clearer explanation of the logic behind what i am trying to construct. Since i am trying to create a demand curve. any price lower than the bestbid price will have an accumulated quantity (look at the demand and supply curve). By doing so, I will have to think of a way to extract the price and arrange them as the bestbid, secondbid..... The best bid from the output from the simple code below will be the first price that is not '.' starting from right to left. The secondbid will be the next accumated quantity change after then bestbid from right to left. At least by doing this, i am not worried about how far the queue will stretch. The datasize is about 10G with about 20 million observations. Firstly, I am wondering if this can be done more efficiently using 'hash'. Secondly, if the code below is a sensible one, from the output below how can i extract the bestbid, secondbid,... and the quantity? I am using proc transpose for testing but i am not sure about using it when the data size increases. All suggestions and criticism are welcome. Thank you in advance! Wing data given; input quote quantity ; cards; 0.6126 3 0.6126 -3 0.611 1 0.611 -1 0.6115 1 0.6115 -1 0.6103 1 0.6103 -1 0.6075 5 0.6075 -5 0.6061 19 0.609 2 0.6118 2 0.6075 5 0.6061 -19 0.6084 19 0.6118 -2 0.6115 1 0.6121 2 0.6121 -2 0.612 1 0.6118 2 0.6121 1 0.6121 -1 0.612 -1 0.612 5 0.6054 2 0.612 1 0.612 -1 0.61205 1 0.6116 5 0.61105 1 0.61105 -1 0.61205 -1 0.6123 1 0.6123 -1 0.612 2 0.612 2 0.612 -2 0.6116 -5 0.6115 -1 0.612 1 ; DATA maxmin; SET given END=lastobs ; IF _N_ = 1 THEN DO; start = quote; finish = quote ; END; RETAIN start finish; start = (MIN(start,quote)); finish = (MAX(finish,quote)) ; IF lastobs THEN DO; start=start*100000-1; finish=finish*100000+1; diffmaxminprice=finish-start+1; CALL SYMPUT('lo',LEFT(PUT(start,8.))) ; CALL SYMPUT('hi',LEFT(PUT(finish,8.))); CALL SYMPUT('diffmaxminprice',LEFT(PUT(diffmaxminprice,8.))); END; RUN; %put lo &lo.; %put hi &hi; %put diffmaxminprice &diffmaxminprice; data roll3(drop = i newpoint); set given (obs=1000); array v(&LO:&HI) v&LO-v&HI; retain v&LO-v&HI id(0) ; if _n_=1 then id=0; id=id+1; bid=int(quote*100000); newpoint = missing(v(bid) ); do i = bid to &lo+1 by -1; if i=bid or not missing(v(i) ) then v(i) = sum(v(i),quantity); if v(i)<=0 then v(i) = .; end; do i = bid+1 to &Hi-1 until(v(i) or i=&hi); end; if v(bid)=v(i) then v(bid)=.; if newpoint then do; do i = bid+1 to &Hi-1 until(v(i) or i=&hi); end; v(bid) = sum(v(bid),v(i)); end; run; proc transpose data=roll3 out=dvector; by id ; var v:; run;quit; data dvector; set dvector; if col1=. or col1=0 then delete; run; ```
Back to: Top of message | Previous page | Main SAS-L page | 1,119 | 3,336 | {"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-2013-20 | latest | en | 0.844875 |
http://www.shmoop.com/quadratic-formula-function/imaginary-complex-number-exercises.html | 1,464,724,576,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464052946109.59/warc/CC-MAIN-20160524012226-00013-ip-10-185-217-139.ec2.internal.warc.gz | 776,428,704 | 15,956 | We have changed our privacy policy. In addition, we use cookies on our website for various purposes. By continuing on our website, you consent to our use of cookies. You can learn about our practices by reading our privacy policy.
Quadratic Formula and Functions
Imaginary and Complex Numbers Exercises
Imaginary and Complex Numbers Exercises
Example 1
Simplify.
3i3 – 4i(6i16) + 2
Simplify.
i2346
Example 3
Simplify.
5 + i – (i7 – 4)(2i + 1)
Simplify.
(2 – i)4
Example 5
Simplify.
-1 + 3i2 + 6(4 – i) | 155 | 517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-22 | longest | en | 0.89772 |
https://mathematica.stackexchange.com/questions/255302/how-to-make-a-calculator-to-display-the-non-dimensional-number | 1,716,382,018,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00628.warc.gz | 327,180,874 | 39,374 | # How to make a calculator to display the Non-Dimensional number
I am a beginner here and want to learn this tool. At first, I want to make one calculator that will display the value of the Non-Dimensional number of Fluid dynamics based on user inputs.
In this code, I need to add one output display box, which will show the value of that number and named that output box name as Reynolds number.
Manipulate[ Rey[v], {{v, 1.5, "Velocity of Fluid(m/s): "}, ControlType -> InputField, FieldSize -> 7},{{roh, 1, "Fluid Density (Kg/m^3): "}, ControlType -> InputField, FieldSize -> 7}, {{dia, 0.01, "Hydraulic Diameter (m) "}, ControlType -> InputField, FieldSize -> 7}, {{muh, 0.00089, "Dynamic Viscosity (Kg/ms): "},ControlType -> InputField, FieldSize -> 7}, Initialization :> {Rey[v_] := (roh*v*dia)/muh}]
• Crossposted here. Sep 6, 2021 at 14:21
Update 2 Improve layout, add Calculate button
CloudDeploy[
Manipulate[ "Reynolds number: " <> ToString@Style[Rey[v], 14],
Text[Style["Reynolds number calculator", 16]],
Delimiter,
{{v, 1.5, "Velocity of Fluid(m/s): "}, ControlType -> InputField,
FieldSize -> 7}, {{roh, 1, "Fluid Density (Kg/m^3): "},
ControlType -> InputField,
FieldSize -> 7}, {{dia, 0.01, "Hydraulic Diameter (m) "},
ControlType -> InputField,
FieldSize -> 7}, {{muh, 0.00089, "Dynamic Viscosity (Kg/ms): "},
ControlType -> InputField, FieldSize -> 7},
Delimiter,
Item[Button["Calculate", Nothing, BaseStyle -> {15}]],
LabelStyle -> {14},
ContentSize -> 310,
Initialization :> {Rey[v_] := (roh*v*dia)/muh},
TrackedSymbols :> All]]
Update More attractive :-)
Manipulate[
ResourceFunction["WolfieSay"][Labeled[Rey[v], "Reynolds number", Top]],
{{v, 1.5, "Velocity of Fluid(m/s): "}, ControlType -> InputField, FieldSize -> 7},
{{roh, 1, "Fluid Density (Kg/m^3): "}, ControlType -> InputField, FieldSize -> 7},
{{dia, 0.01, "Hydraulic Diameter (m) "}, ControlType -> InputField, FieldSize -> 7},
{{muh, 0.00089, "Dynamic Viscosity (Kg/ms): "},ControlType -> InputField, FieldSize -> 7},
Initialization :> {Rey[v_] := (roh*v*dia)/muh},
TrackedSymbols :> All]
Something like this?
Manipulate[
Panel[Row[{"Reynolds number: ", Rey[v]}]],
{{v, 1.5, "Velocity of Fluid(m/s): "}, ControlType -> InputField, FieldSize -> 7},
{{roh, 1, "Fluid Density (Kg/m^3): "}, ControlType -> InputField, FieldSize -> 7},
{{dia, 0.01, "Hydraulic Diameter (m) "}, ControlType -> InputField, FieldSize -> 7},
{{muh, 0.00089, "Dynamic Viscosity (Kg/ms): "},ControlType -> InputField, FieldSize -> 7},
Initialization :> {Rey[v_] := (roh*v*dia)/muh},
TrackedSymbols :> All]
• Thanks, Rohit.. it is close to what I am looking for... Do you know any other way... which shows output more attractive... Sep 6, 2021 at 14:47
• What exactly do you mean by "output more attractive"? The output is a string and a real number. Sep 6, 2021 at 19:58
• I want to build a calculator something like given in this link resources.wolframcloud.com/FormulaRepository/resources/… Sep 7, 2021 at 15:09
• In the above link formula repository, we can input the real number on the corresponding string and after clicking the submit button we can display all values at bottom of it. In the same way, I want to build the calculator in my case. Do you know how I can get the source code of the above link? Sep 7, 2021 at 16:04
• The code in my answer does not require a submit button because it is not hosted on a web page. Change any of the numbers in the input fields and hit enter/return (not shift enter/return) and the computed Reynolds number will update. Are you interested in building a web interface or one that will run in Mathematica/Notebook Player? Sep 7, 2021 at 17:10 | 1,107 | 3,656 | {"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} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.705449 |
http://docplayer.net/17752457-Tutorial-1-introduction-to-using-ansys-fluent-in-ansys-workbench-fluid-flow-and-heat-transfer-in-a-mixing-elbow.html | 1,547,947,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583688396.58/warc/CC-MAIN-20190120001524-20190120023524-00123.warc.gz | 62,326,035 | 37,495 | # Tutorial 1. Introduction to Using ANSYS FLUENT in ANSYS Workbench: Fluid Flow and Heat Transfer in a Mixing Elbow
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1 Tutorial 1. Introduction to Using ANSYS FLUENT in ANSYS Workbench: Fluid Flow and Heat Transfer in a Mixing Elbow Introduction This tutorial illustrates using ANSYS Workbench to set up and solve a three-dimensional turbulent fluid flow and heat transfer problem in a mixing elbow using ANSYS FLUENT fluid flow systems. This tutorial is designed to introduce you to the ANSYS Workbench tool set using a familiar geometry (the first tutorial in the separate Tutorial Guide). Within this tutorial, you will create the elbow geometry and the corresponding computational mesh using the geometry and meshing tools within ANSYS Workbench. You will use ANSYS FLUENT to set up and solve the CFD problem, then visualize the results in both ANSYS FLUENT and the ANSYS CFD-Post postprocessing tool. Some capabilities of ANSYS Workbench (e.g., duplicating fluid flow systems, connecting systems, and comparing multiple data sets) are also examined in this tutorial. This tutorial demonstrates how to do the following: Launch ANSYS Workbench. Create an ANSYS FLUENT fluid flow analysis system in ANSYS Workbench. Create the elbow geometry using ANSYS DesignModeler. Create the computational mesh for the geometry using ANSYS Meshing. Set up the CFD simulation in ANSYS FLUENT, which includes: Set material properties and boundary conditions for a turbulent forced convection problem. Initiate the calculation with residual plotting. Calculate a solution using the pressure-based solver. Visually examine the flow and temperature fields using ANSYS FLUENT (and ANSYS CFD-Post). Create a copy of the original ANSYS FLUENT fluid flow analysis system in ANSYS Workbench. Release 12.1 c ANSYS, Inc. September 10,
2 Change the geometry in ANSYS DesignModeler, using the duplicated system. Regenerate the computational mesh. Recalculate a solution in ANSYS FLUENT. Compare the results of the two calculations in ANSYS CFD-Post. Prerequisites This tutorial assumes that you have little to no experience with ANSYS DesignModeler, ANSYS Meshing, ANSYS FLUENT, or ANSYS CFD-Post, and so each step will be explicitly described. Problem Description The problem to be considered is shown schematically in Figure 1.1. A cold fluid at K flows into the pipe through a large inlet, and mixes with a warmer fluid at K that enters through a smaller inlet located at the elbow. The pipe dimensions, the fluid properties and boundary conditions are given in SI units. Note: Since the geometry of the mixing elbow is symmetric, only half of the elbow needs to be modeled. 1-2 Release 12.1 c ANSYS, Inc. September 10, 2009
3 Density: ρ = 1000 kg/m 3 Viscosity: 4 µ = 8 x 10 Pa s Conductivity: k = W/m K Specific Heat: = 4216 J/kg K Cp 8" 4" U x = 0.4 m/s o T = 20 C I = 5% 4" Dia. 8" 3" 1" 1" Dia. = 1.2 m/s o T = 40 C Uy I = 5% Figure 1.1: Problem Specification Release 12.1 c ANSYS, Inc. September 10,
4 Setup and Solution Preparation 1. Download elbow-workbench.zip from the User Services Center to your working folder. This file can be found by using the Documentation link on the ANSYS FLUENT product page. 2. Unzip elbow-workbench.zip. The.zip file contains a folder (elbow-workbench) that in turn contains two geometry files (elbow-geometry.agdb and elbow-geometry.stp) along with an AN- SYS Workbench project file (elbow-workbench.wbpj) and a corresponding project folder (elbow-geometry files) that contains supporting files for the project (e.g., solution and results files). Note: ANSYS FLUENT tutorials are prepared using ANSYS FLUENT on a Windows system. The screen shots and graphic images in the tutorials may be slightly different than the appearance on your system, depending on the operating system or graphics card. Step 1: Creating a Fluid Flow Analysis System in ANSYS Workbench In this step, you will start ANSYS Workbench and create a new fluid flow analysis system. 1. Start ANSYS Workbench by choosing the Start menu, then select the Workbench option in the ANSYS 12.1 program group. Start All Programs ANSYS 12.1 Workbench This displays the ANSYS Workbench application window, containing the Toolbox on the left and the Project Schematic on the right. Various supported analyses and applications are listed in the Toolbox, while you visualize the components of the analysis in the Project Schematic. When you first start ANSYS Workbench, the Getting Started pop-up window is displayed, offering assistance through the on-line help for using the application. You can keep the window open, or close it by clicking the red X icon in the upper right-hand corner. If you need to access the on-line help at any time, use the Help menu, or press the < F 1 > key. 1-4 Release 12.1 c ANSYS, Inc. September 10, 2009
5 2. Create a new fluid flow analysis system by double-clicking the Fluid Flow (FLUENT) option under Analysis Systems in the Toolbox. You will initially see a green dotted line outline in the Project Schematic, indicating potential locations for the new system. Figure 1.2: Selecting the Fluid Flow (FLUENT) Analysis System in ANSYS Workbench You can also drag-and-drop the analysis system into the Project Schematic: the green dotted outline will turn into a red box, indicating the location of the new system. Release 12.1 c ANSYS, Inc. September 10,
6 This creates a new ANSYS FLUENT-based fluid flow analysis system in the Project Schematic. Figure 1.3: ANSYS Workbench with a New ANSYS FLUENT-Based Fluid Flow Analysis System 3. Name the analysis. (a) Double-click the Fluid Flow (FLUENT) label underneath the analysis system (if it is not already highlighted). (b) Enter elbow for the name of the analysis system. 4. Save the project. (a) Select the Save... option under the File menu in ANSYS Workbench. File Save... This displays the Save As dialog, where you can browse to a specific folder and enter a specific name for the ANSYS Workbench project. (b) In your working folder, enter elbow-workbench as the project File name and click the Save button to save the project. ANSYS Workbench saves the project with a.wbpj extension as well as supporting files for the project. 1-6 Release 12.1 c ANSYS, Inc. September 10, 2009
7 5. View the files generated by ANSYS Workbench. ANSYS Workbench allows you to easily view the files associated with your project using the Files view. To open the Files view, select the Files option under the View menu. View Files Figure 1.4: ANSYS Workbench Displaying the Files View for the Project After Adding a ANSYS FLUENT-Based Fluid Flow Analysis System In the Files view, you will be able to see the name and type of file, the ID of the cell the file is associated with, the size of the file, the location of the file, and other information. For more information about the Files view, see the separate FLUENT in Workbench User s Guide and the ANSYS Workbench on-line help. From here, you will move on to create the geometry described in Figure 1.1, and later set up a fluid flow analysis for the geometry. Release 12.1 c ANSYS, Inc. September 10,
8 You will note that the ANSYS FLUENT- based fluid flow analysis system, for example, is composed of various cells (Geometry, Mesh, etc.) that represent the work flow for performing the analysis. ANSYS Workbench is composed of multiple data-integrated (e.g., ANSYS FLUENT) and native applications into a single, seamless project flow, where individual cells can obtain data from and provide data to other cells. As a result of this constant flow of data, a cell s state can quickly change. ANSYS Workbench provides visual indications of a cell s state at any given time via icons on the right side of each cell. Brief descriptions of the various states are provided below. For more information about cell states, see the ANSYS Workbench on-line help: Unfulfilled ( ) indicates that required upstream data does not exist. For example, when you first create a new Fluid Flow (FLUENT) analysis system, all cells downstream of the Geometry cell appear as Unfulfilled because you have not yet specified a geometry for the system. Refresh Required ( ) indicates that upstream data has changed since the last refresh or update. For example, after you assign a geometry to the geometry cell in your new Fluid Flow (FLUENT) analysis system, the Mesh cell appears as Refresh Required since the geometry data has not yet been passed from the Geometry cell to the Mesh cell. Attention Required ( ) indicates that the current upstream data has been passed to the cell, however, you must take some action to proceed. For example, after you launch ANSYS FLUENT from the Setup cell in a Fluid Flow (FLUENT) analysis system that has a valid mesh, the Setup cell appears as Attention Required because additional data must be entered in ANSYS FLUENT before you can calculate a solution. Update Required ( ) indicates that local data has changed and the output of the cell needs to be regenerated. For example, after you launch ANSYS Meshing from the Mesh cell in a Fluid Flow (FLUENT) analysis system that has a valid geometry, the Mesh cell appears as Update Required because the Mesh cell has all the data it needs to generate a ANSYS FLUENT mesh file, but the ANSYS FLUENT mesh file has not yet been generated. Up To Date ( ) indicates that an update has been performed on the cell and no failures have occurred (or an interactive calculation has been completed successfully). For example, after ANSYS FLUENT finishes performing the number of iterations that you request, the Solution cell appears as Up-to-Date. Interrupted ( ) indicates that you have interrupted an update (or canceled an interactive calculation that is in progress). For example, if you select the Cancel 1-8 Release 12.1 c ANSYS, Inc. September 10, 2009
9 button in ANSYS FLUENT while it is iterating, ANSYS FLUENT completes the current iteration and then the Solution cell appears as Interrupted. Input Changes Pending ( ) indicates that the cell is locally up-to-date, but may change when next updated as a result of changes made to upstream cells. For example, if you change the Mesh in an Up-to-Date Fluid Flow (FLUENT) analysis system, the Setup cell appears as Refresh Required, and the Solution and Results cells appear as Input Changes Pending. Step 2: Creating the Geometry in ANSYS DesignModeler For the geometry of your fluid flow analysis, you can create a geometry in ANSYS DesignModeler, or import the appropriate geometry file. In this tutorial, we will create the geometry from scratch in ANSYS DesignModeler. Note the Attention Required icon ( ) within the Geometry cell for the system. This indicates that the cell is requires data (e.g., a geometry). Once the geometry is defined, the state of the cell will change accordingly. Likewise for the state of the remaining cells in the system. For more information about system cell states, see the separate FLUENT in Workbench User s Guide and the on-line documentation for ANSYS Workbench. Note: If you would rather not create the geometry in ANSYS DesignModeler, you can import a pre-existing geometry by right-clicking the Geometry cell and selecting the Import Geometry option from the context menu. From there, you can browse your file system to locate the elbow-geometry.agdb geometry file that is provided for this tutorial. If you do not have access to ANSYS DesignModeler, you can use the elbow-geometry.stp file instead. Release 12.1 c ANSYS, Inc. September 10,
10 1. Start ANSYS DesignModeler. In the ANSYS Workbench Project Schematic, double-click the Geometry cell in the elbow fluid flow analysis system. This displays the ANSYS DesignModeler application. You can also right-click on the Geometry cell to display the context menu where you can select the New Geometry... option. 2. Set the units in ANSYS DesignModeler. When ANSYS DesignModeler first appears, you are prompted to select the desired system of length units to work from. For the purposes of this tutorial, where you will create the geometry in inches and perform the CFD analysis using SI units, select Inch as the desired length unit and click OK to close the prompt. Figure 1.5: Setting the Units in ANSYS DesignModeler 1-10 Release 12.1 c ANSYS, Inc. September 10, 2009
11 3. Create the geometry. The geometry for this tutorial (Figure 1.1) consists of a large curved pipe accompanied by a smaller side pipe. To create the larger main pipe, you will use the Sweep operation. Sweeping requires the use of two sketches: one that defines the profile to be swept (in this case, a half circle since the symmetry of the problem allows you to not have to generate the entire pipe geometry) and the other that defines the path through which the profile is swept. (a) Create the profile. i. Create a new plane by selecting YZPlane from the Tree Outline and click on New Plane from the Active Plane/Sketch toolbar, near the top of the ANSYS Workbench window. Clicking YZPlane first ensures that the new plane is based on the YZPlane. ii. In the Details View for the new plane (Plane 4), set Transform 1 (RMB) to Offset Global X, and set the Value of the offset to -8 in. iii. Click on Generate ( to create the plane. located in the ANSYS DesignModeler toolbar) Release 12.1 c ANSYS, Inc. September 10,
12 iv. Create a new sketch by selecting Plane4 from the Tree Outline and then click New Sketch from the Active Plane/Sketch toolbar, near the top of the ANSYS Workbench window. Clicking the plane first ensures that the new sketch is based on Plane4. v. On the Sketching tab, open the Settings toolbox, select Grid, and enable the Show in 2D and the Snap options. vi. Set Major Grid Spacing to 1 in and Minor-Steps per Major to 2. vii. Zoom in on the center of the grid, so that you can see the grid lines clearly. You can do this by holding down the right-mouse button and dragging a box over the desired viewing area Release 12.1 c ANSYS, Inc. September 10, 2009
13 viii. On the Sketching tab, open the Draw toolbox and select Arc by Center (you may need to use the arrows in the toolbox to scroll down to see the correct tool). Draw an arc with a radius of 2 in, centered on X = 6 in, Y = 0 in (located below the origin of Plane4). The grid settings that you have just set up will help you to position the arc and set its radius correctly. ix. On the Sketching tab open the Draw toolbox and select Line (you may need to use the arrows in the toolbox to scroll up to see the correct tool). Draw a line from X = 4 in, Y = 0 in to X = 8 in, Y = 0 in. Figure 1.6: Creating the Arc Profile (b) Create the path. i. On the Modeling tab select XYPlane, then click New Sketch and then Generate to create a new sketch based on the XYPlane. ii. On the Sketching tab, open the Settings toolbox, select Grid, and enable the Show in 2D and the Snap options. iii. Set Major Grid Spacing to 1 in and Minor-Steps per Major to 2. iv. In the Draw toolbox, select Line to draw two straight lines on the sketch. For reference, the coordinates of the endpoints of the lines are (X = 8 in, Y = 6 in), (X = 0 in, Y = 6 in) for the horizontal line and (X = 6 in, Y = 0 in), (X = 6 in, Y = 8 in) for the vertical line. v. In the Draw toolbox, select Arc by Center and click once on the origin (center of the arc). Now select one of the end points of the arc, and then move the mouse around to the other end point and click on it to draw the quarter-circle. If the wrong part of the arc is drawn (that is, a 270 degree segment instead of a 90 segment), click Undo from the Undo/Redo Release 12.1 c ANSYS, Inc. September 10,
14 toolbar and try again, making sure that after you click on the first end point, that you move the mouse in the correct direction for the arc that is to be drawn. Figure 1.7: Creating the Elbow Path (c) Create the pipe. i. Select Sweep from the 3D Features toolbar. ii. Set the Profile to be Sketch1: click on Sketch1 in the Tree Outline and then click on Apply in the Details View at the bottom-left of the screen. iii. Set the Path to Sketch2: click on the Not selected text next to Path, click on Sketch2 in the Tree Outline, and then click Apply. iv. Click on Generate to create the pipe. Figure 1.8: Generating the Pipe 1-14 Release 12.1 c ANSYS, Inc. September 10, 2009
15 (d) Create the side pipe. i. Create a new plane based on the ZXPlane: as before, first make ZXPlane active by clicking on it, then use New Plane to create the plane based upon it. ii. In the Details View, set Transform 1 (RMB) to Offset Global X, and set the Value of the offset to 5.5 in. iii. Set Transform 2 to Offset Global Y and the Value of the offset to -9 in. iv. Click on Generate to create the plane. v. With the new plane selected in the Tree Outline, create a New Sketch. vi. On the Sketching tab, open the Draw toolbox and select Arc by Center to create the arc centered on the origin with a radius of 0.5 in. Create the arc initially with any convenient radius, and then open the Dimensions toolbox and select Radius to specify the radius more precisely. vii. On the Sketching tab, open the Draw toolbox and select Line. Draw a line connecting the open ends of the arc. viii. Select Extrude from the 3D Features toolbar. ix. Set Base Object to be the new sketch (Sketch3), and set Operation to Add Material. x. Set Direction to Normal and Extent Type to Fixed. Set Depth to 4 in. xi. Click on Generate to create the side-pipe. Figure 1.9: Generating the Additional Pipe Release 12.1 c ANSYS, Inc. September 10,
16 (e) Specify the geometry as a fluid body. i. In the Tree Outline, open the 1 Part, 1 Body branch and select Solid branch. ii. In the Details View of the body, change the name of the Body from Solid to Fluid. iii. Change the Fluid/Solid property from Solid to Fluid. iv. Click on Generate. 4. Close ANSYS DesignModeler. You can simply close the ANSYS DesignModeler application. ANSYS Workbench automatically saves the geometry and updates the Project Schematic accordingly (the question mark in the Geometry cell is replaced by a check mark, indicating that there is a geometry now associated with the fluid flow analysis system) Release 12.1 c ANSYS, Inc. September 10, 2009
17 5. View the files generated by ANSYS Workbench. Figure 1.10: ANSYS Workbench Displaying the Files View for the Project After Creating the Geometry Note the addition of the geometry file (FFF.agdb, where FFF indicates a FLUENTbased fluid flow system) to the list of files. If you had imported the geometry file provided for this tutorial (rather than creating the geometry from scratch), the elbow-geometry.agdb (or the elbow-geometry.stp) file would be listed instead. Release 12.1 c ANSYS, Inc. September 10,
18 Step 3: Meshing the Geometry in the ANSYS Meshing Application Now that you have created the mixing elbow geometry, you need to generate a computational mesh throughout the flow volume. For this tutorial, you will use the ANSYS Meshing application to create a mesh for your CFD analysis. Note the Refresh Required icon ( ) within the Mesh cell for the system. This indicates that the state of the cell requires a refresh and that upstream data has changed since the last refresh or update (e.g., the geometry). Once the mesh is defined, the state of the cell will change accordingly. Likewise for the state of the remaining cells in the system. For more information about system cell states, see the separate FLUENT in Workbench User s Guide and the on-line documentation for ANSYS Workbench. 1. Open the ANSYS Meshing application. In the ANSYS Workbench Project Schematic, double-click the Mesh cell in the elbow fluid flow analysis system (cell A3). This displays the ANSYS Meshing application with the elbow geometry already loaded. You can also right-click on the Mesh cell to display the context menu where you can select the Edit... option. Figure 1.11: The ANSYS Meshing Application with the Elbow Geometry Loaded 1-18 Release 12.1 c ANSYS, Inc. September 10, 2009
19 The first time you open the ANSYS Meshing application, the Meshing Options are displayed on the right-hand side of the application window where you can set various meshing options. For the purposes of this tutorial, nothing needs be set here, so click OK to close the Meshing Options window. 2. Create named selections for the geometry boundaries. In order to simplify your work later on in ANSYS FLUENT, you should label each boundary in the geometry by creating named selections for the pipe inlets, the outlet, and the symmetry surface (the outer wall boundaries are automatically detected by ANSYS FLUENT). (a) Select the large inlet in the geometry that is displayed in the ANSYS Meshing application. (b) Right-click and select the Create Named Selection option. Figure 1.12: Selecting a Face to Name Release 12.1 c ANSYS, Inc. September 10,
20 This displays the Selection Name dialog box. Figure 1.13: Applying a Name to a Selected Face (c) In the Selection Name dialog box, enter velocity-inlet-large for the name and click OK. (d) Perform the same operations for the smaller inlet velocity-inlet-small, the large outlet (pressure-outlet), and the symmetry planes (symmetry). To select multiple planes, hold down the < CT RL > key and select the individual planes in the graphics window. It is important to note that by using the strings velocity inlet and pressure outlet in the named selections (with or without hyphens or underscore characters), ANSYS FLUENT automatically detects and assigns the corresponding boundary types accordingly. 3. Set some basic meshing parameters for the ANSYS Meshing application. (a) In the Outline view, select Mesh under Project/Model to display the Details view below the Outline view Release 12.1 c ANSYS, Inc. September 10, 2009
21 Note that since the ANSYS Meshing application automatically detects that you are going to perform a CFD fluid flow analysis using ANSYS FLUENT, the Physics Preference is already set to CFD and the Solver Preference is already set to Fluent. (b) Expand the Sizing node to reveal additional sizing parameters, and make sure that Use Advanced Size Function is set to On:Curvature. (c) Expand the Inflation node to reveal additional inflation parameters, and change Use Automatic Tet Inflation to Program Controlled. 4. Generate the mesh. Right-click Mesh in the project Outline tree, and select Update in the context menu. Note: Once the mesh is generated, to view the mesh statistics, open the Statistics node in the Details view to reveal information about the number of nodes, the number of elements, and other details. Figure 1.14: The Computational Mesh for the Elbow Geometry in the AN- SYS Meshing Application Using the Generate Mesh option creates the mesh, but does not actually create the relevant mesh files for the project and is optional if you already know that the mesh is acceptable. Using the Update option automatically generates the mesh and creates the relevant mesh files for your project and updates the ANSYS Workbench cell that references this mesh. Release 12.1 c ANSYS, Inc. September 10,
22 5. Close the ANSYS Meshing application. You can simply close the ANSYS Meshing application. ANSYS Workbench automatically saves the mesh and updates the Project Schematic accordingly (the Refresh Required icon in the Mesh cell is replaced by a check mark, indicating that there is a mesh now associated with the fluid flow analysis system). 6. View the files generated by ANSYS Workbench. View Files Figure 1.15: ANSYS Workbench Displaying the Files View for the Project After Mesh Creation Note the addition of the mesh files (FFF.msh and FFF.mshdb) to the list of files. The FFF.msh file is created when you updated the mesh, and the FFF.mshdb file is generated when you close the ANSYS Meshing application Release 12.1 c ANSYS, Inc. September 10, 2009
23 Step 4: Setting Up the CFD Simulation in ANSYS FLUENT Now that you have created a computational mesh for the elbow geometry, you can proceed to setting up a CFD analysis using ANSYS FLUENT. 1. Start ANSYS FLUENT. In the ANSYS Workbench Project Schematic, double-click the Setup cell in the elbow fluid flow analysis system. You can also right-click on the Setup cell to display the context menu where you can select the Edit option. When ANSYS FLUENT is first started, FLUENT Launcher is displayed, allowing you to view and/or set certain ANSYS FLUENT start-up options. ANSYS FLUENT Launcher allows you to decide which version of ANSYS FLUENT you will use, based on your geometry and on your processing capabilities. Figure 1.16: ANSYS FLUENT Launcher (a) Ensure that the proper options are enabled. Note that the Dimension setting is already filled in and cannot be changed, since ANSYS FLUENT automatically sets it based on the mesh or geometry for the current system. i. Make sure that Serial from the Processing Options list is enabled. ii. Make sure that the Display Mesh After Reading, Embed Graphics Windows, and Workbench Color Scheme options are enabled. Release 12.1 c ANSYS, Inc. September 10,
24 Note: An option is enabled when there is a check mark in the check box, and disabled when the check box is empty. To change an option from disabled to enabled (or vice versa), click the check box or the text. iii. Make sure that the Double-Precision option is disabled. (b) Click OK to launch ANSYS FLUENT. Figure 1.17: The ANSYS FLUENT Application Note: The mesh is automatically loaded and displayed in the graphics window by default Release 12.1 c ANSYS, Inc. September 10, 2009
25 2. Set some general settings for the CFD analysis. Select General in the navigation pane to perform the mesh-related activities and to choose a solver. General Release 12.1 c ANSYS, Inc. September 10,
26 (a) Change the units for length. Since we want to specify and view values based on a unit of length in inches from within ANSYS FLUENT, change the units of length within ANSYS FLU- ENT from meters (the default) to inches. Note that the ANSYS Meshing application automatically converts and exports meshes for ANSYS FLUENT using meters (m) as the unit of length regardless of what units were used to create them. This is so you do not have to scale the mesh in ANSYS FLUENT under ANSYS Workbench. General Units... This displays the Set Units dialog box. i. Select length in the Quantities list. ii. Select in in the Units list. iii. Close the dialog box. Now, all subsequent inputs that require a value based on a unit of length can be specified in inches rather than meters Release 12.1 c ANSYS, Inc. September 10, 2009
27 (b) Check the mesh. General Check ANSYS FLUENT will report the results of the mesh check in the console. Domain Extents: x-coordinate: min (m) = e-001, max (m) = e-001 y-coordinate: min (m) = e-001, max (m) = e-001 z-coordinate: min (m) = e-018, max (m) = e-002 Volume statistics: minimum volume (m3): e-009 maximum volume (m3): e-006 total volume (m3): e-003 Face area statistics: minimum face area (m2): e-006 maximum face area (m2): e-004 Checking number of nodes per cell. Checking number of faces per cell. Checking thread pointers. Checking number of cells per face. Checking face cells. Checking cell connectivity. Checking bridge faces. Checking right-handed cells. Checking face handedness. Checking face node order. Checking closed cells. Checking contact points. Checking element type consistency. Checking boundary types: Checking face pairs. Checking wall distance. Checking node count. Checking nosolve cell count. Checking nosolve face count. Checking face children. Checking cell children. Checking storage. Done. Release 12.1 c ANSYS, Inc. September 10,
28 Note: The minimum and maximum values may vary slightly when running on different platforms. The mesh check will list the minimum and maximum x and y values from the mesh in the default SI unit of meters. It will also report a number of other mesh features that are checked. Any errors in the mesh will be reported at this time. Ensure that the minimum volume is not negative, since ANSYS FLUENT cannot begin a calculation when this is the case. 3. Set up your models for the CFD simulation. Models (a) Enable heat transfer by activating the energy equation. Models Energy Edit... You can also double-click a list item in order to open the corresponding dialog box. i. Enable the Energy Equation option. ii. Click OK to close the Energy dialog box Release 12.1 c ANSYS, Inc. September 10, 2009
29 (b) Enable the k-ɛ turbulence model. Models Viscous Edit... i. Select k-epsilon from the Model list. The Viscous Model dialog box will expand. ii. Select Realizable from the k-epsilon Model list. iii. Click OK to accept the model and close the Viscous Model dialog box. Release 12.1 c ANSYS, Inc. September 10,
30 4. Set up your materials for the CFD simulation. Materials (a) Create a new material called water using the Create/Edit Materialsdialog box (Figure 1.18). Materials i. Enter water for Name. Fluid Create/Edit... ii. Enter the following values in the Properties group box: Property Value Density 1000 kg/m 3 c p Thermal Conductivity Viscosity 4216 J/kg K W/m K 8e-04 kg/m s 1-30 Release 12.1 c ANSYS, Inc. September 10, 2009
31 Figure 1.18: The Create/Edit Materials Dialog Box iii. Click Change/Create. A Question dialog box will open, asking if you want to overwrite air. Click No so that the new material water is added to the list of materials which originally contained only air. Extra: You could have copied the material water-liquid (h2o<l>) from the materials database (accessed by clicking the FLUENT Database... button). If the properties in the database are different from those you wish to use, you can edit the values in the Properties group box in the Create/Edit Materials dialog box and click Change/Create to update your local copy. The original copy will not be affected. iv. Make sure that there are now two materials (water and air) defined locally by examining the Fluent Fluid Materials drop-down list. Both the materials will also be listed under Fluid in the Materials task page. v. Close the Create/Edit Materials dialog box. Release 12.1 c ANSYS, Inc. September 10,
32 5. Set up the cell zone conditions for the CFD simulation. Cell Zone Conditions (a) Set the cell zone conditions for the fluid zone. i. Select fluid in the Zones list in the Cell Zone Conditions task page, then click the Edit... button to open the Fluid dialog box. You can also double-click a list item in order to open the corresponding dialog box Release 12.1 c ANSYS, Inc. September 10, 2009
33 ii. In the Fluid dialog box, select water from the Material Name drop-down list. iii. Click OK to close the Fluid dialog box. Release 12.1 c ANSYS, Inc. September 10,
34 6. Set up the boundary conditions for the CFD analysis. Boundary Conditions 1-34 Release 12.1 c ANSYS, Inc. September 10, 2009
35 (a) Set the boundary conditions at the cold inlet (velocity-inlet-large). Boundary Conditions velocity-inlet-large Edit... Hint: If you are unsure of which inlet zone corresponds to the cold inlet, you can probe the mesh display using the right mouse button or the probe toolbar button ( ) as described in a previous step. The information will be displayed in the ANSYS FLUENT console, and the zone you probed will be automatically selected from the Zone selection list in the Boundary Conditions task page. i. Select Components from the Velocity Specification Method drop-down list. The Velocity Inlet dialog box will expand. ii. Enter 0.4 m/s for X-Velocity. iii. Retain the default value of 0 m/s for both Y-Velocity and Z-Velocity. iv. Select Intensity and Hydraulic Diameter from the Specification Method dropdown list in the Turbulence group box. v. Enter 5% for Turbulent Intensity. Release 12.1 c ANSYS, Inc. September 10,
36 vi. Enter 4 in for Hydraulic Diameter. The hydraulic diameter D h is defined as: D h = 4A P w where A is the cross-sectional area and P w is the wetted perimeter. vii. Click the Thermal tab. viii. Enter K for Temperature. ix. Click OK to close the Velocity Inlet dialog box Release 12.1 c ANSYS, Inc. September 10, 2009
37 (b) In a similar manner, set the boundary conditions at the hot inlet (velocityinlet-small), using the values in the following table: Boundary Conditions velocity-inlet-small Edit... Velocity Specification Method Components X-Velocity 0 m/s Y-Velocity 1.2 m/s Z-Velocity 0 m/s Specification Method Intensity & Hydraulic Diameter Turbulent Intensity 5% Hydraulic Diameter 1 in Temperature K (c) Set the boundary conditions at the outlet (pressure-outlet), as shown in the Pressure Outlet dialog box. Boundary Conditions pressure-outlet Edit... Note: ANSYS FLUENT will use the backflow conditions only if the fluid is flowing into the computational domain through the outlet. Since backflow might occur at some point during the solution procedure, you should set reasonable backflow conditions to prevent convergence from being adversely affected. Release 12.1 c ANSYS, Inc. September 10,
38 7. Set up solution parameters for the CFD simulation. In the steps that follow, you will set up and run the calculation using the task pages listed under the Solution heading in the navigation pane. (a) Change the convergence criteria for the continuity equation residual. Monitors Residuals Edit... i. Make sure that Plot is enabled in the Options group box. ii. Enter 1e-05 for the Absolute Criteria of continuity, as shown in the Residual Monitor dialog box. iii. Click OK to close the Residual Monitors dialog box. Note: By default, all variables will be monitored and checked by ANSYS FLU- ENT as a means to determine the convergence of the solution Release 12.1 c ANSYS, Inc. September 10, 2009
39 (b) Initialize the flow field, using the boundary conditions settings at the cold inlet (velocity-inlet-large) as a starting point. Solution Initialization i. Select velocity-inlet-large from the Compute From drop-down list. ii. Enter 1.2 m/s for Y Velocity in the Initial Values group box. Note: While an initial X Velocity is an appropriate guess for the horizontal section, the addition of a Y Velocity component will give rise to a better initial guess throughout the entire elbow. iii. Click Initialize. Release 12.1 c ANSYS, Inc. September 10,
40 (c) Check to see if the case conforms to best practices. Run Calculation Check Case i. Click the Solver tab and examine the Recommendation in the Manual Implementation group box. The only recommendation for this mesh is to use discretization of a higher order. This recommendation can be ignored for the time being, as it was performed in the first tutorial in the separate Tutorial Guide. ii. Close the Case Check dialog box. Note that, while you are working in the ANSYS FLUENT application, the states of the Setup and Solution cells in the fluid flow FLUENT analysis system in ANSYS Workbench are changing. For example: The state of the Setup cell becomes Up-to-Date and the state of the Solution cell becomes Refresh Required after the Solution Initialization task page is visited and the number of iterations is specified. The state of the Solution cell is Update Required while iterations are taking place. The state of the Solution cell is Up-to-Date when the specified number of iterations are complete (or if convergence is reached). For more information, see the separate FLUENT in Workbench User s Guide Release 12.1 c ANSYS, Inc. September 10, 2009
41 8. Calculate a solution. (a) Start the calculation by requesting 250 iterations. Run Calculation i. Enter 250 for Number of Iterations. ii. Click Calculate. Note that ANSYS FLUENT settings file is written before the calculation begins. For more information about settings files, see the separate FLUENT in Workbench User s Guide. Release 12.1 c ANSYS, Inc. September 10,
42 As the calculation progresses, the residuals will be plotted in the graphics window (Figure 1.19). Figure 1.19: Residuals for the Converged Solution Note: The solution will be stopped by ANSYS FLUENT after approximately 230 iterations, when the residuals reach their specified values. The exact number of iterations will vary, depending on the platform being used. An Information dialog box will open to alert you that the calculation is complete. Click OK in the Information dialog box to proceed. Since the residual values vary slightly by platform, the plot that appears on your screen may not be exactly the same as the one shown here Release 12.1 c ANSYS, Inc. September 10, 2009
43 9. View the files generated by ANSYS Workbench. View Files Figure 1.20: ANSYS Workbench Displaying the Files View for the Project After Generating a Solution Note the addition of the ANSYS FLUENT settings file (FFF.set) to the list of files. Also note that the status of the Solution cell is now up-to-date. Release 12.1 c ANSYS, Inc. September 10,
44 Step 5: Displaying Results in ANSYS FLUENT and ANSYS CFD-Post 1. Display results in ANSYS FLUENT. With ANSYS FLUENT still running, you can perform a simple evaluation of the velocity and temperature contours on the symmetry plane. Later, you will use ANSYS CFD-Post (from within ANSYS Workbench) to perform the same evaluation. (a) Display filled contours of velocity magnitude on the symmetry plane (Figure 1.21). Graphics and Animations Contours Set Up... You can also double-click a list item in order to open the corresponding dialog box. i. Enable Filled in the Options group box. ii. Make sure that Node Values is enabled in the Options group box Release 12.1 c ANSYS, Inc. September 10, 2009
45 iii. Select Velocity... and Velocity Magnitude from the Contours of drop-down lists. iv. Select symmetry from the Surfaces selection list. v. Click Display to display the contours in the active graphics window. Figure 1.21: Velocity Distribution Along Symmetry Plane Release 12.1 c ANSYS, Inc. September 10,
46 (b) Display filled contours of temperature on the symmetry plane (Figure 1.22). Graphics and Animations Contours Set Up... i. Select Temperature... and Static Temperature from the Contours of dropdown lists Release 12.1 c ANSYS, Inc. September 10, 2009
47 ii. Click Display and close the Contours dialog box. Figure 1.22: Temperature Distribution Along Symmetry Plane (c) Close the ANSYS FLUENT application. File Close FLUENT Note that the ANSYS FLUENT case and data files are automatically saved when you exit ANSYS FLUENT and return to ANSYS Workbench. Release 12.1 c ANSYS, Inc. September 10,
48 (d) View the files generated by ANSYS Workbench. View Files Figure 1.23: ANSYS Workbench Displaying the Files View for the Project After Exiting ANSYS FLUENT Note the addition of the compressed ANSYS FLUENT case file (FFF-1.cas.gz) and corresponding data file (FFF dat.gz) to the list of files. Also note that the name of the data file is based on the number of iterations, so your data file name may be different Release 12.1 c ANSYS, Inc. September 10, 2009
49 2. Display results in ANSYS CFD-Post. (a) Start ANSYS CFD-Post. In the ANSYS Workbench Project Schematic, double-click the Results cell in the elbow fluid flow analysis system (cell A6). This displays the ANSYS CFD-Post application. You can also right-click on the Results cell to display the context menu where you can select the Edit option. This displays the ANSYS CFD-Post application with the elbow geometry already loaded (displayed in outline mode). Note that ANSYS FLUENT results (e.g., case and data files) are also automatically loaded into ANSYS CFD-Post. Figure 1.24: The Elbow Geometry Loaded into ANSYS CFD-Post (b) Re-orient the display. Click the blue Z axis on the axis triad in the bottom right hand corner of the graphics display to orient the display so that the view is of the front of the elbow geometry. Release 12.1 c ANSYS, Inc. September 10,
50 (c) Display filled contours of velocity magnitude on the symmetry plane (Figure 1.25). i. Insert a contour object. Insert Contour This displays the Insert Contour dialog box. ii. Keep the default name of the contour (Contour 1) and click OK to close the dialog box. This displays the Details of Contour 1 view below the Outline view in ANSYS CFD-Post. This view contains all of the settings for a contour object. iii. In the Geometry tab, select fluid in the Domains list. iv. Select symmetry in the Locations list. v. Select Velocity in the Variable list. vi. Click Apply. Figure 1.25: Velocity Distribution Along Symmetry Plane 1-50 Release 12.1 c ANSYS, Inc. September 10, 2009
51 (d) Display filled contours of temperature on the symmetry plane (Figure 1.26). i. Deselect the Contour 1 object under User Location and Plots in ANSYS CFD-Post to hide the first contour display. ii. Insert a contour object. Insert Contour This displays the Insert Contour dialog box. iii. Keep the default name of the contour (Contour 2) and click OK to close the dialog box. iv. In the Geometry tab, select fluid in the Domains list. v. Select symmetry in the Locations list. vi. Select Temperature in the Variable list. vii. Click Apply. Figure 1.26: Temperature Distribution Along Symmetry Plane 3. Close the ANSYS CFD-Post application. Note that the ANSYS CFD-Post state files are automatically saved when you exit ANSYS CFD-Post and return to ANSYS Workbench. 4. Save the elbow-workbench project in ANSYS Workbench. Release 12.1 c ANSYS, Inc. September 10,
52 5. View the files generated by ANSYS Workbench. View Files Figure 1.27: ANSYS Workbench Displaying the Files View for the Project After Viewing Results in ANSYS CFD-Post Note the addition of the ANSYS CFD-Post state file (elbow.cst) to the list of files. For more information about ANSYS CFD-Post (and the files associated with it), see the ANSYS CFD-Post documentation Release 12.1 c ANSYS, Inc. September 10, 2009
53 Step 6: Duplicating the ANSYS FLUENT-Based Fluid Flow Analysis System At this point, you have a completely defined fluid flow system that is comprised of a geometry, a computational mesh, a CFD setup and solution, and corresponding results. In order to study the effects upon the flow field that may occur if you were to alter the geometry, another fluid flow analysis is required. One approach would be to use the current system and change the geometry, however you would overwrite the data from your previous simulation. A more suitable and effective approach would be to create a copy, or duplicate, of the current system, and then make the appropriate changes to the duplicate system. In this step, you will create a duplicate of the original ANSYS FLUENT-based fluid flow system. 1. In the Project Schematic, right-click the title cell of the Fluid Flow (FLUENT) system and select Duplicate from the context menu. Figure 1.28: Duplicating the Fluid Flow System Figure 1.29: The Original Fluid Flow System and Its Duplicate Notice that in the duplicated system, the state of the Solution cell indicates that the cell requires an update while the state of the Results cell indicates that the cell requires attention. This is because when a system is duplicated, the case and data Release 12.1 c ANSYS, Inc. September 10,
54 files are not copied to the new system, therefore, the new system does not yet have solution data associated with it. 2. Rename the duplicated system to new-elbow. 3. Save the elbow-workbench project in ANSYS Workbench. Step 7: Changing the Geometry in ANSYS DesignModeler Now that you have two separate, but equivalent, ANSYS FLUENT-based fluid flow systems to work from, you can make changes to the second system without impacting the original system. In this step, you will make a slight alteration to the elbow geometry in ANSYS DesignModeler by changing the diameter of the smaller inlet. 1. Open ANSYS DesignModeler. Double-click the Geometry cell of the new-elbow system (cell B2) to display the geometry in ANSYS DesignModeler. 2. Change the diameter of the small inlet (velocity-inlet-small). (a) Select Sketch3, under Extrude1 to open the Details View of the small inlet extrusion. (b) In the Details View, under Dimensions:1, change the R1 (radius) value from 0.5 in to 0.75 in. (c) Click the Generate button to generate the geometry with your new values. Figure 1.30: Changing the Diameter of the Small Inlet in ANSYS DesignModeler 3. Close ANSYS DesignModeler Release 12.1 c ANSYS, Inc. September 10, 2009
55 4. View the files generated by ANSYS Workbench. View Files Figure 1.31: ANSYS Workbench Displaying the Files View for the Project After Duplicating the System and Changing the Geometry Note the addition of the geometry, mesh, and ANSYS FLUENT settings files now associated with new duplicated system. Release 12.1 c ANSYS, Inc. September 10,
56 Step 8: Updating the Mesh in the ANSYS Meshing Application The modified geometry now requires a new computational mesh. The mesh settings for the duplicated system are retained in the duplicated system, so all you need to do is update the mesh based on the mesh settings from the original system. In the Project Schematic, right-click on the Mesh cell of the new-elbow system (cell B3) and select Update from the context menu. This will update the mesh for the new geometry based on the mesh settings you specified earlier in the ANSYS Meshing application without having to open the editor to regenerate the mesh. Figure 1.32: Updating the Mesh for the Changed Geometry It will take a few moments to update the mesh. Once the update is complete, the state of the Mesh cell is changed to up-to-date, symbolized by a green check mark. For illustrative purposes of the tutorial, the new geometry and the new mesh is displayed below. Inspecting the files generated by ANSYS Workbench reveals the updated mesh file for the duplicated system. View Files 1-56 Release 12.1 c ANSYS, Inc. September 10, 2009
57 Figure 1.33: The Updated Geometry and Mesh in the ANSYS Meshing Application Figure 1.34: ANSYS Workbench Displaying the Files View for the Project After Updating the Mesh for the Altered Geometry Release 12.1 c ANSYS, Inc. September 10,
58 Step 9: Calculating a New Solution in ANSYS FLUENT Now that there is an updated computational mesh for the modified geometry in the duplicated system, a new solution needs to be generated using ANSYS FLUENT. In this step, you will revisit the settings within ANSYS FLUENT, calculate another solution, then view the new results. 1. Open ANSYS FLUENT. In the Project Schematic, right-click on the Setup cell of the new-elbow system (cell B4) and select Edit... from the context menu. Since the mesh has been changed, you are prompted as to whether you want to load the new mesh into ANSYS FLUENT or not. Select Yes to continue, and click OK when FLUENT Launcher is displayed in order to open ANSYS FLUENT. Figure 1.35: ANSYS Workbench Prompt When the Upstream Mesh Has Changed 2. Make sure that the unit of length is set to inches. General Units Check the mesh (optional). General Check 4. Revisit the boundary conditions for the small inlet. Boundary Conditions velocity-inlet-small Edit... Here, you need to set the hydraulic diameter to 1.5 in based on the new dimensions of the small inlet Release 12.1 c ANSYS, Inc. September 10, 2009
59 5. Reinitialize the solution. Solution Initialization Again, use the velocity-inlet-large boundary condition to initialize the solution, adding the Y Velocity component of 1.2 as you did before. 6. Recalculate the solution. Run Calculation Keep the Number of Iterations set to 250. This time, the solution converges after approximately 220 iterations. 7. Close ANSYS FLUENT. 8. Revisit the results of the calculations in ANSYS CFD-Post. Double-click the Results cell of the new-elbow fluid flow system to re-open ANSYS CFD-Post where you can review the results of the new solution. 9. Close ANSYS CFD-Post. 10. Save the elbow-workbench project in ANSYS Workbench. Release 12.1 c ANSYS, Inc. September 10,
60 11. View the files generated by ANSYS Workbench. View Files Figure 1.36: ANSYS Workbench Displaying the Files View for the Project After Viewing the New Results in ANSYS CFD-Post Note the addition of the solution and state files now associated with new duplicated system Release 12.1 c ANSYS, Inc. September 10, 2009
61 Step 10: Comparing the Results of Both Systems in ANSYS CFD-Post In this step, you will create a new Results system in ANSYS Workbench, and use that system to compare the solutions from each of the two ANSYS FLUENT-based fluid flow analysis systems in ANSYS CFD-Post at the same time. 1. Create a Results system. In ANSYS Workbench, drag a Results system from the Component Systems section of the Toolbox and drop it into the Project Schematic, next to the fluid flow systems. Figure 1.37: The New Results System in the Project Schematic Release 12.1 c ANSYS, Inc. September 10,
62 2. Add the solutions of each of the systems to the Results system. (a) Select the Solution cell in the first Fluid Flow analysis system (cell A5) and drag it over the Results cell in the Results system (cell C2). This creates a transfer data connection between the two systems. Figure 1.38: Connecting the First Fluid Flow System to the New Results System (b) Select the Solution cell in the second Fluid Flow analysis system (cell B5) and drag it over the Results cell in the Results system (cell C2). This creates a transfer data connection between the two systems. Figure 1.39: Connecting the Second Fluid Flow System to the New Results System 1-62 Release 12.1 c ANSYS, Inc. September 10, 2009
63 3. Open ANSYS CFD-Post to compare the results of the two fluid flow systems. Now that the two fluid flow systems are connected to the Results system, doubleclick the Results cell in the Results system (cell C2) to open ANSYS CFD-Post. Within ANSYS CFD-Post, both geometries are displayed side by side. Figure 1.40: ANSYS CFD-Post with Both Fluid Flow Systems Displayed (a) Re-orient the display. In each view, click the blue Z axis on the axis triad in the bottom right hand corner of the graphics display to orient the display so that the view is of the front of the elbow geometry. Alternatively, you can select the synchronization tool ( ) in the 3D Viewer Toolbar to synchronize the views, so that when you re-orient one view, the other view is automatically updated. Release 12.1 c ANSYS, Inc. September 10,
64 (b) Display filled contours of velocity magnitude on the symmetry plane. i. Insert a contour object. Insert Contour This displays the Insert Contour dialog box. ii. Keep the default name of the contour (Contour 1) and click OK to close the dialog box. This displays the Details of Contour 1 view below the Outline view in ANSYS CFD-Post. This view contains all of the settings for a contour object. iii. In the Geometry tab, select fluid in the Domains list. iv. Select symmetry in the Locations list. v. Select Velocity in the Variable list. vi. Click Apply. In each view, the velocity contours are displayed. Figure 1.41: ANSYS CFD-Post Displaying Velocity Contours for Both Geometries 1-64 Release 12.1 c ANSYS, Inc. September 10, 2009
65 (c) Display filled contours of temperature on the symmetry plane. i. Deselect the Contour 1 object under User Location and Plots in ANSYS CFD-Post to hide the first contour display. ii. Insert another contour object. Insert Contour This displays the Insert Contour dialog box. iii. Keep the default name of the contour (Contour 2) and click OK to close the dialog box. This displays the Details of Contour 2 view below the Outline view in ANSYS CFD-Post. This view contains all of the settings for a contour object. iv. In the Geometry tab, select fluid in the Domains list. v. Select symmetry in the Locations list. vi. Select Temperature in the Variable list. vii. Click Apply. In each view, the temperature contours are displayed. Figure 1.42: ANSYS CFD-Post Displaying Temperature Contours for Both Geometries Release 12.1 c ANSYS, Inc. September 10,
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### Chapter 11. Assembly Modeling. Evaluation chapter. Logon to for more details. Learning Objectives
Chapter 11 Assembly Modeling Learning Objectives After completing this chapter you will be able to: Insert components into an assembly file. Create bottom-up assemblies. Insert components into a product | 17,985 | 77,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-04 | latest | en | 0.811927 |
https://devforum.roblox.com/t/help-to-calculate-stiffnessdamping-on-suspension/1044816 | 1,713,872,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00838.warc.gz | 184,086,989 | 5,482 | # Help to calculate Stiffness/Damping on Suspension
Hi guys, I am having some trouble while making a car suspension, it’s very bumpy, here’s code;
``````local Mass = 0
for i,v in pairs(script.Parent.Parent:GetDescendants()) do
if (v:IsA("BasePart") or v:IsA("MeshPart")) then
Physics:SetPartCollisionGroup(v,"Chassis")
if not (v.Massless) then
Mass += v:GetMass()
end
end
end
Mass = Mass * workspace.Gravity
Configurations.SpringConstraint.MaxForce = Mass
Configurations.SpringConstraint.Stiffness = Mass/2
for i,v in pairs(Constraints:GetChildren()) do
if (v:IsA("SpringConstraint")) then
v.Stiffness = Configurations.SpringConstraint.Stiffness
v.Damping = Configurations.SpringConstraint.Damping
v.LimitsEnabled = true
v.MinLength = 1
v.MaxForce = Configurations.SpringConstraint.MaxForce
v.FreeLength = Configurations.SpringConstraint.FreeLength
end
end
``````
And here’s configurations
``````return {
Gears = {
[1] = {Speed = 40; Torque = 5000;};
[2] = {Speed = 0; Torque = 0;};
[3] = {Speed = -60; Torque = 1000;};
[4] = {Speed = -120; Torque = 900;};
[5] = {Speed = -180; Torque = 700;};
[6] = {Speed = -240; Torque = 500;};
[7] = {Speed = -300; Torque = 350;};
[8] = {Speed = -360; Torque = 200;};
};
BrakePower = 2000;
SteeringSpeed = 1;
SpringConstraint = {
Stiffness = 0; -- Changed later
Damping = 40;
MaxForce = 0; -- Changed later
FreeLength = 2.25;
};
};
``````
Result;
Can someone help me out?
I can give a repro file if needed | 451 | 1,451 | {"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-2024-18 | latest | en | 0.576651 |
https://uk.mathworks.com/matlabcentral/cody/problems/2022-find-a-pythagorean-triple/solutions/1754069 | 1,579,840,539,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00088.warc.gz | 694,167,477 | 15,724 | Cody
# Problem 2022. Find a Pythagorean triple
Solution 1754069
Submitted on 19 Mar 2019 by Tran Hiep Dinh
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 = 1; b = 2; c = 3; d = 4; flag_correct = false; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
subA = 1 2 3 1 2 4 1 3 4 2 3 4
2 Pass
a = 2; b = 3; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
subA = 2 3 4 2 3 5 2 4 5 3 4 5 flag = logical 1
3 Pass
a = 3; b = 4; c = 5; d = 6; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
subA = 3 4 5 3 4 6 3 5 6 4 5 6 flag = logical 1
4 Pass
a = 3; b = 4; c = 4.5; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
subA = 3.0000 4.0000 4.5000 3.0000 4.0000 5.0000 3.0000 4.5000 5.0000 4.0000 4.5000 5.0000 flag = logical 1
5 Pass
a = 3; b = 3.5; c = 4; d = 5; flag_correct = true; assert(isequal(isTherePythagoreanTriple(a, b, c, d),flag_correct))
subA = 3.0000 3.5000 4.0000 3.0000 3.5000 5.0000 3.0000 4.0000 5.0000 3.5000 4.0000 5.0000 flag = logical 1 | 558 | 1,250 | {"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-2020-05 | latest | en | 0.374085 |
https://inspiria.edu.in/ncert-solutions-for-class-12-maths-relations-and-functions/ | 1,619,036,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00336.warc.gz | 427,296,813 | 28,040 | NCERT Solutions for Class 12 Maths All Chapters Mathematics is indubitably the most easy to score yet one of the toughest subjects in Class 12. And the subject only demands practice. While there are a lot of study materials out there to practice class 12 maths, NCERT solutions is one of the most essential one. NCERT Solutions for class 12 maths has not only proven to be useful in CBSE Board examinations but also helps when appearing for various engineering and other entrance exams like JEE.
NCERT has divided class 12 mathematics in two books, Part I and Part II with a total of 13 chapters. Mathematics Part I comprises 6 chapters and Mathematics Part II has 7 chapters. Most of these chapters pick up the concepts introduced in class 11 with addition of introduction to some new concepts and tools of mathematics. We have given chapter wise details and NCERT Solutions below.
Relations and Functions
This chapter is an expansion and ascension of the concepts of Relations and Functions introduced in Class 11. This chapter talks more about domain, codomain, real valued functions and their graphs. It includes four exercises which unify the concepts explained in the chapter.
## Topics and Sub Topics Class 12 Maths Chapter 1- Relations and Functions:
1 – Relations and Functions
1.1 Introduction
1.2 Types of Relations
1.3 Types of Functions
1.4 Composition of Functions and Invertible Function
1.5 Binary Operations | 296 | 1,429 | {"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} | 2.828125 | 3 | CC-MAIN-2021-17 | longest | en | 0.913812 |
https://www.omnicalculator.com/physics/resultant-force | 1,669,646,241,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00438.warc.gz | 964,502,367 | 100,551 | # Net Force Calculator
Created by Purnima Singh, PhD
Reviewed by Steven Wooding
Last updated: Oct 19, 2022
Omni's net force calculator allows you to determine the resultant force on a body when several forces act simultaneously on it.
Continue reading this article to know what net force is and how to find the net force on an object. You will also see some examples of net force calculations.
If you need help calculating force from Newton's second law of motion, check out the force calculator.
## What is net force? – Net force definition
When a number of forces act simultaneously on an object or a system, the net force or resultant force is the vector sum of all these forces.
We know that the effect of a force $F$ acting on an object of mass $m$ is to accelerate the object according to the equation:
$\quad a = \frac{F}{m}$
Our magnitude of acceleration calculator is a handy tool in case you want to explore more about acceleration.
The effect of net force would accelerate the object by the same amount as all the actual forces acting on the object. So we can say that the net force is a single force that would produce the same effect as all the forces working together.
Now that we know what net force is, let us understand how to find the net force on an object.
## How to find net force – Net force equation
We know that force is a vector quantity, i.e., we need to specify both the magnitude and direction of a force to give its complete description. This means that we should add the individual forces to find the net force just like we add other vectors (check out the vector addition calculator to learn all about vector addition).
To understand this, let us consider a simple scenario where two forces $F_1$ and $F_2$ are acting on a body from two different directions. We can represent these two forces as two vectors $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ acting at angles $\theta_1$ and $\theta_2$ (see figure 1).
Using the triangle law of vector addition, we know that if two vectors acting simultaneously on a body can be represented in magnitude and direction by the two sides of a triangle taken in one order, the resultant of these two vectors can be represented in magnitude and direction by the third side of the triangle taken in the opposite order (see the net force diagram in figure 1).
Hence, we get the resultant vector:
$\ \ \overrightarrow{F} = \overrightarrow{F_1} + \overrightarrow{F_2}$
If several forces are acting together (see figure 2), we can apply the polygon rule and write a more general net force formula as:
\begin{aligned} \overrightarrow{F} &= \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3} + ... + \overrightarrow{F_n}\\ \text{ or:}\\ \overrightarrow{F} &= \sum_{i=1}^\infty \overrightarrow{F_i} \end{aligned}
In the next section, we will see a step-by-step guide to how to find the magnitude and direction of the resultant force?
## How to calculate net force
To get the magnitude of the resultant force $F$, we will use the net force equation. We will first resolve each of the forces $F_1$ and $F_2$ into their respective rectangular components:
$\quad F_{1x} = F_1\cos \theta_1 \\\ \\ \quad F_{1y} = F_1\sin \theta_1$
Similarly, for the force $F_2$, we can write:
$\quad F_{2x} = F_2\cos \theta_2\\\ \\ \quad F_{2y} = F_2\sin \theta_2$
Since the components of a vector are scalar quantities, we can now add them. To find the horizontal component $F_x$ of the resultant force, we will sum all the horizontal components of the individual forces:
$\quad F_x = F_{1x} + F_{2x}$
In the same way, the vertical component will be:
$\quad F_y = F_{1y} + F_{2y}$
Finally, we will calculate the magnitude of the resultant force using:
$\quad F = \sqrt{F_x^2 + F_y^2}$
And the angle of the resultant force with respect to the horizontal axis with:
$\quad \theta = \tan^{-1}\frac{F_y}{F_x}$
For $n$ number of forces, we can write a general formula as:
$\quad F_x = \sum_{i=1}^\infty F_{ix}\\\ \\ \quad F_y = \sum_{i=1}^\infty F_{iy}$
Using the net force formula, we can calculate the magnitude of the resultant force using:
$\quad F = \sqrt{F_x^2 + F_y^2}$
And the angle of the resultant force with respect to the horizontal axis using:
$\quad \theta = \tan^{-1}\frac{F_y}{F_x}$
## Examples of net force calculation
To further understand how to find the resultant force, we will consider two simple examples.
First we will consider a simple case where two forces $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ are applied to an object in opposite directions such that;
• $|F_1| = 10\ N$ and $\theta_1 = 0 \degree$; and
• $|F_2| = 15\ N$ and $\theta_2 = 180 \degree$.
An example of this case would be when you and your friend are sitting in front of each other and trying to push a book towards each other.
Let us see how to calculate the net force.
1. First, we will find the vertical and horizontal components for both the forces:
• $F_{1x} = 10 \cdot \cos 0 \degree$ and $F_{1y} =10 \cdot \sin 0 \degree$.
$\implies F_{1x} = 10$ and $F_{1y} =0$
• $F_{2x} = 15 \cdot \cos 180 \degree$ and $F_{2y} =15 \cdot \sin 180 \degree$.
$\implies F_{1x} = -15$ and $F_{1y} =0$
2. As the vertical components, $F_{1y}$ and $F_{2y}$ are zero, the vertical component of the resultant force will be zero. We can calculate the horizontal component as:
• $F_x = 10 + (-15)$, or
$F_x = -5\ N$
3. According to the definition of the net force, the magnitude of the resultant force will be:
• $F = \sqrt{F_x^2} = 5\ N$
and the direction will be along the direction of the larger force, i.e., $180 \degree$.
Now let us consider another case where the magnitude of the forces is the same as in the previous example, but now both the forces act in the same direction, i.e., $\theta_1 = \theta_2 = 180 \degree$. For example, when you and your friend try to push a heavy box together. In this case, the net force would be $25\ N$ along the direction of both the forces, i.e., $180 \degree$.
In the next section, we will see how to solve the same problem using our resultant force calculator.
## How to use the net force calculator
Let us see how to use the net force calculator:
1. Enter the magnitude ($|F_1| = 10\ N$ and $|F_2| = 15\ N$) and direction ($\theta_1 = 0 \degree$ and $\theta_2 = 180 \degree$) of different forces acting on the body. The directions are measured with respect to the positive x-axis.
2. You can add data for up to 10 forces; fields will appear as you need them.
3. The resultant force calculator will display the magnitude ($F = 5\ N$) and direction ($\theta = 180 \degree$) of the net force. It will also show the values of the horizontal and vertical components of the resultant force.
To convert between different units of force, head on to Omni's force converter.
## FAQ
### How do I find the resultant force acting on an object?
To find the resultant force or net force acting on an object, follow the given instructions:
1. Determine the horizontal and vertical components of all the individual forces by using the formula;
• Horizontal component –
Fₓ = F cos θ
• Vertical component –
Fᵧ = F sin θ
2. Add the individual horizontal and vertical components to get the horizontal and vertical components of the resultant force.
3. Sum the square of the horizontal and vertical components of the resultant force and take the square root of the result. You will get the magnitude of the resultant force.
### When the net force on an object is zero?
According to Newton's first law of motion, if an object is either at rest or moving in a straight line with constant velocity (no acceleration), the net force on the object is zero.
### What are the units of net force?
The units for net force are the same as the unit of force. The SI unit of the net force is the newton (N), and the cgs unit of the net force is dyne.
### How does net force affect the speed of an object?
From Newton's second law of motion, we know that the acceleration or rate of change of speed of an object is directly proportional to the net force acting on the object.
Purnima Singh, PhD
Force 1 (F₁)
N
Angle 1 (θ₁)
deg
Force 2 (F₂)
N
Angle 2 (θ₂)
deg
You can add up to 10 different forces. Each angle is measured with respect to the positive x-axis.
Resultant force
Horizontal component (Fx)
0.97
N
Vertical component (Fy)
0.26
N
Magnitude of resultant force (F)
1
N
Direction of resultant force (θ)
15
deg
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# Game 1: Mr. Happy’s Quest - PowerPoint PPT Presentation
Game 1: Mr. Happy’s Quest. For our original canvas, we had designed a child’s background using simple shapes. For this project, we will add objects that a game player (child) can move around using the arrow keys on the keyboard. In the example, a smiley face has begun to be created.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about 'Game 1: Mr. Happy’s Quest' - enrico
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Game 1: Mr. Happy’s Quest
• For our original canvas, we had designed a child’s background using simple shapes.
• For this project, we will add objects that a game player (child) can move around using the arrow keys on the keyboard.
• In the example, a smiley face has begun to be created
The Draw Function and Mr. Happy
size(800,600);
var x = 400;
var t = 5;
varhx = 200;
varhy = 200;
//draw mr happy
noStroke();
fill(255,255,0);
ellipse(hx,hy,100,100);
stroke(0,0,0);
ellipse(hx-20,hy-20,10,10);
ellipse(hx+20,hy-20,10,10);
//ADD A MOUTH WITH A BLACK ELIPSE AND A RECTANGLE TO ERASE THE TOP HALF OF THE ELLIPSE
}
Bouncy Obstacles Example
void draw() {
background(100,100,100);
//bouncy ball designs
fill(255,255,255);
ellipse(x,150,200,200);
fill(0,0,0);
ellipse(x,450,200,200);
x=x+t;
if(x>800) t = -t;
if(x<0) t = -t;
text("Hello!", 50, 50);
keyPressed() Function Example
void keyPressed() {
if (keyCode == UP) hy = hy - 20;
if (keyCode == DOWN) hy = hy + 20;
}
Practice 1: 20%
• Take what has been done in project 31 and expand upon it to add a figure, character and/or face that moves around the screen with the keyboard arrow keys.
• Make certain that the screen:
• Has moving objects that the character must avoid
• An aesthetically pleasing background
• A goal to get to (golden treasure?)
Guiding the Movements of the Star
• For the second example, the end user (game player) will use keys to adjust the path of the star.
• The game will need something for the star to attack, gather, and/or avoid.
• Use the work done in our previous unit and expand upon the canvas with examples and original designs.
var x = 300;
var y = 0;
var t = 5;
void draw() {
background(255,255,255);
//top left part of star
stroke(255,0,0);
line(x,0,300,300);
line(x,20,280,300);
line(x,40,260,300);
line(x,60,240,300);
x=x+t;
if(x>600) t = -t;
if(x<0) t = -t;
}
void keyPressed() {
if (keyCode == LEFT) x = x - 25;
if (keyCode == RIGHT) x = x + 25;
}
Practice 2: 20%
• Implement an improved and fully capable keyPress function that will enable the star to move in all 4 directions based on the game players use of the arrow keys.
• Add game objects that the player should move the star towards and/or avoid so that there is an eventual obtainable goal for the game player.
Review
• Guiding the Movements of the Star
• Code Example
• Code Example
• Practice 2
Car Game
• The car has already been designed to be drawn with simple shapes and the illusion of movement based on the roadway’s movement in the opposite direction.
• Our goal for this part of the project will be to enable the car to move around the road in 4 directions and to add objects that can be obtained and/or avoided for points (to be implemented in a later version).
size(300, 800);
varkeyX = 0;
varkeyY = 0;
var y = 0;
var x = 0;
var t = .1;
void draw(){
background(100,100,100);
stroke(255,255,0);
strokeWeight(10);
line(150,y+0,150,y+50);
y = y + 3;
if(y>50) y = 0;
//car vrooom
noStroke();
fill(0,0,0);
rect(keyX+x+20,50,90,200);
fill(255,255,255);
rect(keyX+x+50,50,30,200);
x = x + t;
if(x>8) t = -t;
if(x<0) t = -t;
}
void keyPressed() {
if (keyCode == LEFT) keyX = keyX - 15;
if (keyCode == RIGHT) keyX = keyX + 15;
}
Practice 3: 20%
• Create a keypress function that enables the game player to move the car around the screen in a way that looks realistic.
• Add moving objects that the car can obtain and objects that the car must avoid to create a game.
Review
• Car Game
• Example Code
• Example Code
• Practice 3: 20%
Making the Plane Fly (WITH LASERS!)
• In the previous project animations were created that enabled an object to appear to fly across the screen.
• In this section, the animation will be expanded upon to have the keyboard move the plane.
• The plane will also have a LASER that is fired with the spacebar.
size(800,600);
varkeyY = 0;
varlaserOn = false;
varlaserTimeOut = 20;
vartimeCount = 0;
tx = 400;
x = 0;
void draw(){
background(0,255,255);
//tree
noStroke();
fill(150,75,0);
rect(tx-10,500,25,200);
fill(0,255,0);
ellipse(tx,475,150,100);
tx = tx - 10;
if (tx < 0) tx = 1200;
//cloud
noStroke();
fill(255,255,255);
ellipse(tx-250,90,166,127);
//jet
strokeWeight(5);
noStroke();
fill(255,0,0);
//LASERS!!!!!
stroke(255,0,0);
if(laserOn)
{
line(x+300,300+keyY, x+300+233,300+keyY);
timeCount++;
if (timeCount > laserTimeOut) {
timeCount =0;
laserOn = false;
}
}
x= x+3;
if (x>800) x = -200;
}
void keyPressed() {
if (keyCode == UP) keyY = keyY - 15;
if (keyCode == DOWN) keyY = keyY + 15;
if (keyCode == LEFT) laserOn = true;
if (keyCode == 32) laserOn = true;
}
Practice 4: 20%
• The airplane should look like it is really flying. It should be controllable in all 4 directions.
• The airplane should have a functional and aesthetically pleasing weapon (the laser).
• Objects should move around the screen that can either help the plane or provide something for the plane to shoot at.
Review
• Making the Plane Fly WITH LASERS!
• Example Code
• Example Code
• Practice 4: 20%
Improving Canvas 1
• Make certain that the canvas has something to get, a place to take it and a nice design of the avatar (thing the player moves).
• The canvas should have items moving around that are both good for the avatar and bad for the avatar such as treasure and slime.
• Choose your own ideas and make it into an original looking game.
Improving Canvas 2
• The star has to be part of a game that can enable the user to interact in a meaningful way.
• Make the star bigger or smaller.
• Make more than one star.
• There should be objects for the game player to obtain and/or avoid.
Improving Canvas 3
• The car game has a lot of room for improvement.
• The car its self can be more realistic.
• More drawings can accompany the side of the road to move by quickly and more drawings can be on the road.
• The car will need moving objects to obtain and avoid such as other cars and “power ups.”
Improving Canvas 4
• The aircraft its self can be made to look more modern, sleek or realistic.
• The items in the sky and ground can become more varied and more detailed.
• The aircraft has a laser: give it something to shoot at!
• Objects to avoid and/or shoot and/or obtain should be added.
Review
• Improving Canvas 1
• Improving Canvas 2
• Improving Canvas 3
• Improving Canvas 4
• <title>Gabrielle's Fun Games</title>
• <script type="text/javascript" src="http://www.scottbunin.com/processing.js"></script>
• <script type="text/javascript" src="http://www.scottbunin.com/init.js"></script>
• <body><center>
• <script type="application/processing">
• size(800,600);
• var x = 400;
• var t = 5;
• void draw() {
• background(0,255,255);
• fill(0,0,0);
• ellipse(x,450,200,200);
• x=x+t;
• if(x>800) t = -t;
• if(x<0) t = -t;
• stroke(255,0,0);
• strokeWeight(5);
• line(0,0,800,600);
• stroke(255,255,0);
• strokeWeight(5);
• line(800,0,0,600);
• }
• </script><canvas tabindex="0" id="__processing3" width="800" height="600" style="image-rendering: -webkit-optimize-contrast !important;"></canvas>
• <br>
• <br>
• <script type="application/processing">
• size(600, 600);
• strokeWeight(1);
• var x = 300;
• var t = 5;
• void draw() {
• background(255,255,255);
• //top left part of star
• stroke(255,0,0);
• line(x,0,300,300);
• line(x,20,280,300);
• line(x,40,260,300);
• line(x,60,240,300);
• line(x,80,220,300);
• line(x,100,200,300);
• line(x,120,180,300);
• line(x,140,160,300);
• line(x,160,140,300);
• line(x,180,120,300);
• line(x,200,100,300);
• line(x,220,80,300);
• line(x,240,60,300);
• line(x,260,40,300);
• line(x,280,20,300);
• line(x,300,0,300);
• line(x,0,300,300);
• //top right part of star
• stroke(255,255,0);
• line(x,20,320,300);
• line(x,40,340,300);
• line(x,60,360,300);
• line(x,80,380,300);
• line(x,100,400,300);
• line(x,120,420,300);
• line(x,140,440,300);
• line(x,160,460,300);
• line(x,180,480,300);
• line(x,200,500,300);
• line(x,220,520,300);
• line(x,240,540,300);
• line(x,260,560,300);
• line(x,280,580,300);
• line(x,300,600,300);
• x=x+t;
• if(x>600) t = -t;
• if(x<0) t = -t;
• }
• </script><canvas tabindex="0" id="__processing2" width="600" height="600"></canvas>
• <br>
• <br>
• <script type="application/processing">
• size(300, 800);
• var y = 0;
• var x = 0;
• var t = .05;
• void draw(){
• background(100,100,100);
• stroke(255,255,0);
• strokeWeight(10);
• line(150,y+0,150,y+50);
• line(150,y+100,150,y+150);
• y = y + 3;
• if(y>80) y = 0;
• noStroke();
• fill(50,50,50);
• rect(x+15,100,100,30);
• fill(0,0,0);
• rect(x+20,50,90,200);
• fill(255,255,255);
• rect(x+50,50,30,200);
• x = x + t;
• if(x>5) t = -t;
• if(x<0) t = -t;
• }
• </script><canvas tabindex="0" id="__processing1" width="300" height="800" style="image-rendering: -webkit-optimize-contrast !important;"></canvas>
• <br>
• <br>
• <script type="application/processing">
• size(800,600);
• tx = 400;
• x = 0;
• void draw(){
• background(0,255,255);
• //tree
• noStroke();
• fill(150,75,0);
• rect(tx-10,500,25,200);
• fill(0,255,0);
• ellipse(tx,475,150,100);
• tx = tx - 10;
• if (tx < 0) tx = 1200;
• //cloud
• noStroke();
• fill(255,255,255);
• ellipse(tx-250,90,166,127);
• ellipse(tx-150,100,95,77);
• ellipse(tx-350,100,95,77);
• ellipse(tx-350,90,166,127);
• ellipse(tx-275,100,95,77);
• ellipse(tx-450,100,95,77);
• //jet
• noStroke();
• fill(255,0,0);
• triangle(x+100,100,x+100,300,x+300,200);
• fill(150,75,0);
• triangle(x+200,100,x+200,300,x+400,200);
• fill(255,192,203);
• triangle(x+300,100,x+300,300,x+500,200);
• x= x+3;
• if (x>800) x = -300;
• }
• </script><canvas tabindex="0" id="__processing0" width="800" height="600" style="image-rendering: -webkit-optimize-contrast !important;"></canvas>
• <br>
• <br>
• </center><span style="position: absolute; top: 0px; left: 0px; opacity: 0; font-family: PjsEmptyFont, fantasy;">AAAAAAAA</span></body></html> | 3,478 | 10,918 | {"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-2018-47 | latest | en | 0.77211 |
http://www.lmfdb.org/Genus2Curve/Q/1142/a/2284/1 | 1,563,716,640,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00157.warc.gz | 219,951,394 | 23,864 | # Properties
Label 1142.a.2284.1 Conductor 1142 Discriminant -2284 Sato-Tate group $\mathrm{USp}(4)$ $$\End(J_{\overline{\Q}}) \otimes \R$$ $$\R$$ $$\End(J_{\overline{\Q}}) \otimes \Q$$ $$\Q$$ $$\overline{\Q}$$-simple yes $$\mathrm{GL}_2$$-type no
# Related objects
Show commands for: Magma / SageMath
## Minimal equation
magma: R<x> := PolynomialRing(Rationals()); C := HyperellipticCurve(R![-2, -1, 1, -1, -1], R![0, 0, 1, 1]);
sage: R.<x> = PolynomialRing(QQ); C = HyperellipticCurve(R([-2, -1, 1, -1, -1]), R([0, 0, 1, 1]))
$y^2 + (x^3 + x^2)y = -x^4 - x^3 + x^2 - x - 2$
## Invariants
magma: Conductor(LSeries(C)); Factorization($1); $$N$$ = $$1142$$ = $$2 \cdot 571$$ magma: Discriminant(C); Factorization(Integers()!$1); $$\Delta$$ = $$-2284$$ = $$-1 \cdot 2^{2} \cdot 571$$
### G2 invariants
magma: G2Invariants(C);
$$I_2$$ = $$1888$$ = $$2^{5} \cdot 59$$ $$I_4$$ = $$-46016$$ = $$-1 \cdot 2^{6} \cdot 719$$ $$I_6$$ = $$-27370048$$ = $$-1 \cdot 2^{6} \cdot 53 \cdot 8069$$ $$I_{10}$$ = $$-9355264$$ = $$-1 \cdot 2^{14} \cdot 571$$ $$J_2$$ = $$236$$ = $$2^{2} \cdot 59$$ $$J_4$$ = $$2800$$ = $$2^{4} \cdot 5^{2} \cdot 7$$ $$J_6$$ = $$46521$$ = $$3^{3} \cdot 1723$$ $$J_8$$ = $$784739$$ = $$227 \cdot 3457$$ $$J_{10}$$ = $$-2284$$ = $$-1 \cdot 2^{2} \cdot 571$$ $$g_1$$ = $$-183020620544/571$$ $$g_2$$ = $$-9200979200/571$$ $$g_3$$ = $$-647758404/571$$
Alternative geometric invariants: G2
## Automorphism group
magma: AutomorphismGroup(C); IdentifyGroup($1); $$\mathrm{Aut}(X)$$ $$\simeq$$ $$C_2$$ (GAP id : [2,1]) magma: AutomorphismGroup(ChangeRing(C,AlgebraicClosure(Rationals()))); IdentifyGroup($1); $$\mathrm{Aut}(X_{\overline{\Q}})$$ $$\simeq$$ $$C_2$$ (GAP id : [2,1])
## Rational points
magma: f,h:=HyperellipticPolynomials(C); g:=4*f+h^2; HasPointsEverywhereLocally(g,2) and (#Roots(ChangeRing(g,RealField())) gt 0 or LeadingCoefficient(g) gt 0);
This curve is locally solvable everywhere.
magma: [C![-2,2,1],C![-1,0,1],C![1,-1,0],C![1,0,0]];
All rational points: (-2 : 2 : 1), (-1 : 0 : 1), (1 : -1 : 0), (1 : 0 : 0)
magma: #Roots(HyperellipticPolynomials(SimplifiedModel(C)));
Number of rational Weierstrass points: $$2$$
## Invariants of the Jacobian:
Analytic rank: $$0$$
magma: TwoSelmerGroup(Jacobian(C)); NumberOfGenerators($1); 2-Selmer rank: $$1$$ magma: HasSquareSha(Jacobian(C)); Order of Ш*: square Tamagawa numbers: 2 (p = 2), 1 (p = 571) magma: TorsionSubgroup(Jacobian(SimplifiedModel(C))); AbelianInvariants($1);
Torsion: $$\Z/{10}\Z$$
### Sato-Tate group
$$\mathrm{ST}$$ $$\simeq$$ $\mathrm{USp}(4)$ $$\mathrm{ST}^0$$ $$\simeq$$ $$\mathrm{USp}(4)$$
### Decomposition
Simple over $$\overline{\Q}$$
### Endomorphisms
Not of $$\GL_2$$-type over $$\Q$$
Endomorphism ring over $$\Q$$:
$$\End (J_{})$$ $$\simeq$$ $$\Z$$ $$\End (J_{}) \otimes \Q$$ $$\simeq$$ $$\Q$$ $$\End (J_{}) \otimes \R$$ $$\simeq$$ $$\R$$
All $$\overline{\Q}$$-endomorphisms of the Jacobian are defined over $$\Q$$. | 1,217 | 2,952 | {"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} | 3.15625 | 3 | CC-MAIN-2019-30 | latest | en | 0.338212 |
http://nrich.maths.org/public/leg.php?code=-377&cl=1&cldcmpid=5715 | 1,481,255,692,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542680.76/warc/CC-MAIN-20161202170902-00078-ip-10-31-129-80.ec2.internal.warc.gz | 199,152,233 | 6,275 | # Search by Topic
#### Resources tagged with Enrichment similar to Working Effectively with All Learners:
Filter by: Content type:
Stage:
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##### Other tags that relate to Working Effectively with All Learners
Group worthy. Selecting and using information. Rich Tasks. Problem solving. Mathematical Thinking. Video. Enrichment. Pedagogy. Learning mathematics. Questioning.
### There are 18 results
Broad Topics > Mathematics Education and Research > Enrichment
### Integrating Rich Tasks - Activity 1.4
##### Stage: 1 and 2
This professional development activity encourages you to investigate how rich tasks and problem solving link together.
### Integrating Rich Tasks - Activity 2.1
##### Stage: 1 and 2
This professional development activity looks at what teachers can do to support learners engaging with rich tasks
### Integrating Rich Tasks - Activity 4.2
##### Stage: 1 and 2
This professional development activity is designed to help you assess your embedding of rich tasks into the curriculum through evaluating a theme
### Integrating Rich Tasks - Activity 4.3
##### Stage: 1 and 2
This professional development activity is designed to help you assess your embedding of rich tasks into the curriculum and, in particular, think about what to do next
### Working with Highly Able Mathematicians
##### Stage: 1 and 2
In this article for teachers, Bernard describes ways to challenge higher-attaining children at primary level.
### Integrating Rich Tasks - Activity 1.5
##### Stage: 1 and 2
This professional development activity encourages you to investigate what pupils are doing when they problem solving.
### Integrating Rich Tasks - Activity 1.1
##### Stage: 1 and 2
This is activity 1.1 in the series of activities designed to support professional development through integrating rich tasks. This activity looks specifically at what makes an activity "rich".
### Integrating Rich Tasks - Activity 1.3
##### Stage: 1 and 2
This professional development activity encourages you to investigate what is meant by higher-order thinking skills.
### Integrating Rich Tasks - Activity 1.2
##### Stage: 1 and 2
These two tasks are designed to support professional development on integrating rich tasks. You are asked to think about what problems that encourage Higher Order Thinking Skills look like.
### Outside the Box
##### Stage: 2 and 3
This article explores the links between maths, art and history, and suggests investigations that are enjoyable as well as challenging.
### Two Heads Are Better Than One
##### Stage: 1, 2 and 3
An article that reminds us about the value and importance of communication in the mathematics classroom.
### MEI 2005
##### Stage: 1, 2, 3, 4 and 5
Presentation given at the MEI conference in Reading 2005
### Crossing Bridges
##### Stage: 1, 2, 3, 4 and 5
An article for teachers based on a lecture and workshop activities at the NZAMT conference in New Zealand 2007
### Enriching Patterns
##### Stage: 1, 2, 3 and 4
Following on from a workshop at an MA Easter conference, Jennifer and Jenni talked about the way in which the website is made more accessible to teachers who want to plan threads of. . . .
### Mathematical Ways to Spend Your Summer
##### Stage: 2, 3, 4 and 5
Here we look back at the year with NRICH and suggest mathematical summer holiday activities for students, parents and teachers.
### What's Inside/outside/under the Box?
##### Stage: 2 and 3
This article describes investigations that offer opportunities for children to think differently, and pose their own questions, about shapes.
### Interacting with the Geometry of the Circle
##### Stage: 1, 2, 3 and 4
Jennifer Piggott and Charlie Gilderdale describe a free interactive circular geoboard environment that can lead learners to pose mathematical questions.
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails. | 901 | 4,031 | {"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} | 3.015625 | 3 | CC-MAIN-2016-50 | longest | en | 0.894246 |
https://www.skytowner.com/explore/difference_between_sort_and_sorted_in_python | 1,721,299,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00687.warc.gz | 860,731,120 | 9,037 | search
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# Difference between sort and sorted in Python
schedule Aug 12, 2023
Last updated
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The difference between sort and sorted in Python is that while `list.sort(~)` permanently modifies the order of the original list, `sorted(~)` returns your list in a sorted order but does not actually modify the original list.
# sort method
To permanently modify order of elements in `animals` according to reverse alphabetical order:
``` animals = ['cat', 'doge', 'bird']print("Using sort: ", animals.sort(reverse=True))print("Original list: ", animals) Using sort: NoneOriginal list: ['doge', 'cat', 'bird'] ```
Notice that nothing is returned from `animals.sort(reverse=True)`. However, the order of elements in the original list is modified.
# sorted method
To return the list `animals` in reverse alphabetical order:
``` animals = ['cat', 'doge', 'bird']print("Using sorted: ", sorted(animals, reverse=True))print("Original list: ", animals) Using sorted: ['doge', 'cat', 'bird']Original list: ['cat', 'doge', 'bird'] ```
Notice that `sorted(animals, reverse=True)` returns the sorted list. However, the original list is unmodified.
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Ryan collected 43 candy bars on Halloween he gave 15 of them to his father andhe gave his mother ten less candy bars than he gave his father. Then Ryan ate 4 of the candy bars
. Hwo many Andy bars does he have left ?
I got 19
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I agree.
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84.56
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He got diabetes
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https://cutiumum.net/gas-laws-worksheet-boyle-s-and-charles-law-review/ | 1,569,163,735,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575515.93/warc/CC-MAIN-20190922135356-20190922161356-00061.warc.gz | 429,793,970 | 27,940 | # Gas Laws Worksheet Boyle S And Charles Law Review
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Gas Laws Worksheet Boyle S And Charles Law Review Gas laws finally an app for that teacher doug ragan describes two apps that he uses to investigate gas properties and the relationship between boyle s law and charles law classroom this link He quotes a former intelligence official who puts the situation very bluntly we ve seen a growing body of law from the court quot surveillance types make a distinction between secrecy of laws The interaction of pressure volume and temperature of a gas are governed by the perfect gas laws courtesy of boyle charles and gay lussac combining these is the general gas law this.
Gas Laws Worksheet Boyle S And Charles Law Review This chapter from basic principles and calculations in process technology discusses gas laws including boyle s law charles s law does not interfere with the relationships of pressure volume Choose an answer and hit next you will receive your score and answers at the end the volume doubles it decreases to half its original volume the volume stays the same it increases by a factor The gas laws are a set of laws that describe the relationship between thermodynamic temperature t pressure p and volume v of gases three of these laws boyle s law charles s law and.
Gas Laws Worksheet Boyle S And Charles Law Review Also inserted at appropriate intervals within each chapter are sections entitled quot reflection quot and quot projection quot that review graham s law of effusion has been left out as have the individual Exercises 1 1 ideal and van der waals gas laws 2 boyle s law 3 ideal gas law 8 density of gases igl 10 dalton s law of partial pressures mole fractions igl 13 van der waals igl Because so many pressure measurement applications involve gases it is important to consider the gas laws in the ideal gas law is a combination of the boyle s law and charles law and it is.
3 when air flows through a smaller section of a duct it increases in velocity and decreases in pressure and temperature gas turbine engines have advantages and disadvantages evaluate the following The gas particles take up more volume relative to the overall volume the gas particles are compressed so they take up less volume relative to the overall volume the gas particles become hotter so After boyle s law was published by robert boyle in it was postulated by jacques charles in and confirmed in that the volume of a given mass of a gas would vary according to the.
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Top | 1,482 | 7,430 | {"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} | 3 | 3 | CC-MAIN-2019-39 | longest | en | 0.901616 |
https://www.mechamath.com/author/jefferson/page/42/ | 1,675,725,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00131.warc.gz | 880,332,097 | 23,809 | ## Apothem of a Pentagon – Formulas and Examples
The apothem of a pentagon is the perpendicular distance from the center of the pentagon to the center of one … Read more
## Area and Perimeter of a Pentagon – Formulas and Examples
The perimeter of a pentagon is the length of the outline of the pentagon. On the other hand, its area … Read more
## Isosceles Obtuse Triangle – Characteristics and Examples
Isosceles obtuse triangles are triangles that have two sides of the same length and an angle greater than 90 degrees. … Read more
## Isosceles Acute Triangle – Characteristics and Examples
An acute isosceles triangle is a triangle that has two sides of equal length and whose interior angles are acute. … Read more
## What are the characteristics of isosceles triangles?
An isosceles triangle is a triangle consisting of two equal sides. Since these triangles have two equal sides, it also … Read more
## Height of an Isosceles Triangle – Formulas and Examples
The height of an isosceles triangle is the perpendicular distance from the base of the triangle to the opposite vertex. … Read more
## Area and Perimeter of an Isosceles Triangle – Formulas and Examples
The perimeter of an isosceles triangle represents the total length around the triangle. On the other hand, the area represents … Read more
## Isosceles Right Triangle – Formulas and Examples
A right triangle is a triangle in which exactly one internal angle measures 90 degrees. Since the sum of the … Read more
## What are the characteristics of right triangles?
A right triangle is one of the most important figures in geometry and forms the basis for trigonometry. These triangles … Read more
## Hypotenuse of a Right Triangle – Formulas and Examples
The hypotenuse of a right triangle is the side opposite the 90-degree angle. We can calculate the hypotenuse by using … Read more | 418 | 1,870 | {"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} | 3.75 | 4 | CC-MAIN-2023-06 | latest | en | 0.84314 |
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# Solve the empty gas tank of a truck needs to be completely filled. the tank is shaped like a cylinder that is 3.5 ft long with a diameter of 2.4 ft. suppose gas is poured into the tank at a rate of 1.7 ft3 per minute. how many minutes does it take to fill the empty tank
+1
1. 16 August, 18:13
0
The gas tank of a truck cylinder is 4 ft long with a diameter of 2.5ft.
At the gas station, a pump pours gas at the rate of 3ft cubed per minute.
How many minutes does it take to fill the empty tank with that pump?
:
Find vol of a cylinder
r = 1.25 ft
V = 19.635 cu ft
:
time =
t = 6.545 minutes | 189 | 621 | {"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} | 3.5625 | 4 | CC-MAIN-2023-23 | latest | en | 0.938577 |
https://kidzone.ws/math/wp/thankswp4.asp | 1,679,671,500,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00492.warc.gz | 404,890,990 | 3,928 | Mrs. Brown is going to purchase groceries for Thanksgiving dinner. She has found a turkey for \$22.44 and picked all the potatoes and other vegetables she needs for \$12.72. Then she found some pies and desserts for \$11.73. Write the math equation: ______________________ How much did Mrs. Brown spend at the store? _______ | 74 | 326 | {"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-2023-14 | latest | en | 0.942405 |
https://puzzling.meta.stackexchange.com/questions/5484/what-exactly-are-the-scopes-of-tagalgorithm-and-tagstrategy | 1,723,391,988,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00819.warc.gz | 378,277,945 | 28,371 | # What exactly are the scopes of [tag:algorithm] and [tag:strategy]?
If really is meant only for puzzles "related to algorithmics", as its tag description states, then I expect that most questions actually tagged had better be tagged instead. I see several questions related to Rubik's cube and none asking whether some task is $NP$-complete or the like (which would in any case belong better at https://cs.stackexchange.com/, I think).
Strictly speaking, algorithm and strategy are different concepts, and not every strategy can be formulated as an algorithm.
For example, in the Prisoners Problem with infinitely many prisoners and accepting the axiom of choice, the prisoners have a winning strategy, but no winning algorithm.
As it often happens, the mathematical distinction isn't mirrored in everyday language (I'd expect the average SE user to think of a strategy as a type of video game, and an algorithm as a piece of code to make random numbers or sorted arrays) and that's the reason why the tags seem to be used interchangeably.
That said, I agree with the OP that would be a better tag for most question currently tagged with .
If you look further in the algorithm tag wiki, it defines an algorithm as
a process or set of rules to be followed in calculations or other problem-solving operations (especially by a computer).
Really, an algorithm is just a strategy for solving a problem which can be described by a discrete list of precise rules. The concept of "algorithmics" is not constrained to theoretical CS concepts like complexity, computability, languages, etc. It is about anything which can be solved by an algorithm, which does encompass Rubik's cubes, water measuring puzzles, lights out, etc. For these problems, the tags algorithm and strategy are equally valid.
Furthermore, puzzles don't always require finding an algorithm, but simply involve using "algorithmic reasoning." This usually boils down to finding some sort of invariant or non increasing quantity. In the chameleon puzzle, the invariant is the mod 3 difference between the numbers of each pair of colors. Also, this includes information theoretic concepts that show up in liar and weighing puzzles, like, "if there are $n$ possibilities, then it takes $\log_2 n$ binary tests to find the correct one," which comes from the pigeonhole principle.
In summary, I think the scope of the algorithm tag is
• Any puzzle whose solution is an algorithm
• Any puzzle which uses algorithmic reasoning, like invariants, the pigeonhole principle, the extreme value principle (any finite set has a largest element), modular arithmetic, etc.
While the scope of the strategy tag is
• Any puzzle whose solution is a strategy (this further includes "escape from this situation" puzzles)
Neither tag is a subset of the other. | 571 | 2,811 | {"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} | 3.25 | 3 | CC-MAIN-2024-33 | latest | en | 0.95767 |
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Question
# There are two wires, A and B, made of same material. Both the wires have the same length. It is observed that the resistance of wire A is four times the resistance of wire B, then find the ratio of their cross sectional areas.
A
4:1
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B
1:4
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C
1:2
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D
2:1
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Solution
## The correct option is B 1:4We know that R=ρ×lAGiven the material of both wires A and B is same and their lengths are also given to be same. ρandl are constant.So, we have the relation:RARB=ABAA=41So, AAAB=14
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Join BYJU'S Learning Program | 291 | 979 | {"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-2024-18 | latest | en | 0.893548 |
https://discourse.shapr3d.com/t/scale-model-to-specified-dimension/21909 | 1,725,844,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00743.warc.gz | 195,277,499 | 6,829 | # Scale model to specified dimension
Hi,
Somewhat new to Shapr3D and so far has been amazing + intuitive to pick up. Keep running into this problem with scaling though and haven’t been able to figure it out.
I would like to take a model (including all its various parts) that is 4"L x 5"W x 10"H and scale proportionally to 12.625" tall.
How would I do this?
Right now the only way I can figure is by manually calculating the percentage difference and plugging into the scale tool. Which seems odd. I know it’s not the most complicated thing to calculate but it is kind of a pain and really slows things down. Am I missing something?
Appreciate the help!
5 Likes
I have the same question and didn’t find an answer. It seems odd to me, that while constructing things the only option I have is to relatively scale my objects.
This seems pretty easy and obvious, since it’s already available for specific shapes like lines or rectangles - if you select them, you have the option to enter values for length or width.
It would be great to have this for multiple selections as well. Every slicer offers this option to scale the complete object to a specific size, so it seems not really a big deal to implement.
2 Likes
Here’s how. I used the dimensions as noted in @Wally 's original post.
3 Likes
Thanks @TigerMike Yes, currently that seems to be the only way to do it. I guess I was just hoping there was a way to skip the extra input, or external calculating and just type in the target dimension needed. (hopefully a future update!)
Where I find this get’s tricky/cumbersome is when I have a group of objects/parts where furthest points of the group aren’t parallel or it’s harder to determine what the furthest boundary points of group are. Short of making a a new object that is the length of that boundary area, I’d have no way to input a new scale for the group. (to be honest I’m not sure how to measure this in Shaper3D even.) See quick example attached w/ some random forms.
1 Like | 454 | 2,003 | {"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-2024-38 | latest | en | 0.960437 |
http://spssx-discussion.1045642.n5.nabble.com/TURF-rating-scale-consider-frequencies-td5733760.html | 1,495,939,856,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609409.62/warc/CC-MAIN-20170528024152-20170528044152-00416.warc.gz | 425,261,016 | 14,034 | # TURF - rating scale / consider frequencies
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## TURF - rating scale / consider frequencies
Hi, really hope that anybody can help. How to consider e.g. different usage frequencies in SPSS TURF? Lets say I have 4 different products (P1 to P4), 5 respondenst and 4 different possible usage frequenies. How to consider rel usage frequenies in SPSS and not only reach? Also: What's wrong with the Syntax. It handles CRITERIA as if 1 ,2 ,3 are the same as "1". Many thank in advance! values 0 "no usage" 1 "usage 1 time per week" 2 "usage 2 times per week" 3 "usage 3 times per week". data p1 P2 P3 P4 1 1 3 3 2 1 3 3 3 1 3 0 0 0 0 0 0 0 1 0 Syntax: SPSSINC TURF VARIABLES = P1 P2 P3 P4 /OPTIONS BESTN = 5 NUMBERTODISPLAY = 2 THRESHOLD = 0 CRITERIA = 1 2 3 REMOVALCYCLES = 0. MAXSETS=5000000 /IMPORTANCE STRICT=YES.
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## Re: TURF - rating scale / consider frequencies
TURF is designed for ratings, not frequencies. The logic is to find bundles of products that maximize the probability that a customer will like at least one item. Also, I hope your actual data is much larger than what you posted.The CRITERIA keyword takes a list of values that are considered positive responses, so CRITERIA = 1 2 3 means that all of those are positive responses. It does not distinguish among them. This is consistent with the probability statement above. In effect, in your case, you are looking for the bundle with the most positive value as they all indicate a favorable response.HTH,On Thu, Jan 26, 2017 at 8:11 AM, nocheinschmidt wrote:Hi, really hope that anybody can help. How to consider e.g. different usage frequencies in SPSS TURF? Lets say I have 4 different products (P1 to P4), 5 respondenst and 4 different possible usage frequenies. How to consider rel usage frequenies in SPSS and not only reach? Also: What's wrong with the Syntax. It handles CRITERIA as if 1 ,2 ,3 are the same as "1". Many thank in advance! values 0 "no usage" 1 "usage 1 time per week" 2 "usage 2 times per week" 3 "usage 3 times per week". data p1 P2 P3 P4 1 1 3 3 2 1 3 3 3 1 3 0 0 0 0 0 0 0 1 0 Syntax: SPSSINC TURF VARIABLES = P1 P2 P3 P4 /OPTIONS BESTN = 5 NUMBERTODISPLAY = 2 THRESHOLD = 0 CRITERIA = 1 2 3 REMOVALCYCLES = 0. MAXSETS=5000000 /IMPORTANCE STRICT=YES. -- View this message in context: http://spssx-discussion.1045642.n5.nabble.com/TURF-rating-scale-consider-frequencies-tp5733760.html Sent from the SPSSX Discussion mailing list archive at Nabble.com. ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD -- Jon K Peck[hidden email] ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD
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## Re: TURF - rating scale / consider frequencies
Hi Jon, many thanks for that quick response! Just started dealing with TUrf and before gathering data I wanted to understand how it works...so it was just a short example to describe the issue. Ok, no way to consider real (usage) frequencies by using SPSS Turf...so maby I have to search for another method...guess I am looking for some kind of portfolio optimizer including costs, revenue....and reach. Have a nice weekend! Thanks again.Sven Jon Peck [via SPSSX Discussion] <[hidden email]> schrieb am 18:50 Donnerstag, 26.Januar 2017: TURF is designed for ratings, not frequencies. The logic is to find bundles of products that maximize the probability that a customer will like at least one item. Also, I hope your actual data is much larger than what you posted.The CRITERIA keyword takes a list of values that are considered positive responses, so CRITERIA = 1 2 3 means that all of those are positive responses. It does not distinguish among them. This is consistent with the probability statement above. In effect, in your case, you are looking for the bundle with the most positive value as they all indicate a favorable response.HTH,On Thu, Jan 26, 2017 at 8:11 AM, nocheinschmidt wrote:Hi, really hope that anybody can help. How to consider e.g. different usage frequencies in SPSS TURF? Lets say I have 4 different products (P1 to P4), 5 respondenst and 4 different possible usage frequenies. How to consider rel usage frequenies in SPSS and not only reach? Also: What's wrong with the Syntax. It handles CRITERIA as if 1 ,2 ,3 are the same as "1". Many thank in advance! values 0 "no usage" 1 "usage 1 time per week" 2 "usage 2 times per week" 3 "usage 3 times per week". data p1 P2 P3 P4 1 1 3 3 2 1 3 3 3 1 3 0 0 0 0 0 0 0 1 0 Syntax: SPSSINC TURF VARIABLES = P1 P2 P3 P4 /OPTIONS BESTN = 5 NUMBERTODISPLAY = 2 THRESHOLD = 0 CRITERIA = 1 2 3 REMOVALCYCLES = 0. MAXSETS=5000000 /IMPORTANCE STRICT=YES. -- View this message in context: http://spssx-discussion. 1045642.n5.nabble.com/TURF- rating-scale-consider- frequencies-tp5733760.html Sent from the SPSSX Discussion mailing list archive at Nabble.com. ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD -- Jon K Peck[hidden email] ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD If you reply to this email, your message will be added to the discussion below: http://spssx-discussion.1045642.n5.nabble.com/TURF-rating-scale-consider-frequencies-tp5733760p5733761.html To unsubscribe from TURF - rating scale / consider frequencies, click here. NAML | 1,850 | 6,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-22 | longest | en | 0.872554 |
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Multimeter and pulsed DC current (PWM)
Multimeter and pulsed DC current (PWM)
A multimeter usual can measure DC and AC but what happens when measuring a pulsing voltage or current?
For this article I will be using 3. meters:
Fluke 289: A very expensive meter with just about everything.
BSide ZT301: A cheap meter with all the common function.
UNI-T UT210E: A cheap clamp meter, I will only use this when looking at current.
Contents
DC voltage
DC voltage BSide ZT301
DC voltage Fluke 289
What voltage do I use (DC / RMS / DC+AC)?
DC current
DC current BSide ZT301
DC current Fluke 289
DC current UNI-T UT210E
What current do I use (DC / RMS / DC+AC)?
Can a DMM Measuring maximum and minimum voltage/current in PWM
Power
Conclusion
Notes
DC voltage
In DC a multimeter will not use RMS circuit, but a averaging circuit.
I will use 3 waveforms for this:
A 50% duty cycle waveform from 0V to 10V, this means the average voltage is 5.00V, a RMS meter will show 5.00V (It removes the DC component) and the DC+AC voltage is 7.07V, because it is a square it will contain much higher frequencies than the nominal frequency.
A 10% duty cycle waveform from 0V to 10V, this means the average voltage is 1.00V, a RMS meter will show 3.00V and DC+AC is voltage 3.16V, because it is a square with a 10% duty cycle the frequencies in this curve will be much higher than above.
A 90% duty cycle waveform from 0V to 10V, this means the average voltage is 9.00V, a RMS meter will show 3.00V and the DC+AC voltage is 9.49, this will have the same high frequency content as the 10% curve.
DC voltage BSide ZT301
The meter shows average voltage correctly from 10Hz and up to 1MHz, but RMS is already wrong at 1kHz, because the meter do not have bandwidth enough to handle a 1kHz square wave.
At 10% and 90% the DC voltage is shown correctly and the RMS is more wrong due to the more higher frequencies in the PWM signal.
DC voltage Fluke 289
This high end meter has much higher bandwidth, but it is not very good a 10Hz (It is too fast).
The average voltage is again correctly shown and the RMS works at higher frequencies. The DC+AC only works as far up in frequencies as the RMS works, because it is partially based on it.
With different duty cycle the DC is again correct, but the RMS and DC+AC will show wrong results at a lower frequencies.
What voltage do I use (DC / RMS / DC+AC)?
The interesting question is now what value to use? The answer is that it depends on application.
For a resistive heater it is the DC+AC value that is interesting.
An example:
1ohm heater with 10V: power -> sqr(voltage)/resistance -> 10*10 / 1 -> 100 Watt
When using a 50% PWM the power will 50% of that or 50 Watt, this could be calculates as sqr(DC+AC)/1 -> 7*7/1 -> 50 Watt
For 10% the power will 10% of that or 10 Watt, this could be calculates as sqr(DC+AC)/1 -> 3.16*3.16/1 -> 10 Watt
If the PWM goes to a filter to be converted to a DC voltage, then it is the DC that is interesting and the voltage will be proportional to the duty cycle.
DC current
For current I only use the 50% and 10% waveforms, but adds a meter more.
For the clamp meter I shorted the output of my generator, because it has a 50ohm output impedance and delivers 10V the current is 200mA. For the multimeters I added a 1kOhm series resistance, i.e. total resistance was 1050ohm, in addition to between 1 and 2 ohm in the multimeter
This average the current is 4.75mA for multimeters
This average the current is 0.95mA for multimeters
DC current BSide ZT301
The DC works at all frequencies and the RMS is limited by the RMS converter in the meter, i.e. it cannot handle 1kHz correctly.
At 10% duty cycle the DC again works and the RMS drops a bit faster, at higher frequencies some of the signal is running around the converter inside the meter.
DC current Fluke 289
The DC works perfectly and the RMS can handle higher frequencies on this meter, the AC+DC reading requires RMS to be correct, before it can be correct.
No surprise here, the DC again works and the RMS drops of at higher frequencies.
DC current UNI-T UT210E
This is a clamp meter and I had to use higher current for it.
The behaviour is exactly the same as the other meters, DC works and RMS cannot handle higher frequencies.
Again the same.
What current do I use (DC / RMS / DC+AC)?
When working with DC the answer here will usual be the DC value, this is the current that will be drained from a battery or capacitor and also the current devices will use.
The exception is if you want to calculate power for resistors.
Another way to look at it:
DC measures the average current going to or from a device.
AC+DC (RMS) measures the total current going back and forth in a cable and weights it according to how much heat it generates.
With a clean DC the two values will be exactly the same.
With a pulsing DC they will be equal with a resistive load, but with anything else they will be different.
Can a DMM Measuring maximum and minimum voltage/current in PWM
This is mostly a job for an oscilloscope, multimeters are not fast enough.
If you can control the PWM it is easiest to measure DC voltage at highest and lowest PWM setting.
The min/max function on multimeteres usual requires about 0.3s or 300ms to measure a value, this means it cannot be used even at 10Hz.
Some meters has a “peak” function, this may be able to measure 1ms or even 0.1ms peaks.
From the above table it is fairly obvious that even the peak function is too slow when the PWM frequency gets much above 100Hz.
Power
When working with pulsed current or PWM you cannot just multiply voltage and current to get power, here the RMS (or rather DC+AC) values can be useful.
Example:
0V to 10V with 10% duty cycle will delivering 0A at 0V and 1A 10V.
It is easy to calculate that it will deliver 10 watt when (Voltage x Current -> 10 × 1) on and with 10% duty cycle the power is 1 Watt.
But using the DC values that is 1V (10% of 10V) and 0.1A (10% of 1) gives 0.1 Watt, i.e. very wrong.
What about DC+AC values, that is 3.16V and 0.316A and gives the correct 1 Watt.
But there is a caveat on that, the RMS part will include current running both ways, this is fine for dimensioning the cable and calculating power for resistors, but when the circuit has a inductor or capacitor that returns some current each cycle, the above power calculations will fail!
Conclusion
When measuring on PWM it is important to know what the result must be used for, average is useful for some measurement and AC+DC is required for other (If meter is missing AC+DC use a calculator, see below).
The bandwidth on multimeters is for AC measurements including DC+AC, but DC measurement usual works at much higher frequencies.
Notes
For meters without DC+AC it can be calculated this way: DC+AC = sqrt(sqr(DC)+sqr(AC))
In this article I assumes DC+AC is based on the true RMS value of the AC (I have not seen AC+DC on a non-true-RMS meter)
My website with reviews of many chargers and batteries (More than 1000): https://lygte-info.dk/ | 1,881 | 7,181 | {"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} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.869612 |
https://convertertoolz.com/conv/212-cm-to-inches/ | 1,726,453,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00129.warc.gz | 158,123,193 | 11,786 | # 212 cm to inches
## The importance of understanding metric and imperial systems
When it comes to understanding measurements, having a clear grasp of both the metric and imperial systems is crucial. These systems are widely used across the globe, and being able to convert between them accurately can prevent errors and misunderstandings. Whether you are working in a scientific field, traveling internationally, or even just following a recipe from a foreign cookbook, having a solid understanding of metric and imperial units will undoubtedly come in handy.
The metric system, also known as the International System of Units (SI), is the most widely used system of measurement in the world. It is based on powers of ten, providing a logical and consistent way to measure various quantities. The imperial system, on the other hand, is primarily used in the United States and a few other countries. It originated from the ancient English system of measurement and is known for its use of units such as inches, feet, and pounds. While the metric system is based on a decimal system, the imperial system is more fragmented and can often be challenging to work with for those who are accustomed to the metric system. Understanding both systems allows for seamless communication and cooperation across different regions and disciplines.
## Exploring the differences between centimeters and inches
Centimeters and inches are two commonly used units of measurement that serve different purposes in various industries and everyday life. One of the key differences between centimeters and inches is their origin. Centimeters belong to the metric system, which is used in most countries around the world and is based on units of ten. On the other hand, inches are part of the imperial system primarily used in the United States, which is based on historical English measurements.
Another important distinction lies in their relative sizes. A centimeter is smaller than an inch, with 2.54 centimeters equating to one inch. This means that if you were to compare a ruler in centimeters to one in inches, the centimeter ruler would have more markings within the same length. This variance can make conversions between the two systems challenging, particularly when accuracy is crucial. The reasons for using one unit over the other often depend on cultural traditions, geographic location, and the specific field of application.
## Understanding the conversion factor between centimeters and inches
Of paramount importance is having a clear comprehension of the conversion factor between centimeters and inches. Understanding this conversion allows for seamless transition between the metric and imperial systems of measurement. With knowledge of the conversion factor, one can easily convert centimeters to inches and vice versa accurately and efficiently.
The conversion factor between centimeters and inches is 2.54. This means that one inch is equivalent to 2.54 centimeters, and conversely, one centimeter is equal to 0.3937 inches. This conversion factor is derived from the exact definition of an inch, which is equal to 2.54 centimeters. By multiplying or dividing a measurement by this conversion factor, one can easily convert between the two systems. For example, to convert 10 centimeters to inches, one would multiply 10 by 0.3937 to get 3.937 inches. This simple formula ensures accurate conversions and facilitates the effective use of both systems in various applications.
## Why 212 cm is a commonly used measurement
The measurement of 212 cm holds a significant place in various industries and fields. One reason for its common usage is its relevance in clothing manufacturing and sizing. The fashion industry relies on accurate measurements to design and produce garments that fit well. With 212 cm being approximately 83.5 inches, it corresponds to a commonly found tall height range for individuals. Therefore, this measurement is frequently used as a reference point for creating clothing sizes suitable for taller individuals.
Another factor contributing to the popularity of 212 cm as a common measurement lies in the realm of architectural and interior design. When it comes to designing spaces such as doorways and ceiling heights, standard measurements are often followed to ensure consistency across buildings. In certain regions, 212 cm has become a standard door height due to its practicality and ease of installation. By using this measurement as a general guideline, contractors and designers can ensure that doors are of an appropriate height, accommodating most people without excessive alterations.
## The step-by-step process to convert 212 cm to inches
One method to convert 212 cm to inches is through multiplication. To get started, you will need to know that there are 2.54 centimeters in one inch. The conversion factor here is 1 inch equals 2.54 centimeters.
Follow these simple steps to convert 212 cm to inches:
Step 1: Take the given measurement, which is 212 cm.
Step 2: Multiply the given measurement by the conversion factor. In this case, you would perform the calculation: 212 cm × (1 inch/2.54 cm).
Step 3: Simplify the equation by canceling out the centimeters units. This will leave you with the measurement in inches.
Step 4: Perform the multiplication: 212 × 1 inch = 212 inches.
Thus, the final result of the conversion is 212 centimeters equals 212 inches.
## Examples of real-life situations where knowing this conversion is useful
When it comes to real-life situations, having knowledge of the conversion between centimeters and inches can prove to be extremely useful. For instance, consider a DIY enthusiast who is following a tutorial to build a piece of furniture. The tutorial may provide measurements in inches, but if the enthusiast is more comfortable working with centimeters, knowing the conversion can help them quickly and accurately convert the measurements to suit their preferred metric.
In another example, the fashion industry often uses both centimeters and inches when designing clothing. Models and designers alike need to have a solid understanding of both systems in order to accurately determine measurements and ensure the perfect fit. Knowing the conversion between centimeters and inches is crucial in this scenario as it allows for seamless communication and precise execution of measurements, ultimately resulting in well-fitting garments. | 1,220 | 6,434 | {"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-2024-38 | latest | en | 0.964747 |
http://physicsformom.blogspot.com/2010_09_01_archive.html | 1,529,796,629,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865438.16/warc/CC-MAIN-20180623225824-20180624005824-00447.warc.gz | 245,125,301 | 10,103 | ## Monday, September 6, 2010
### Angular scales from the CMB
The physics of the CMB is extremely rich, and I won't do it justice in this series of entries. However, I do want to give one example of how the spectrum of the CMB fluctuations and in particular the location and size of the peaks gives us information about the universe. Let's look at the results from analyzing the CMB maps using the spherical harmonic functions one more time:
As I said last time, this plot tells us about the correlations between different regions of the sky. What exactly does that mean? Well, to the naked eye, the CMB map looks fairly random - some parts are blue, some parts are green, but there's no obvious pattern. What the above graph tells us is that regions of the map separated by 1 degree are actually related to each other. The amplitude of the spherical harmonic with a "frequency" of 1 degree is very high compared to other frequencies, just as the amplitude of the sine curve with the same frequency as the A note was very high when I was decomposing the A chord using Fourier analysis.
The next question to be asked is, "so what?" And the answer relates to stuff I was talking about a while ago in the posts on Gravitational Potential Wells. There, I talked about how early fluctuations in the gravitational potential created oscillations, like balls rolling in and out of a divot. In the early universe, small gravitational fluctuations of all shapes and sizes were created. These fluctuations expanded and contracted, interacting with photons to make hot and cold photons, up until the moment of last scatter when the universe became neutral and no longer interacted with photons at all, creating the CMB. What the large peak at 1 degree in the CMB spectrum tells us is precisely the size of the fluctuation that oscillated one time before the moment of last scatter.
Let me see if I can come up with an analogy for how that works that makes sense. Imagine watching a swimming race where the swimmers do laps in the pool. Let's suppose this race includes swimmers of all abilities, so some are very slow and some are like Michael Phelps. The race starts with all swimmers along the starting line, but as time passes, the swimmers spread out according to their abilities. However, because they have to swim laps, the absolute distance between the swimmers and the starting distance is always less than or equal to the length of the pool - in other words, you can't tell the fast and slow swimmers by how far they are from the start, since there will be times when the fast swimmer is heading back to the start line while the slow swimmer is still at the far end of the pool and vice versa.
Now let's stop the race and have each swimmer stop exactly where they are. They will be spread out all over the pool. But, a few of them will be near the far end of the pool. And some of them, the ones whose pace was exactly right will be exactly at the far end of the pool. If we measure the maximum fluctuation in the position of the swimmers, we find that all the swimmers who exactly swam the full length of the pool have that maximum fluctuation - they are farthest from the start. The CMB measurement is making a similar type of measurement. It measures the size of the gravitational potential well that was maximally expanded when we stopped the race, or when the CMB decoupled from the universe. Therefore, we can calculate the size of the pool - or the size of the universe at the time of last scatter. The location of the first peak in the CMB is like a ruler for the early universe. And that ruler helps us find all that other information I've been talking about.
As usual, I fear I have not fully done the physics justice in this rather slow developing and superficial treatment of the problem. However, I think I'm done with the CMB for now. If you want to learn more about it, I do recommend Professor Wayne Hu's excellent website at http://background.uchicago.edu/~whu/ which explains the CMB much better than I could hope to. | 864 | 4,038 | {"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-2018-26 | latest | en | 0.963463 |
https://cracku.in/112-the-least-number-which-is-a-perfect-square-and-is--x-snap-2015 | 1,722,642,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00201.warc.gz | 150,199,560 | 25,497 | Question 112
# The least number which is a perfect square and is divisible by each of the numbers 14, 16, 18 is
Solution
The least number that is divisible by 14,16,18 will be the LCM of three which is 1008 but it is not a perfect square .
$$1008=2^4\cdot3^2\cdot7^{ }$$ To make it a perfect square you need to multiply it by 7.
1008*7=7056
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• Top 5000+ MBA exam Solved Questions for Free | 141 | 474 | {"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} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.890758 |
https://viderime.com/civil-engineering/geo-technical-engg/seepage-velocity-and-seepage-pressure/ | 1,726,236,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00803.warc.gz | 566,803,124 | 24,415 | # Seepage velocity and Seepage pressure
Seepage velocity:
The velocity of water obtained using the total cross-sectional area of the soil is known as discharge velocity or superficial velocity.
Since water flows only through voids and not through the total cross-sectional area, the actual velocity of water is much higher than the discharge velocity, that is known as seepage velocity (vs).
By continuity equation used in fluid mechanics:-
q = vA = v× Av •••••••••••• (1)
where vs is the seepage velocity and Av is the area of voids in the cross-section.
From Equation (1),
We have vs = v A/Av.
We know that:
(Av/A) = [(Av x L)/(A x L)] = (Vv/V) = n
where Vv is the volume of voids
V is the total volume of the soil, and n is the porosity of the soil. Hence:
v= v/n
Seepage velocity is always more than discharge velocity. The coefficient of percolation is always more than the coefficient of permeability.
Seepage Pressure:
The pressure exerted by the flowing water on the soil due to viscous friction between the water and soil particles is known as seepage pressure. Mathematically, it is given by –
Seepage pressure, σs = γwh •••••••••••(2)
where h is the hydraulic head lost due to the viscous friction and
γw is the density of water.
Seepage pressure always acts in the direction of flow. When the flow takes place in the vertical direction –
σ= γwh = γ(h/z). z = γw.i.z ••••••••••••(3)
where i is the hydraulic gradient. The vertical effective pressure may be increased or decreased due to the seepage, depending upon whether the flow takes place in the downward or the upward direction, respectively.
Effective pressure, σ’ = γ’z ± γw iz••••••••••••(4)
When the flow takes place in the upward direction, the seepage pressure therefore acts in the upward direction and the effective pressure is reduced. If the flow takes place at high hydraulic gradient, the net effective pressure, defined by Eq. (4)), is reduced to zero.
In such a case, a cohesionless soil mass becomes weightless and cannot support any load, because its shear strength is reduced to zero. This phenomenon of loss of shear strength of soil in the upward flow condition in cohesionless soils is known as the quick condition, boiling condition, or quicksand.
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# Chapter 10
Standards
14.0 Students solve a quadratic equation by factoring or completing the
square.
19.0 Students know the quadratic formula and are familiar with its proof
by completing the square.
20.0 Students use the quadratic formula to find the roots of a second-
degree polynomial and to solve quadratic equations.
21.0 Students graph quadratic functions and know that their roots are the
x-intercepts.
23.0 Students apply quadratic equations to physical problems, such as the
motion of an object under the force of gravity.
The human cannonball
act is a performance in which a
person (the "cannonball") is ejected
from a specially designed "cannon".
It reaches a height of 70 ft and
travels a distance of 160 ft before
landing in a safety net.
Equation of the path flown:
f(x) = -7 (x 80)
2
+ 70
640
for 0x160.
All objects that are projected in the
Completing the Square
Lesson 10.1
Objectives:
1. Solve quadratic equations by taking the square root
of both sides.
2. Solve quadratic equations by completing the square.
3. Use quadratic equations to solve for missing parts of
right triangles.
Think & Share
Suppose a deposit of
\$3,000 in a savings
account that paid an
annual interest rate r
(compounded yearly) is
Worth \$3,456 after 2
years. Using the formula
A = P(1+r)
t
, we have
3, 456 = 3,000 ( 1 + r)
2
Solve for r to find the
annual interest rate.
Solution:
456 = 3,000 (1 + r)
2
456 = 3,000 (1 +r)
2
3000 3,000
1.152 = (1 +r)
2
1.73 = 1 + r (take the square root of both sides)
0.73 = r
Therefore the interest rate is 735%.
Think & Share
An escalator in a
department store is to
carry people a vertical
distance of 20 feet
between floors. How long
is the escalator if it
makes an angle of 45
with the ground?
Solutions
c
2
= 20
2
+20
2
c
2
= 400
+400
c
2
= 800
(take the square root of both sides)
c = 20 2
Therefore the escalator is about 20 ft long.
45
45
90
20 ft
20 ft
c?
The square root Property
If b is a real number and a
2
= b,
then
b a =
Solve x
2
= 49
2 = x
Solve (y 3)
2
= 4
Solve 2x
2
= 4
x
2
= 2
7 49 = = x
y = 3 2
y = 1 or 5
2 4 3 = = y
Square Root Property
Solve x
2
+ 4 = 0
x
2
= 4
There is no real solution because the square
root of 4 is not a real number.
Square Root Property
Solve (x + 2)
2
= 25
x = 2 5
x = 2 + 5 or x = 2
5
x = 3 or x = 7
5 25 2 = = + x
Square Root Property
Solve (3x 17)
2
= 28
7 2 17 3 = x
3
7 2 17
= x
7 2 28 = 3x 17 =
Square Root Property
What is completing the square used for?
Completing the square is used for all
those not factorable problems!!
It is used to solve these equations for
the variable.
Rule for Completing the Square
bx x +
2
2
2
2
|
.
|
\
|
+ +
b
bx x
2
2
|
.
|
\
|
+ =
b
x
This is now a
Perfect Square
Trinomial!
So, it factors into
this!
Example: Find the value of c that makes this
a PST, then write the expression as the
square of a binomial. x
2
-3x+c
b=-3
2 2
2
3
2
|
.
|
\
|
=
|
.
|
\
|
=
b
c
4
9
=
2
2
3
|
.
|
\
|
x
4
9
3
2
+ x x
Example: Solve by completing the
square. x
2
+6x-8=0
x
2
+6x-8=0
x
2
+6x=8
x
2
+6x+___=8+___
x
2
+6x+9=8+9
(x+3)
2
=17
( ) 9 3
2
6
2
2
2 2
= =
|
.
|
\
|
=
|
.
|
\
|
b
17 3 = + x
17 3 = x
Dont forget: Whatever
you add to one side of an
to the other side!
Completing a Square
1) If the coefficient of x
2
is NOT 1, divide
both sides of the equation by the
coefficient.
2) Isolate all variable terms on one side of
the equation.
3) Complete the square (half the coefficient
of the x term squared, added to both sides
of the equation).
4) Factor the resulting trinomial.
5) Use the square root property.
More Examples!
5x
2
-10x+30=0
x
2
-2x+6=0
x
2
-2x=-6
x
2
-2x+__=-6+__
x
2
-2x+1=-6+1
(x-1)
2
=-5
3x
2
-12x+18=0
x
2
-4x+6=0
x
2
-4x=-6
x
2
-4x+__=-6+__
x
2
-4x+4=-6+4
(x-2)
2
=-2
( ) 1 1
2
2
2
2
2 2
= =
|
.
|
\
|
=
|
.
|
\
|
b
5 1 = x
5 1 i x =
( ) 4 2
2
4
2
2
2 2
= =
|
.
|
\
|
=
|
.
|
\
|
b
2 2 = x
2 2 i x =
Last Example! Write the quadratic function
y=x
2
+6x+16 in vertex form.
What is the vertex of the functions graph?
y=x
2
+6x+16
y-16=x
2
+6x
y-16+__=x
2
+6x+__
y-16+9=x
2
+6x+9
y-7=(x+3)
2
y=(x+3)
2
+7
If the equation, in vertex
form, is y=(x+3)
2
+7,
then the vertex must
be (-3,7).
( ) 9 3
2
6
2
2
2 2
= =
|
.
|
\
|
=
|
.
|
\
|
b
And ye are complete in Him
[Jesus Christ] which is the head
of all principality and power." -
Colossians 2:10
IFL IFL IFL IFL IFL
Homework
Problem Set 10.1
TB pp. 603-605, Multiples of 4 (4, 8, , 76)
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#### Resources tagged with biology similar to Twizzles Venture Forth:
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### There are 21 results
Broad Topics > Applications > biology
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Core Scientific Mathematics
##### Stage: 4 and 5 Challenge Level:
This is the area of the advanced stemNRICH site devoted to the core applied mathematics underlying the sciences.
### Ratios and Dilutions
##### Stage: 4 Challenge Level:
Scientists often require solutions which are diluted to a particular concentration. In this problem, you can explore the mathematics of simple dilutions
##### Stage: 4 Challenge Level:
Which units would you choose best to fit these situations?
### The Genes of Gilgamesh
##### Stage: 4 Challenge Level:
Can you work out the parentage of the ancient hero Gilgamesh?
### Bigger or Smaller?
##### Stage: 4 Challenge Level:
When you change the units, do the numbers get bigger or smaller?
### Chemnrich
##### Stage: 4 and 5 Challenge Level:
chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study. . . .
### Robot Camera
##### Stage: 4 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Bionrich
##### Stage: 4 and 5 Challenge Level:
bioNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of the biological sciences, designed to help develop the mathematics required to get the most from your. . . .
### Alternative Record Book
##### Stage: 4 and 5 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Mixed up Mixture
##### Stage: 4 Challenge Level:
Can you fill in the mixed up numbers in this dilution calculation?
### Exact Dilutions
##### Stage: 4 Challenge Level:
Which exact dilution ratios can you make using only 2 dilutions?
### Dilution Series Calculator
##### Stage: 4 Challenge Level:
Which dilutions can you make using 10ml pipettes and 100ml measuring cylinders?
### Packing 3D Shapes
##### Stage: 4 Challenge Level:
What 3D shapes occur in nature. How efficiently can you pack these shapes together?
### Conversion Sorter
##### Stage: 4 Challenge Level:
Can you break down this conversion process into logical steps?
### Investigating the Dilution Series
##### Stage: 4 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Epidemic Modelling
##### Stage: 4 and 5 Challenge Level:
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths. | 695 | 3,196 | {"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-2017-47 | latest | en | 0.853518 |
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### Messages - Xinyu Jing
Pages: [1]
1
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 06:30:34 AM »
(𝑎)
𝑊=$𝑐𝑒^{−∫𝑝(𝑡)𝑑𝑡}$=$𝑐𝑒^{−∫4𝑑𝑡}$=$𝑐𝑒^{−4𝑡}$
(𝑏)
$𝑟^{3}+4𝑟^{2}+𝑟−6$=0
(𝑟−1)($𝑟^{2}$+5𝑟+6)=0
𝑟=1,𝑟=−2,𝑟=−3
∴𝑦(𝑡)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}$
W=\begin{vmatrix}
e^{t} & e^{-2t} & e^{-3t} \\
e^{t} & -2e^{-2t} & -3e^{-3t} \\
e^{t} & 4e^{-2t} & 9e^{-3t} \\
\end{vmatrix}=−12𝑒−4𝑡
∴𝑐=−12 compare with (a)
(𝑐)
𝑦𝑝(𝑡)=$𝐴𝑡𝑒^{𝑡}$
𝑦′𝑝(𝑡)=$𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
𝑦′′𝑝(𝑡)=$2𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
𝑦′′′𝑝(𝑡)=$3𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
$3𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}+8𝐴𝑒^{𝑡}+4𝐴𝑡𝑒^{𝑡}+𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}−6𝐴𝑡𝑒^{𝑡}=12𝐴𝑒^{𝑡}$
12𝐴=24
𝐴=2
∴𝑦(𝑡)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}+2𝑡𝑒^{𝑡}$
2
##### Quiz-5 / LEC0101 QUIZ5
« on: October 31, 2019, 08:35:16 PM »
Given
$x^{2}y''+xy'+(x^{2}-0.25)y=3x^{3/2}sinx$,x>0;
$y_{1}(x)=x^{-1/2}sinx$, $y_{2}(x)=x^{-1/2}cosx$
Step 1
The equation is written in standard form as:
$y''+\frac{1}{x}y'+\frac{(x^{2}-0.25)}{x}y=3x^{-1/2}sinx$
$g(x)=3x^{-1/2}sinx$
Now further, Wronskian is evaluated as:
W($x^{-1/2}sinx$,$x^{-1/2}cosx$)=$\left | {x^{-1/2}sinx \qquad \qquad \qquad \qquad \qquad \qquad x^{-1/2}cosx} \right |$
$\left | \frac{-1}{2}x^{-1/2}sinx+x^{-1/2}cosx \qquad \frac{-1}{2}x^{-3/2}cosx-x^{-1/2}sinx \right |$
=$\frac{-1}{2}x^{-2}sinxcosx-x^{-1}sin^{2}x+\frac{1}{2}x^{-2}sinxcos-x^{-1}cos^{2}x$
=$-x^{-1}(sin^{2}x+cos^{2}x)$=$-x^{-1}$
Step 2
The parameters U1 and U2 are evaluated as:
$u_{t}=-\int{\frac{y_{2}g(x)}{W(y_{1},y_{2})}}dx$
$u_{t}=-\int{\frac{x^{-1/2}cosx*3x^{-1/2}sinx}{-x^{-1}}dx}$
$u_{t}=\int{3cosxsinxdx}$
$u_{t}=\frac{-3}{2}cos^{2}x$
$u_{2}=-\int{\frac{y_{1}g(x)}{W(y_{1},y_{2})}dx}$
$u_{2}=-\int{\frac{x^{-1/2}sinx*3x^{-1/2}sinx}{-x^{-1}dx}}dx$
$u_{2}=\int3sinxcosx-\frac{3}{2}x$
STEP 3
Futher,
Y(x)=$y_{1}u_{1}$+$y_{2}u_{2}$
Y(x)=$x^{-1/2}sinx*\frac{-3}{2}cos^{2}x+x^{-1/2}cosx(\frac{3}{2}sinxcosx-\frac{3}{2}x)$
Y(x)=$\frac{-3}{2}x^{-1/2}cos^{2}xsinx+\frac{3}{2}x^{-1/2}cos^{2}xsinx-\frac{3}{2}x^{1/2}cosx$
Y(x)=$\frac{-3}{2}x^{1/2}cosx$
Hence, the solution is $Y(x)=\frac{-3}{2}x^{1/2}cosx$
3
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 24, 2019, 01:08:06 PM »
𝑀𝑦=$1+6𝑦𝑒^{3𝑥}$
𝑁𝑥=$4𝑦𝑒^{2𝑥}$
𝑀𝑦≠𝑁𝑥,it is not exact
$𝑅_{2}$=(𝑀𝑦−𝑁𝑥)/𝑁=$\frac{1+2𝑦𝑒^{2𝑥}}{1+2𝑦𝑒^{2𝑥}}$=1
μ=$𝑒∫𝑅_{2}𝑑𝑥$=𝑒∫1𝑑𝑥=$𝑒^{𝑥}$
Multiplying both sides by 𝜇, we get
$𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}+(𝑒𝑥+2𝑦𝑒^{3𝑥})𝑦′=0$
$𝑀′𝑦=𝑒𝑥+6𝑦𝑒^{3𝑥}$
$𝑁′𝑥=𝑒𝑥+6𝑦𝑒^{3𝑥}$
𝑀′𝑦=𝑁′𝑥,it is exact
∃φ(𝑥,𝑦)𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 φ𝑥=𝑀′,φ𝑦=𝑁′
φ(𝑥,𝑦)=∫𝑀′𝑑𝑥=$∫𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}𝑑𝑥=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}+ℎ(𝑦)$
$φ𝑦=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}+ℎ(𝑦)′=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}$
Then ℎ(𝑦)′=0
Hence h(y)=c
$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=𝑐$
Since y(0)=1
$1⋅𝑒^{0}+12⋅𝑒^{0}=2=𝑐$
$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=2$
4
##### Term Test 1 / Re: Problem 3 (morning)
« on: October 24, 2019, 12:45:24 PM »
𝑦′′−6𝑦′+8𝑦=48sinh(2𝑥)
𝑦′′−6𝑦′+8𝑦=$24𝑒^{2𝑥}−24𝑒^{−2𝑥}$
Let $𝑦=𝑒^{𝑟𝑥(2)}$
𝑦′=sec(𝑟𝑥) 𝑦′′=$𝑟^{2}𝑒^{𝑟𝑥(3)}$
$r^{2}−r𝑥+8=0$
$𝑟_{1}=2, 𝑟_{2}=4$
$𝑦=𝑐_{1}𝑒^{4𝑥}+𝑐_{2}𝑒^{2x}$
let 𝑦=$𝐴𝑥𝑒^{2𝑥}$
𝑦′=$𝐴𝑒^{2𝑥}+2𝐴𝑥𝑒^{2𝑥}$
𝑦″=$4𝐴𝑒^{2𝑥}𝑥+2𝐴𝑒^{2𝑥}+2𝐴𝑒^{2𝑥}=4𝐴𝑒^{2𝑥}𝑥+4𝐴𝑒^{2𝑥}$
$(8𝐴−12𝐴+4𝐴)𝑡⋅𝑒^{2𝑥}+(−6𝐴+4𝐴)𝑒^{2𝑥}=24𝑒^{2𝑥}−2𝐴𝑒^{2𝑥}=24𝑒^{2𝑥}$
𝐴=−12 $𝑦=−12𝑥𝑒^{2𝑥}$
Let 𝑦=$B𝑒^{−2𝑥}$
𝑦′=$−2B𝑒^{−2𝑥}$
𝑦′′=$4B𝑒^{−2𝑥}$
$4𝐵𝑒^{−2𝑥}+12𝐵𝑒^{−2𝑥}+8𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$
$24𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$
𝐵=−1
∴$𝑦𝑝(𝑥)=−𝑒^{−2𝑥}$
so
$𝑦=𝑎𝑒^{2}+𝑐_{2}𝑒^{4𝑥}−12𝑒^{2𝑥}−𝑒^{−2𝑥}$
$𝑦′=2𝑐_{1}𝑒^{2𝑡}+4𝑐_{2}𝑒^{4𝑡}−24𝑟𝑒^{2𝑥}+2𝑒^{−2𝑥}−12𝑒^{2𝑥}$
𝑦(0)=𝑦′(0)=0
$𝑦=−3𝑒^{2𝑥}+4𝑒^{4𝑥}−12𝑥𝑒^{2𝑥}−𝑒^{−2𝑥}$
5
##### Quiz-4 / QUIZ4 TUT 0502
« on: October 19, 2019, 10:47:18 AM »
Question: 1 + [x/y - sin(y)]y’ = 0
We firstly simplify the equation into
dx + [x/y - sin(y)] dy= 0
Then we have M and N
M(x,y) = 1
N(x,y) = [x/y - sin(y)]
Then, we find the derivative of M with respect to y and N with respect to x
My = 0
Nx = 1/y
Since My is not equal to Nx, it is not exact.
Thus, we need to multiply a factor 𝓾 that satisfies the equation
R1 = [ (My - Nx)/ M] = [(0-1/y)] = -1/y
𝓾 = e-∫R1dy = e-∫(-1/y)dy = elny = y
We multiply the 𝓾 on both sides of the equation to find an exact equation
𝓾 dx + 𝓾[x/y - sin(y)] dy= 0
y dx + y [x/y - sin(y)] dy= 0
Then we have our new M’ and N’
M’(x,y) = y
N’(x,y) = y[x/y - sin(y)] = x - ysin(y)
Thus, there exist a function 𝒞(x,y) such that
𝒞x = M’
𝒞y = N’
By Integrating M’ with respect to x
𝒞x = M’
𝒞 = ∫ M’ dx = ∫ y dx = xy + h(y)
By differentiating with respect to y and equating to 𝒞y = N’
We get x + h’(y) = x - ysin(y)
Therefore, h’(y) = - ysin(y)
By integrating on both sides
h(y) =∫ - ysin(y) dy = ycos(y) - sin(y)
Now, we have
𝒞 = xy + ycos(y) - sin(y)
Thus, the solutions of differential equation are given implicitly by
xy + ycos(y) - sin(y) = C
6
##### Quiz-4 / QUIZ4 TUT 0502
« on: October 19, 2019, 01:23:47 AM »
Find the general solution of the differential equation
𝑦″+2𝑦′+2𝑦=0
The characteristic equation of the given equation is:
$𝑟^{2}$+2𝑟+2=0
𝑟=$\frac{−𝑏±\sqrt{𝑏^{2}−4𝑎𝑐}}{2𝑎}$=$\frac{−2±\sqrt{-4}}{2}$=−1±𝑖
Then,
$𝑟_{1}$=−1+𝑖 $𝑟_{2}$=−1−𝑖
Therefore, the general solution of the given differential equation is:
𝑦=$𝑐_{1}𝑒^{−𝑡𝑐𝑜𝑠𝑡}+𝑐_{2}𝑒^{−𝑡𝑠𝑖𝑛𝑡}$
7
##### Quiz-3 / QUIZ3 TUT 0502
« on: October 12, 2019, 12:20:07 AM »
Question: 𝑐𝑜𝑠(𝑡)𝑦″+𝑠𝑖𝑛(𝑡)𝑦′−𝑡𝑦=0
Find the Wronskian of two solutions of the given differential equation without solving the equation.
Solution:
Divide both sides by 𝑐𝑜𝑠(𝑡)
𝑦″+𝑡𝑎𝑛(𝑡)𝑦′−𝑡𝑐𝑜𝑠(𝑡)𝑦=0
𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑒−∫𝑝(𝑡)𝑑𝑡
𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑒−∫𝑡𝑎𝑛(𝑡)𝑑𝑡=𝑐𝑒−(−𝑙𝑛|𝑐𝑜𝑠(𝑡)|)
𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑒𝑙𝑛|𝑐𝑜𝑠(𝑡)|=𝑐𝑐𝑜𝑠(𝑡)
Therefore, the Wronskian of any pair of solutions of the given equation is 𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑐𝑜𝑠(𝑡)
8
##### Quiz-3 / QUIZ3 TUT 0502
« on: October 11, 2019, 08:06:25 PM »
Question:Find the general solution of the given differential equation.
𝑦″−2𝑦′−2𝑦=0
Solution:
𝑦=$𝑒^{𝑟𝑡}$
and it follows that r must be a root of characteristic equation
$𝑟^{2}$−2𝑟−2=0
𝑟=$\frac{−𝑏±\sqrt{𝑏^{2}−4𝑎𝑐}}{2𝑎}$
$𝑟_{1}$=1+\sqrt{3} $𝑟_{2}$=1−\sqrt{3}
Therefore, the general solution of the given differential equation is:
𝑦=$C_{1}𝑒^{(1+\sqrt{3})𝑡}+C_{2}𝑒^{(1−\sqrt{3})𝑡}$
9
##### Quiz-2 / QUIZ2 TUT 0204
« on: October 07, 2019, 09:26:17 AM »
Question: (𝑥+2)𝑠𝑖𝑛(𝑦)+𝑥𝑐𝑜𝑠(𝑦)𝑦′=0 𝑢=𝑥𝑒𝑥
Solution:
𝑀=(𝑥+2)𝑠𝑖𝑛(𝑦) 𝑁=𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑐𝑜𝑠(𝑦) 𝑁𝑥=𝑐𝑜𝑠(𝑦)
therefore 𝑀𝑦≠𝑁𝑥 , therefore the equation is not exact.
multiplies the given integrating factor
(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑦′=0
𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦) 𝑁=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑥𝑒𝑥𝑐𝑜𝑠(𝑦) 𝑁𝑥=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦) + 2𝑥𝑒𝑥𝑐𝑜𝑠(𝑦)
then 𝑀𝑦=𝑁𝑥, therefore it becomes exact.
𝜙(𝑥,𝑦) s.t 𝜙𝑥=𝑀 𝜙𝑦=𝑁
𝜙=∫𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑑𝑦
𝜙=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)
𝜙𝑥=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦) + 2𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ′(𝑥)=𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+0
ℎ′(𝑥)=0
ℎ(𝑥)=𝑐
𝜙(𝑥,𝑦)=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)=𝑐
10
##### Quiz-2 / QUIZ2 TUT 0502
« on: October 07, 2019, 12:12:42 AM »
Question: (3𝑥+6𝑦)+(𝑥2𝑦+3𝑦𝑥)𝑑𝑦𝑑𝑥=0
Solution: We want to find an integrating factor 𝜇 as a function of 𝑥𝑦 such that
(𝜇𝑀)𝑦=(𝜇𝑁)𝑥, Let 𝑧=𝑥𝑦. Thus, 𝜇(𝑥𝑦)=𝜇(𝑧(𝑥,𝑦)) Then
𝜇𝑥(𝑥𝑦)=𝑑𝜇𝑑𝑧∂𝑧∂𝑥=𝑦𝑑𝜇𝑑𝑧
𝜇𝑦(𝑥𝑦)=𝑑𝜇𝑑𝑧∂𝑧∂𝑦=𝑥𝑑𝜇𝑑𝑧
Therefore,
(𝜇𝑀)𝑦=(𝜇𝑁)𝑥
𝜇𝑀𝑦+𝑥𝑀𝑑𝜇𝑑𝑧=𝜇𝑁𝑥+𝑦𝑁𝑑𝜇𝑑𝑧
𝜇(𝑀𝑦−𝑁𝑥)=𝑑𝜇𝑑𝑧(𝑦𝑁−𝑥𝑀)
d𝜇d𝑧=𝜇(𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁)
Therefore,
𝜇(𝑧)=exp(∫𝑅(𝑧)d𝑧)
Then
𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁=2𝑥𝑦−3𝑦𝑥2+6𝑦2𝑥(3𝑥+6𝑦)−𝑦(𝑥2𝑦+3𝑦𝑥)
=2𝑥𝑦−3𝑦𝑥2+6𝑦22𝑥2+6𝑥𝑦−3𝑦2𝑥
=2𝑥𝑦−3𝑦𝑥2+6𝑦2𝑥𝑦(2𝑥𝑦−3𝑦𝑥2+6𝑦2)=1𝑥𝑦
Let 𝑥𝑦=𝑧
𝜇(𝑥𝑦)=exp(∫1𝑧d𝑧)=𝑒log|𝑧|=𝑧=𝑥𝑦
(3𝑥2𝑦+6𝑥)+(𝑥3+3𝑦2)𝑑𝑦𝑑𝑥=0
∂∂𝑦(3𝑥2𝑦+6𝑥)=3𝑥2=∂∂𝑥(𝑥3+3𝑦2)
𝜓𝑥(𝑥,𝑦)=3𝑥2𝑦+6𝑥(1)
𝜓𝑦(𝑥,𝑦)=𝑥3+3𝑦2(2)
𝜓(𝑥,𝑦)=𝑥3𝑦+3𝑥2+ℎ(𝑦)
𝜓𝑦(𝑥,𝑦)=𝑥3+ℎ′(𝑦)
Therefore,
ℎ′(𝑦)=3𝑦2
ℎ(𝑦)=𝑦3
𝜓(𝑥,𝑦)=𝑥3𝑦+3𝑥2+𝑦3
𝑥3𝑦+3𝑥2+𝑦3=𝐶
11
##### Quiz-1 / TUT 0502 Quiz1
« on: September 27, 2019, 11:19:51 PM »
The answer to my quiz question is attached.
Pages: [1] | 6,192 | 7,541 | {"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} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.316642 |
http://scienceblogs.com/dotphysics/2009/04/ | 1,539,751,763,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510998.39/warc/CC-MAIN-20181017044446-20181017065946-00070.warc.gz | 324,147,588 | 14,472 | # Let them have computers
Announcement: There is a ton of extra money for education of teachers. You should submit a grant proposal. Hey, let’s submit a proposal and get laptops for a whole bunch of students. To me, this is like giving morphine to people with the flu. It will make them feel better, but doesn’t really fix the…
# Super Fast Motorcycle
ZapperZ already beat me to this, but if it is worth doing once it is worth doing again. Here is a video of an electric motorcycle that can apparently go from 0 to 60 mph in 1 second. What acceleration would this have? If you want, you can look at my stuff on kinematics, but…
# The Cat in the Hat comes back and gets small
In the second Cat in the Hat book (I think it is the second one), the Cat reveals that he has more smaller cats under his hat. They are labeled A – Z with Z being so small you can’t even see. Question: What is the sequence of sizes for successive cats? How big would…
# When the centrifugal force is the centripetal force
It’s odd that I have talked about these forces so much. First, I talked about how centrifugal forces were not real and the difference between centrifugal and centripetal forces. Then I talked about how sometimes, fake forces are good. Finally, I talked about the origin of the words centrifugal and centripetal. (note: “talked about” means…
# Centripetal vs. Centrifugal (word origins)
A student in my office temporarily confused the words centripetal and centrifugal. This started me thinking about these two words. They mean different things, but do sound and look similar. I have previously talked about the difference between fake and non-fake forces, but let me quickly define these two: Centripetal: This is the force needed…
# Explain Physics – use diagrams and arrows
I have already indirectly talked about this before, but I see it more and more. Say you have a show. It has to be more than entertaining, it has to educational. How do you make things educational? Describe the physics behind the concepts – right? Fine. How do you explain stuff? It’s simple. Just make…
# Cool Quote from an old Scientific American
Some time ago I was looking for materials to possibly build a foucault pendulum. Of course the first step is google. There was a site that suggested two old issues of Scientific American, and it happens that we have tons of old Scientific Americans in the store room. I found the two that I needed.…
# Why are sparks blue?
I am not sure if First Excited State posted this as a blog entry, but it was mentioned on twitter. Question: why are sparks blue? My first gut response was that this is the blackbody color. Wrong for several reasons. The short answer is that sparks are blue because of the colors given off from…
# Gas Mileage and Car Accessories (from Car Talk)
First: Car Talk is awesome. I wish I could come up with some class activities that help students become as good at trouble shooting and critical thinking as Tom and Ray are. Anyway, they are quite entertaining. So, my Dad called and told me he heard a discussion on Car Talk about the effect DC… | 687 | 3,102 | {"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-2018-43 | latest | en | 0.960695 |
https://hilbertthm90.wordpress.com/2009/02/ | 1,526,934,381,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00166.warc.gz | 578,068,337 | 23,794 | # Wrapping up the Jacobson Radical
We now have the following equivalent definitions of the Jacobson radical. Remember right now we assume R is commutative with 1.
1) Intersection of all maximal ideals
2) Intersection of the annihilators of all simple left R-modules
3) The set of non-generators of R
4) The set of elements, x, such that $1-rx$ has a left inverse for all r.
I think I already pointed out that from at least two of these definitions we automatically get that $J(R)$ is a two-sided ideal. Two basic examples are now that if R is any field, then $J(R)=\{0\}$. And if K is a field, and $R=K[[x_1, \ldots x_n]]$, then $J(R)=\{f\in R : f \ has \ 0 \ constant \ term\}$. An important generalization is that in any local ring the Jacobson radical is the set of non-units.
An important result called Nakayama’s Lemma is that if $M$ is finitely generated, then $M=\Phi(M)+N$ implies that $M=N$. Special case: If $M= J(R)M+N$, then $M=N$. Corollary to that special case: If $M=J(R)M$, then $M=\{0\}$ (this last form is what is sometimes called Nakayama’s Lemma).
Proof: Suppose $M=\langle x_1+n_1, x_2+n_2, \ldots, x_m+n_m\rangle$. Where $x_j\in \Phi (M)$, and $n_j\in N$ for all j. Define $S=\{n_1, \ldots, n_m\}$.
Then with this setup, we exploit the non-generator definition. Note that
$M=\langle x_1, n_1, x_2, n_2, \ldots, x_m, n_m\rangle$
$= \langle S, x_1, \ldots, x_m\rangle$
$= \langle S, x_1, \ldots, x_{m-1}\rangle$
… etc
$= \langle S\rangle \subset N$.
And we are done! It may have seemed a little roundabout to go through the “Frattini submodule” in developing the Jacobson radical, but it certainly pays off to have lots of definitions as we see here.
The last little bit I wanted to say was that we can define the Jacobson radical for a ring without identity. I don’t want to go through the details, but a standard trick is to define a new ring (with identity) $S=\mathbb{Z}\times R$ with the standard addition, and then $(a,b)(c,d)=(ac, ad+cb+bd)$. It is pretty basic to check that $J(S)=\{0\}\times I$ where $I$ is some ideal in R (by the fact that $J(\mathbb{Z})=\{0\}$). It is also just algebraic manipulation to check that $I$ is the largest ideal in R such that for every $x\in I$ there is a $y\in I$ such that $x+y-yx=0$. This then is our definition. $J(R)=\cap_{\mathfrak{I}} I$ where $\mathfrak{I}$ is the collection of ideals satisfying that property.
# The Jacobson Radical Part II
First recall that we showed $J(R)=\Phi (R)$, and hence is a submodule of R as a module over itself. Thus $J(R)$ is a left ideal of R. Next recall that we showed $J(R)=J(R)R$, and hence is a right ideal. i.e. $J(R)$ is a two-sided ideal.
Let’s now work towards the annihilator definition. Define an equivalence relation the set of maximal ideals of R by $I$ ~$J$ if there is a simple left R-module M with elements $a,b\in M$ such that $I=ann_R(a)$ and $J=ann_R(b)$. We see that this is an equivalence relation, since $I$ ~ $J$ iff $R/I$ and $R/J$ are isomorphic as R-modules. Examine the module homomorphisms $r\mapsto ra$ and $r\mapsto rb$ to see that if $I$ ~ $J$ then $R/I\cong M \cong R/J$. Also, if $R/I\cong R/J$ by the iso $\varphi$, then $J=ann_R(\varphi^{-1}(1+J))$, so $I$ ~ $J$ since $\phi^{-1}(1+J), 1+J\in M$ with $I=ann_R(1+J)$ and $J=ann_R(\varphi^{-1}(1+J))$.
Now let $\mathfrak{I}$ be an equivalence class of maximal left ideals. I claim that $\cap_{I\in\mathfrak{I}} I=ann_R(M)$, where M is a simple left R-module isomorphic to each $R/I$, for $I\in\mathfrak{I}$. By definition and the property above we get that $\mathfrak{I}=\{ann_R(a): a\in M, \ a\neq 0\}$. Thus if $J\in\mathfrak{I}$, then $J=ann_R(1+J)$ which means that $J=ann_R(a)$ where $\varphi: R/J\to M$ satisfies $\varphi(1+J)=a$. But now this gives precisely $cap_{I\in \mathfrak{I}}I=ann(M)$.
Now just intersect over all the maximal left ideals. We get $\displaystyle J(R)=\cap_{J \ maximal} J=\cap_{\mathfrak{I}}\cap_{i\in \mathfrak{I}} I=\cap_{\mathfrak{I}}ann_R(R/I)=\cap_{M \ simple}ann_R(M)$. And voila, we have it. This was a rather terse run-through and assumed a working knowledge of some facts about modules, but I find it to be a rather fascinating take on the development.
Next we’ll exploit some of these definitions to get some properties of the Jacobson radical, and develop it in a method that doesn’t require our ring to have 1.
# An Approach to the Jacobson Radical
I’ve decided that in order to better understand the concepts in algebra this quarter, I should probably start posting several times a week on them. The quarter system is fast, and so we only have three weeks left.
I want to develop the Jacobson radical (this is really the next step in all that stuff I was presenting before anyway). So at first we’ll assume our ring R has 1 (which was the case before). The typical development seems to define $\displaystyle J(R)=\bigcap_{M \ simple \ R-module}ann_R(M)$. The Jacobson radical is the intersection of the annihilators of all simple left R-modules (note that the $ann_R(M)=\{r\in R : rm=0 \ \forall m\in M\}$.
There are many, many alternative formulations, but I want to develop this from one of the more obscure angles.
First, let $\Phi(M)$ be the “Frattini submodule” i.e. the set of nongenerators. $x$ is a nongenerator if we have any subset of M, say S, and $R\langle S, x\rangle= M$, then $R \langle S\rangle = M$. So a nongenerator means that if you ever have a set which generates your module, then that set without the nongenerator will still generate it.
Step 1: $\Phi(M)=\bigcap_{N.
Suppose $x\notin \Phi(M)$. Claim: there is a maximal submodule N such that $x\notin N$. Let S be such that $M=R\langle S, x \rangle$ and $M\neq R\langle S\rangle$. Then since x is not a nongenerator we can choose S so that $x\notin R\langle S\rangle$. Now $R\langle S\rangle \in P=\{H : H. We have a non-empty partially ordered set, and unioning gives an upper bound to any chain. Thus we apply Zorn’s Lemma to get a maximal element, N. Thus $\displaystyle \Phi(M)\supset\bigcap_{N.
For the reverse, just note that if x is not a member of the right side, then there is a particular maximal submodule, N, such that $x\notin N$. Thus $M=R\langle N, x\rangle$, but $R\langle N\rangle=N$. So $x\notin \Phi(M)$. And we get equality $\Phi(M)=\bigcap_{N. So the Frattini submodule equals the intersection of all maximal left R-modules.
Step 2: Define the Jacobson radical now to be $J(R)=\{x\in R: 1-rx \ has \ left \ inverse \ \forall r\in R\}$. Claim: $\Phi(M)\subset J(R)$.
Suppose $x\notin J(R)$. Then there is some $r\in R$ such that $1-rx$ has no left inverse. i.e. $1\notin R(1-rx)\subset N$ where N is a maximal left ideal. But $1-rx\in N$ since $1\in R$. Thus $x\not in N$ because otherwise would mean that $1=(1-rx)+rx\in N$ a contradiction. So $x\notin \Phi(R)$. Note the subtle shift in usage here. We are considering R to be an R-module over itself.
Step 3: $J(R)M\subset \Phi(M)$.
Suppose $x\in J(R)$ and $m\in M$. Claim: $xm\in \Phi(M)$. Suppose that $M=R\langle S, xm\rangle$ and $m\in M$. Then we have that $m=\sum (r_js_j) +rxm$. i.e. $m-rxm=\sum r_js_j\Rightarrow (1-rx)m=\sum r_js_j$.
Suppose $b(1-rx)=1$ (since $x\in J(R)$). Now $m=b(1-rx)m=\sum br_js_j\in R\langle S\rangle$. Thus $xm\in R\langle S\rangle\Rightarrow R\langle S\rangle=R\langle S, xm\rangle = M$. And we are done.
Now notice from the above steps that $\Phi(R)\subset J(R)=J(R)\cdot \{1\}\subset J(R)\cdot R\subset \Phi(R)$. And so we have equality all the way through. Namely, $\Phi(R)=J(R)$. So the set of all nongenerators, the intersection of all maximal submodules, and the set of elements such that $1-rx$ has a left inverse for all $r\in R$ are all the same (namely, the Jacobson radical).
Next time we will bank on these to show that these are all equivalent to the first listed and more common definition, the intersection the annihilators of all simple left R-modules.
(I claim no responsibility for the invention of this approach. This is the way my professor sees the world).
# Lost in the Funhouse
I’m in the position where I’m too tired to start thinking about a new algebra problem, but it is too early to legitimately call it a night, so I’ll just shift gears and do a new post. It has been one of those days where about everything that can go wrong does. So how about thinking about a little humor.
A very important work of literature in the “post-modern canon” is John Barth’s Lost in the Funhouse. I was first led to it by the last story in DFW’s collection Girl With Curious Hair. He does a sort of parody/hommage to it. Anyone that has taken a 20th century American lit course has probably had to read something by Barth, and it was most likely the title story in this collection.
Barth is known for his excessive meta-fictional devices and influence on writers mentioned previously like Pynchon, Wallace, and probably any serious post-modernist. Despite the term “excessive meta-fictional devices,” I find him quite easy and fun to read. The devices serve a purpose and are usually humorous. Unlike some post-modernists that came after him, Barth is very much concerned with art expressing a human experience (mostly love). Although DFW ultimately rejected Barth, he very much agrees on this point…but we’ll get to that later.
I think I’ll mostly do this as an analysis of the title story. Lost in the Funhouse is a short story “about” a boy’s (Ambrose) trip to Ocean City where he enters a funhouse, and yes gets lost. He enters with a girl Magda, but she continues on with his brother while he is left alone. But really it is not about this at all. The funhouse serves as a metaphor for Ambrose’s first sexual experience. In my reading, I actually don’t believe the trip happens at all, and the whole entire trip is a metaphor.
In the typical Barth fashion, the funhouse is a multi-layered metaphor. A funhouse has mirrors all around. This means that Ambrose must see himself reflected in all shapes and sizes. This represents his fractured subconscious about the experience. His own head is also in the way of ever directly seeing the image in the mirror behind him. This aspect of the metaphor is actually extensively rejected by Wallace. Wallace interprets that aspect as Barth’s way of saying that literature can never directly make it to the reader. It always will hit the reader’s head first and be obscured and never directly viewed. “…that the necessity for an observer makes perfect observation impossible, …” Wallace changes the metaphor and says it is like a bow and arrow. Your arm will always be in the way of shooting directly, but the writer can take this into account and directly hit the reader.
This is all boring, though. Let’s get to the truly interesting aspects of the story. Barth as a narrator sometimes narrates, sometimes talks directly to the reader, and sometimes comments on the narration. It is these comments that are the humorous meta-fictional devices. The story becomes self-aware. It understands and points out the devices it is using. Here is one of my favorite devices:
En route to Ocean City he sat in the back seat of the family car with his brother Peter, age fifteen, and Magda G____, age fourteen, a pretty girl and exquisite young lady, who lived not far from them on B_____ Street in the town of D____, Maryland. Initials, blanks, or both were often substituted for proper names in nineteenth-century fiction to enhance the illusion of reality. It is as if the author felt it necessary to delete the names for reasons of tact or legal liability. Interestingly, as with other aspects of realism it is an illusion that is being enhanced, by purely artificial means.
The story is continually interrupted to go off on tangents like this. He wants to point out, explain, and make fun of the traditional devices he is using. In doing this he is actually creating new and original devices. He doesn’t want the reader to become absorbed in the story and think that it is real for the duration. He wants the reader to be painfully aware that they are reading a story.
Another aspect of the verbal trickery of the story is to somehow assert the primacy of language to experience. All experience must be filtered through language. Thus, instead of ever explicitly describing Ambrose’s experience, we only live on the verbal thoughts flowing through his head throughout the experience. In fact, in searching for a certain quote just now, I came across another that reinforces my reading that the entire story is a metaphor.
With incredible nerve and to everyone’s surprise he invited Magda, quietly and politely, to go through the funhouse with him. ‘I warn you, I’ve never been through it before,’ he added, laughing easily, ‘but I reckon we can manage somehow. The important thing to remember, after all, is that it’s meant to be a funhouse; that is, a place of amusement. If people really got lost or injured or too badly frightened in it, the owner’d go out of business.
Or even the famous opening lines, “For whom is the funhouse fun? Perhaps for lovers. For Ambrose it is a place of fear and confusion.” Don’t read that as “funhouse.” Ambrose is really talking about the fact that it is his first sexual experience. He is trying to convince Magda that it can’t be too scary painful, since people continue to have sex. The funhouse is for lovers? It is scary and confusing for Ambrose? Come on, of course this is what it is talking about.
Alright. Let’s get back from that tangent to Ambrose’s head. He starts telling all of these scenarios of how his being lost gets played out. In one he actually dies. “This can’t go on much longer; it can go on forever. He died telling stories to himself in the dark; years later, when that vast unsuspected area of the funhouse came to light, the first expedition found his skeleton in one of its labyrinthine corridors and mistook it for part of the entertainment. He died of starvation telling himself stories in the dark;…” This all emphasizes the main effect Barth is striving for. All human experience is mediated by language. Language is so primary and important that a mind preoccupied by other stories could completely miss the experience itself.
It seems I’ve gone on longer than I should have, but I feel like I haven’t done the story the slightest bit of justice. It is so great and packed full of interesting things. And this is just one of many stories in the book. I highly recommend this to anyone who aspires to understand modern literature.
# Andrew Bird/Duncan Sheik Joint Post
This amazing and awful thing happened this year. The two people that I feel a strong need to buy an album of on the day it releases both released their new albums at the end of January. These two people are Duncan Sheik and Andrew Bird.
Let’s start with the Bird. I am firmly convinced that this is his best album yet. This is a bit of a relief, since I felt that his previous one took a step down. I thought that his previous one fell massively in lyric creativity, and was much more mainstream musically. Although I was very happy in the direction he was heading with overall sound. That’s why I always recommend Mysterious Production of Eggs for someone interested in getting to know his music.
This album did the perfect combination. It kept going in the interesting sound that I liked from the last one, but brought back the lyrical inventiveness and quirkiness of his earlier stuff. In fact, I might argue that this album places Bird as the nation’s greatest creator of word play and association. Even the New York school of poets didn’t invent stuff this great. Here is something from “Anonanimal”:
I see a sea anemone
The enemy
See a sea anemone
And that’ll be the end of me.
Underneath the stalactites
The troglobites lost their sight
Sometimes the associations are much deeper than just in sound, though. I tend to classify Bird associations in the categories: juxtaposing two words that sound similar and have different meaning, juxtaposing two words that have similar meaning but don’t sound the same, or the most subtle of all, juxtaposing one word that conjures the other through societal associations. It is utterly amazing to hear this constant flow of associations that sometimes linger on an idea or flow through ideas that are completely unrelated except through sound. I tend to walk around as happy as can be.
Duncan Sheik brings a different side to things. If Bird is uplifting, nothing can be a bigger downer than this new Sheik album. Let’s just give two examples:
Lily keeps the lighthouse
She’s afraid of the unknown
She’s no ray of sunshine
So mostly she’s alone
No one cares about her longing
Or the dreams on which she’s fed
And though I’m sad to say it
etc…as it runs through many scenarios in which people would be better off dead. The harshness doesn’t end there, though:
Life is like a sinking ship
And you are at the wheel
See the hulls filling up
You know your fate is sealed
Still you keep on trying
To steer the ship to shore
It’s time to learn a lesson
You should’ve learned before
There’s nothing you can do
And we don’t believe in you
I must admit that at first I was so engulfed by this album that I couldn’t even bring myself to listen the Bird for the first time. I was in some sort of shock that someone could write music like this. It moved and disturbed me. In the long run this album may end up getting a thumbs down, though. It is technically not a “Duncan Sheik” album, but him singing songs he wrote for some sort of musical or something.
I do like the on going narrative of the album, but the lyrics and music get a bit too music theater cliché at points. The album before this one was the cast recording of his musical Spring Awakening. The one before that wasn’t all that impressive to me either. That being said, it is without a doubt worth spending time with. I just have a feeling that it will get old fast and won’t be extremely friendly to re-listens. For the person unfamiliar with Sheik I’d start with either Humming (the most original and creative of his works in my opinion) or Daylight (the most polished and probably most listener friendly). Proceed from there as you wish. | 4,701 | 18,192 | {"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": 114, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.84311 |
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# Compass Trigonometry Placement test ACT community college
Xuất bản 29/06/2016
For more cool math videos visit my site at http://mathgotserved.com or http://youtube.com/mathsgotserved For more cool math videos visit my site at http://mathgotserved.com or http://youtube.com/mathsgotserved COMPASS Mathematics Tests The COMPASS Mathematics Tests are organized around five principal content domains: numerical skills/prealgebra, algebra, college algebra, geometry, and trigonometry. To ensure variety in the content and complexity of items within each domain, COMPASS includes mathematics items of three general levels of cognitive complexity: basic skills, application, and analysis. A basic skills item can be solved by performing a sequence of basic operations. An application item involves applying sequences of basic operations to novel settings or in complex ways. An analysis item requires students to demonstrate a conceptual understanding of the principles and relationships relevant to particular mathematical operations. Items in each of the content domains sample extensively from these three cognitive levels. Students are permitted to use calculators on all current Windows® and Internet versions of COMPASS Mathematics Tests. Calculators must, however, meet ACT's specifications, which are the same for COMPASS and the ACT Assessment. These specifications are updated periodically and can be found at ACT's website at http://www.act.org/aap/taking/calculator.html Geometry Placement Test Primary content areas included in the Geometry Placement Test include: Triangles (perimeter, area, Pythagorean theorem, etc.) Circles (perimeter, area, arcs, etc.) Angles (supplementary, complementary, adjacent, vertical, etc.) Rectangles (perimeter, area, etc.) Three-dimensional concepts Hybrid (composite) shapes Source ACT.org Trigonometry Placement Test The primary content areas assessed by the Trigonometry Placement Test include: Trigonometric functions and identities Right-triangle trigonometry Trigonometric equations and inequalities Graphs of trigonometric functions Special angles (multiples of 30 and 45 degrees) | 434 | 2,205 | {"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 | 0.869709 |
https://discusstest.codechef.com/t/copr16g-editorial/12567 | 1,627,536,002,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00524.warc.gz | 226,114,971 | 4,280 | # COPR16G: Editorial
Practice
Contest
Author: Bhuvnesh Jain
Tester: Bhuvnesh Jain
Editorialist: Bhuvnesh Jain
EASY
### PREREQUISITES:
Diophantine Equations, GCD
### PROBLEM:
You are provided with coins of denominations a and b.
You are required to find the least value of n, such that all currency values greater than or equal to n can be made using any number of coins of denomination a and b and in any order. If no such number exists, print -1 in that case.
### EXPLANATION:
Let us first write the statement as mathematical equations and then procede furthur.
ax + by = n, where x and y are integers.
and au + bv = n+1, where u and v are integers.
The above equations hold for all integers >= n, if solution exists. Let us subtract the 2 equations. We get,
ar + bs = 1, where r and s are integers.
Hence we can easily see that solutions exists iff gcd(a, b)==1. (Using the condition for solvability of linear Diophantine equation).
Now we need to find the value of least possible n. We can also see the above problem as finding the largest interger which can be expressed as ax + by for some integers, x and y. A simple google search could have helped you, hence the name “GET AC IN ONE GO”. The answer is simply (a.b - a - b). You can see a detailed proof here.
Hence, the solution is (a.b - a - b + 1), when gcd(a, b) = 1 else -1.
### COMPLEXITY
O(log(max(a, b))) per test case
### AUTHOR’S SOLUTION:
Author’s solution can be found here.
1 Like
// | 392 | 1,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-31 | latest | en | 0.861896 |
https://nz.education.com/resources/multiplying-fractions/ | 1,611,314,105,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00157.warc.gz | 491,024,160 | 30,176 | # Multiplying Fractions Resources
64 filtered results
64 filtered results
Multiplying Fractions
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How to Multiply Fractions
Worksheet
How to Multiply Fractions
Use this helpful guide to learn and practice multiplying fractions.
Math
Worksheet
Fun with Fractions
Workbook
Fun with Fractions
Fractions are a fundamental math skill for 5th graders to master! In this series your child will review adding mixed fractions, and converting improper fractions to mixed ones.
Math
Workbook
Fractions 4
Guided Lesson
Fractions 4
Students are going to take a deeper dive into fractions in this unit! Learners will apply previous understanding of finding equivalent fractions, and converting between fractions and mixed numbers to work with fractions in more complex ways. Students will continue to use visual models to learn and practice adding, subtracting, multiplying and dividing fractions.
Math
Guided Lesson
Fifth Grade Fall Review Packet - Week 2
Workbook
Fifth Grade Fall Review Packet - Week 2
Prepare learners for their fifth grade debut with Week 2 of our Fifth Grade Fall Review Packet, complete with five more days of engaging activities that will review key skills and concepts.
Workbook
Find the Missing Numerator or Denominator
Worksheet
Find the Missing Numerator or Denominator
On this worksheet, your child will fill in the blanks to find the missing numerator or denominator in each fraction multiplication problem.
Math
Worksheet
Multiplying Fractions Word Problems #1
Worksheet
Multiplying Fractions Word Problems #1
Give your students the chance to interpret, set up, and solve problems involving multiplying fractions.
Math
Worksheet
Multiply Mixed Numbers with Whole Numbers
Worksheet
Multiply Mixed Numbers with Whole Numbers
Multiplying mixed numbers and whole numbers: it looks rather complicated, but it's actually a matter of learning just a few steps.
Math
Worksheet
Multiplying Fractions Word Problems #2
Worksheet
Multiplying Fractions Word Problems #2
Give your students the chance to interpret, set up, and solve problems involving multiplying fractions.
Math
Worksheet
Restaurant Math
Worksheet
Restaurant Math
Hungry for math practice? Get your fill of fractions with this restaurant-themed worksheet.
Math
Worksheet
Percentage Practice: Winter Sale
Worksheet
Percentage Practice: Winter Sale
Learners practice calculating the percentage of two- and three-digit numbers in this winter clothing–themed worksheet.
Math
Worksheet
Recipe Conversion
Worksheet
Recipe Conversion
Introduce your little math champ to fraction multiplication in the kitchen! He'll get a taste of measurements using a recipe to calculate the fractions.
Math
Worksheet
Liquid and Linear Measurement
Workbook
Liquid and Linear Measurement
Count and convert measurements for liquids, lengths and time with this supplemental book! So many practical skills come from knowing how to measure lengths, tell time and convert units of measurement.
Math
Workbook
Order of Operations with Fractions and Decimals
Lesson Plan
Order of Operations with Fractions and Decimals
This lesson is designed for fourth graders to practice their skills with the order of operations along with multiplying fractions and adding or subtracting decimals.
Math
Lesson Plan
Multiplying Fractions by Whole Numbers
Exercise
Multiplying Fractions by Whole Numbers
Multiplying Fractions by Whole Numbers will help students practice this key fifth grade skill. Try our free exercises to build knowledge and confidence.
Math
Exercise
Halloween Patterns
Worksheet
Halloween Patterns
Are you up for a Halloween math challenge? Take a stab at these spooky math problems as you try to complete the number pattern.
Math
Worksheet
Order of Operations: Fractions
Worksheet
Order of Operations: Fractions
Challenge your students to solve expressions with fractions using the order of operations.
Math
Worksheet
Multiplying Mixed Number Fractions and Arrays
Exercise
Multiplying Mixed Number Fractions and Arrays
Multiplying Mixed Number Fractions and Arrays will help students practice this key fifth grade skill. Try our free exercises to build knowledge and confidence.
Math
Exercise
Multiplying Fractions Word Problems #3
Worksheet
Multiplying Fractions Word Problems #3
Give your students the chance to interpret, set up, and solve problems involving multiplying fractions.
Math
Worksheet
Dry Measure Conversion
Worksheet
Dry Measure Conversion
Give recipe conversion practice a try by multiplying the ingredients in this cupcake recipe. Liquid and dry measure conversion are included; cups,…
Math
Worksheet
Math Review Part 4: Phenomenal Fractions
Worksheet
Math Review Part 4: Phenomenal Fractions
Math
Worksheet
Scaling Practice
Worksheet
Scaling Practice
Give your students some practice scaling numbers when multiplying a whole number and a fraction.
Math
Worksheet
Recipe Fractions #2
Worksheet
Recipe Fractions #2
Use food and math together to learn fractions! Help your child multiply fractions to convert these recipes to smaller or larger portions.
Math
Worksheet
Multiplication of Fractions and Scaling
Exercise
Multiplication of Fractions and Scaling
Students who complete this exercise will understand how to scale up their newfound ability in multiplying fractions.
Math
Exercise
How to Multiply Mixed Numbers by Mixed Numbers
Worksheet
How to Multiply Mixed Numbers by Mixed Numbers
This valuable worksheet teaches your little mathematician how to convert mixed numbers to fractions, and then multiply them.
Math
Worksheet
Multiply Mixed Numbers with Fractions
Worksheet
Multiply Mixed Numbers with Fractions
This colorful worksheet shows your child how to multiply a mixed number with a fraction, and gives him practice with the method.
Math
Worksheet
### Multiplying Fractions Resources
Multiplying fractions is a great starting place if your child is just starting to learn how to do arithmetic with fractions. This skill is usually taught starting in fourth grade, so if your student has already learned about the different types of fractions, dive into our resources to help them master multiplying fractions before they move on to more complex concepts.
Multiplying fractions is one of the easier tasks to complete because there are not as many rules. Below is a step by step guide in how to multiply proper, improper, and mixed fractions.
General Rules
If both fractions have a numerator and a denominator, the rules are very simple:
1. Multiply the numerators (top numbers)
2. Multiply the denominators (bottom numbers)
3. Simplify if needed
For a refresher on how to multiply numbers, check out our Multiplication page.
Fractions and Whole Numbers
If you are trying to multiply a fraction by a whole number, convert the whole number into a fraction by giving it a denominator of 1 (3 would become 31 ), and then follow the general rules of multiplying fractions.
Mixed Fractions
In order to multiply mixed fractions, you will have to know how to convert back and forth between a mixed fraction and improper fraction. If your child needs help with this concept, take a look at Fractions Basics at the bottom of the Fractions page and work through some of the worksheets.
1. Convert the mixed fraction into an improper fraction
2. Multiply the fractions by using the general rules
3. Convert the product back to a mixed fraction
Use these rules to work through the various worksheets and exercises on this page until your child is a professional fraction multiplier! | 1,475 | 7,466 | {"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} | 4.03125 | 4 | CC-MAIN-2021-04 | latest | en | 0.869832 |
http://www.science.ca/askascientist/viewquestion.php?qID=2867 | 1,527,038,592,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865023.41/warc/CC-MAIN-20180523004548-20180523024548-00388.warc.gz | 459,822,690 | 7,640 | ## Physics Question #2867
George Chronis, a 34 year old male from Toronto asks on August 4, 2005,
I have a few questions regarding Space-Time that have always confused me. I would very much appreciate if you could provide some insight: 1.Expanding Universe It has been discovered that the universe is expanding at an accelerating rate. This conclusion was based by studying the light of a special class of supernovae in different galaxies. Is it possble that we could come to a different conclusion if we were in another region of space-time? For instance, would not someone situated in/near a black hole see the universe evolving at a fantastic rate? Therefore, is it not possible that the observation that the universe is expanding at an accelerating rate is biased by our local and our motion in/near an intense gravitational field? 2.Cosmos Cartography The universe has been observed as matter existing in "thin filaments" with "large pockets" of empty space. Is this an acurate picture of the universe "now" (with respect to the observer)? When creating such a cartography of the cosmos, was the correlation between distance and time taken into account?; namely, the more distant an object the further back in time the object is observed. For instance, suppose someone on earth was to create a map of the surface of the earth. Furthermore, suppose he travels 1million years back in time per 1000km of radius travelled (from an arbitrary starting point). By the time he has traveled a distance equal to the earth's radius, he would have travelled over six million years into the past. If he were to create a map of the earth based on what he observed at every circle, his map would be one of geography and history (at the "equator" the solar system had not formed, at the midpoint life would just be starting on earth). Therefore, would this map represent the earth the way it is "now"? Similarly, is the "map" of the universe as based on "old data" (light seen from the night sky) the "true picture" of the universe "now". 3.Galaxy Dynamics It has been observed that the outer portions of a galaxy spin faster than inner core of a galaxy. Several theories have been proposed: a ring of dark matter circling the galaxy, modifying the 1/R2 term in Newton's gravitation equation. Could the observed slower spin at the galaxy core be explained by the General Theory of Relativity? The intense gravitational field at the galaxy core would make events at the core appear, to observers further away, slower. 4.The Big Bang A few moments after the big bang, is some times described in spatial terms. For example, at 10^-47 seconds after the big bang, the size of the universe was smaller than an atom. If at 10^-47 seconds after the big bang all of space and time existed at that instant including all matter and energy, can one asign a size at that point in time drawing from the state of the universe as it is today (i.e. what exactly does "smaller than an atom" mean when the universe is just 10^-47 seconds old?) ? Furthermore, can one talk of time near the big bang? By asigning a time, do you not in essence establish absolute time (i.e. time zero becomes an absolute time reference) in contradiction to Theory of Relativity.
viewed 13838 times
### Doug Hube answered on August 17, 2005
1.Expanding Universe
It is true that the strength of a gravitational field affects measurements of space and time. However, except at discrete points occupied by black holes, and perhaps other extreme objects if one is very close to a surface, our location is in a 'weak field situation' as is most of the Universe. We are not in a region where the suggested biases could play a role.
2.Cosmos Cartography
The 'filamentary' and 'bubbly' structure of normal matter has been detected within the relatively 'small' distance of a few billion light years. The scale of the filaments and bubbles/voids is such that, the relative motions of the galaxies which define those structures are too small to have significantly altered the structure during the equivalent few billion years.
3.Galaxy Dynamics
Yes, most spiral galaxies that have been studied appropriately do not follow a Keplerian velocity distribution in their outer regions. That is, rather than the rotational velocity decreasing with increasing distance from the galaxy's centre (as the orbital speeds of planets decrease with increasing distance from the Sun) the velocities either remain roughly constant or, in some cases, increase with distance from the centre of the galaxy. The usual explaination is that there is a large halo of dark matter.
There do seem to be massive black holes at the centres of most galaxies, but they are very much smaller than what we call the 'core'. Relativistic effects will be signifivant very close to those nuclear black holes but very insignificant throughout the much greater volume of a typical core. The cores of galaxies constitute 'weak field' regions.
4.The Big Bang
Yes, the Big Bang event does define a zero point in time ... cosmic time.
Special Relativity says that a clock in one inertial reference frame runs 'slowly' as measured with respect to a clock in another inertial reference frame. A full discussion leads one into a consideration of the meaning of 'simulaneity', etc.
We are not observing the 'cosmic clock' from another reference frame, i.e. from outside the Universe. The 'clocks' and the measurement of their rates are fundamentally different in those two situations. We are integral components of the cosmic structure that is evolving in step with the 'ticks' on the cosmic clock which began to 'tick' 13**9 years ago, just as my 'ticking heart clock' began at a well-defined time as measured by clocks in my local not-relativistically-affected corner of the Universe.
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If you found this answer useful, please consider making a small donation to science.ca. | 1,271 | 5,998 | {"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-2018-22 | latest | en | 0.968839 |
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Jul 6, 2013
In my previous post, I wrote about Adding two numbers in C. Let's learn how to subtract two numbers in C this time. It is…
Jul 6, 2013 | 418 | 1,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.945438 |
https://www.physicsforums.com/threads/messy-fourier-transform-integral.360449/ | 1,722,906,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00211.warc.gz | 728,790,728 | 18,488 | # Messy Fourier Transform Integral
• TimH
In summary, the Fourier transform of a function is given by g(a)=-sin(a)/(pi*a), where g(a) is the Fourier representation of f(x). The Fourier representation can be found by integrating over the function from - to + infinity, using the formula g(a)*e^(iax). The second term in the integral is a function of a and x, but is dropped if it's an odd function.
TimH
I'm teaching myself some basic Fourier analysis from Boas's "Mathematical Methods in the Physical Sciences." I'm a little stuck on Example Problem 1 on p.382. This is a basic example of getting the g(a) Fourier transform of a certain function (pictured on the page) and then plugging it into the formula to get the FT of f(x). The problem has two typos in it according to the online errata, which I've taken into account.
Anyway, you get the g(a) function which is g(a)=sin(a)/(pi*a) (I'm using "a" for the alpha that appears in the book). I understand this.
Then you want to get your Fourier representation of f(x) so you plug g(a) into the integral from - to + infinity of g(a)*e^(iax) da. This I understand.
Now for this integral she expands e^(iax) using Euler to get cos(ax)+i*sin(ax). If you multiply this by g(a), i.e. by sin(a)/(pi*a), you get two terms in your integral, i.e. the integral is of (sin(a)(cos ax))/(pi*a) + (sin(a)(i*sin(ax))/(pi*a) da.
But she does not show these two terms of the multiplication. Rather, she says that the integral is equal to two times the integral of the first term, because the function "sin(a)/a is an even function." So the complex second term is dropped. This I don't understand. The second term is a function of a and x, times i. I can't really tell if its even or odd. If it's odd, I can see dropping it from the integral since the integral is symmetric around the origin so the integral of the term will be = 0. But the fact that its got an "i" in it is throwing me, too.
Could somebody please help me with this second term in the integral? Thanks.
A function of two variables has two parity properties. For example y*cos(x) is even wrt x but odd wrt y. So when you integrate over the function wrt to only one variable, it's the parity wrt to that variable which counts.
For example sin(a*x) is odd wrt a (even wrt x) so integrating over a symmetric interval wrt a will result in zero.
Okay I think I get it. The integral is of i*(sin(a)*sin(ax))/a. We're integrating wrt a. Now sin(a)/a is even. But sin(ax) is odd if we're integrating wrt a. So its an even function times an odd function which gives an odd, and so its integral is zero over the symmetric interval. And so the "i" nicely goes away.
Thanks!
## What is a Messy Fourier Transform Integral?
A Messy Fourier Transform Integral is a mathematical tool used to analyze data in the frequency domain. It is a complex integral that involves the Fourier transform, which transforms a signal from the time domain to the frequency domain.
## How is a Messy Fourier Transform Integral different from a regular Fourier Transform?
A Messy Fourier Transform Integral is more complex and involves additional mathematical operations, such as integration. It is typically used to analyze signals that are not well-behaved or have noise, while a regular Fourier Transform is used for more simple and well-behaved signals.
## What is the purpose of using a Messy Fourier Transform Integral?
The purpose of using a Messy Fourier Transform Integral is to analyze signals in the frequency domain, which can provide insight into the underlying patterns and structures of the data. It is particularly useful for analyzing signals that are not smooth or have noise, as it can help identify important features and filter out noise.
## What types of signals are suitable for a Messy Fourier Transform Integral?
A Messy Fourier Transform Integral is suitable for signals that are not well-behaved or have noise, such as signals with sudden changes or spikes, non-periodic signals, and signals with missing data points. It is also useful for analyzing signals with non-stationary behavior, meaning the signal changes over time.
## How is a Messy Fourier Transform Integral used in scientific research?
A Messy Fourier Transform Integral is commonly used in fields such as physics, engineering, and biology to analyze data and extract important information in the frequency domain. It is also used in signal processing to filter out noise and improve the quality of signals. In scientific research, it can provide valuable insights into the underlying patterns and structures of data, allowing for a deeper understanding of the phenomena being studied.
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10K | 1,194 | 5,035 | {"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} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.920506 |
http://wikivirgil.wikidot.com/2346:online-resources | 1,550,273,221,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479627.17/warc/CC-MAIN-20190215224408-20190216010408-00221.warc.gz | 299,547,905 | 6,068 | Online Resources
List here online resources (websites and videos) that you find.
## Octave
The windows binary (executable) for Octave may be difficult to find so here's the URL: [http://octave.sourceforge.net/]
Install the front end by extracting files to its own directory and navigate to bin > qtoctave.exe
While opening, there will be an error, but go up to Config menu > General Configuration; click on Octave and find the '.exe' for Octave
Another front end is GUI Octave. It might actually be more user friendly. You can find it here: [http://sites.google.com/site/guioctave/]
- Justin K
## Matlab Tutorials
The following site has an excellent tutorial that covers everything from simple functions and variables to linear equations and beyond. [http://www.isr.umd.edu/~austin/ence202.d/matlab-tutorial.html]
- Justin K
Getting started with Matlab. Part1,2,3,4,5 and 6 i think these tutorials are very straight forward
[http://www.math.siu.edu/matlab/tutorial1.pdf]
[http://www.math.siu.edu/matlab/tutorial2.pdf]
[http://www.math.siu.edu/matlab/tutorial3.pdf]
[http://www.math.siu.edu/matlab/tutorial4.pdf]
[http://www.math.siu.edu/matlab/tutorial5.pdf]
[http://www.math.siu.edu/matlab/tutorial6.pdf]
Atte: Abel Barraza
## Linear Algebra Lectures
Lecture on Elimination:
[http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-2-elimination-with-matrices]
Lecture on Independence, Basis, and Dimension:
[http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-9-independence-basis-and-dimension]
Lecture on Properties of Determinants:
[http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-18-properties-of-determinants]
Lecture on Eigenvalues and Eigenvectors:
[http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-21-eigenvalues-and-eigenvectors]
page revision: 8, last edited: 13 May 2012 02:27 | 531 | 1,975 | {"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-2019-09 | longest | en | 0.674678 |
http://mathhelpforum.com/algebra/809-sin-cos-print.html | 1,527,008,414,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864798.12/warc/CC-MAIN-20180522151159-20180522171159-00431.warc.gz | 178,264,615 | 3,032 | # sin and cos
• Aug 28th 2005, 10:11 PM
yesdnil
sin and cos
i can't figure out how to do this problem....
-4+11 sin x = 6 cos^2 x
solve for x....
• Aug 28th 2005, 10:20 PM
ticbol
What do you want to do? Rearrange the equation? Solve for x? What?
• Aug 29th 2005, 12:02 AM
ticbol
So it is solve for x.
-4+11 sin x = 6 cos^2 x
-4 +11sinX = 6cos^2(X) ------(i)
There are two variables in (i). But to the beauty of Trigonometry, by trig identities, we can express one in terms of the other.
It is easier to express cosX in terms of sinX in this case.
Since sin^2(X) +cos^2(X) = 1, then,
cos^2(X) = 1 -sin^2(X)
Substitute that into (i),
-4 +11sinX = 6[1 -sin^2(X)]
-4 +11sinX = 6 -6sin^2(X)]
6sin^2(X) +11sinX -4 -6 = 0
6sin^2(X) +11sinX -10 = 0 ----------(ii)
That (ii) is a quadratic equation in sinX.
Solve for sinX using the Quadratic Formula.
Or factor it if it can be factored---and you know how to
factor it.
Factoring (ii),
(3sinX -2)(2sinX +5) = 0
3sinX -2 = 0
3sinX = 2
sinX = 2/3 = 0.6666667 ------positive.
X = arcsin(0.666667) = 41.81 degrees, in the 1st quadrant
or X = (180 -41.81) = 138.19 deg, in the 2nd quadrant.
That is, sine is positive in the 1st and 2nd quadrants.
Therefore, X = 41.81 or 138.18 deg ------answer.
2sinX +5 = 0
2sinX = -5
sinX = -5/2 = -2.5 ------cannot be because minimum value of sine of any angle is -1.0 only.
--------------
Another way, by using the Quadratic Formula in (ii).
sinX = {-11 +,-sqrt[(11)^2 -4(6)(-10)]} / (2*6)
sinX = {-11 +,-sqrt[361]} / (12)
sinX = {-11 +,-19} / 12
sinX = {-11 +19}/12 = 8/12 = 2/3 = 0.666667
etc....(see above for continuation)....
sinX = {-11 -19}/12 = -30/12 = -10/4 = -2.5
etc... | 664 | 1,672 | {"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-2018-22 | latest | en | 0.786568 |
https://encyclopedia2.thefreedictionary.com/Inhomogeneous+differential+equation | 1,624,275,265,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00262.warc.gz | 214,543,583 | 11,199 | # ordinary differential equation
(redirected from Inhomogeneous differential equation)
## ordinary differential equation
[′ȯrd·ən‚er·ē ‚dif·ə′ren·chəl i′kwā·zhən]
(mathematics)
An equation involving functions of one variable and their derivatives.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
The regular component U(x) is a sufficiently smooth solution of an inhomogeneous differential equation; its first-order derivative is [epsilon]-uniformly bounded.
We consider the transformation f(r) = 1/[lambda](r) and obtain a first-order linear inhomogeneous differential equation as
4: Solve the inhomogeneous differential equation (2.14) for [[phi].sub.1] with the setting [[psi].sub.1] = [v.sub.1,k], ..., [[psi].sub.k] = [v.sub.k,k] and p = k.
Note that (4.4) is just a second order inhomogeneous differential equation with one Dirichlet and one Robin boundary condition.
Equation (2) is an inhomogeneous differential equation, where the function V(R) is called the perturbed scattering potential of the patterned structure relative to the unpatterned stratified structure with the dielectric constant of [[epsilon].sub.f] (R).
The corresponding axial displacement field is shown to be governed by a fourth-order inhomogeneous differential equation. Boundary conditions are naturally inferred by performing a standard localization procedure of a variational problem formulated by making recourse to thermodynamic restrictions see, for example, [24-26], according to the geometric approach illustrated in [27-30].
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Open / Close | 380 | 1,652 | {"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-2021-25 | latest | en | 0.888974 |
https://rdrr.io/cran/adephylo/man/moranIdx.html | 1,643,317,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00301.warc.gz | 519,074,739 | 9,376 | # moranIdx: Computes Moran's index for a variable In adephylo: Exploratory Analyses for the Phylogenetic Comparative Method
## Description
This simple function computes Moran's index of autocorrelation given a variable and a matrix of proximities among observations.
## Usage
`1` ```moran.idx(x, prox, addInfo = FALSE) ```
## Arguments
`x` a numeric vector whose autocorrelation is computed. `prox` a matrix of proximities between observations, as computed by the `proxTips`. Off-diagonal terms must be positive or null, while diagonal terms must all equal zero. `addInfo` a logical indicating whether supplementary info (null value, minimum and maximum values) should be returned (TRUE) or not (FALSE, default); if computed, these quantities are returned as attributes.
## Value
The numeric value of Moran's index.
## Author(s)
Thibaut Jombart tjombart@imperial.ac.uk
## References
Moran, P.A.P. (1948) The interpretation of statistical maps. Journal of the Royal Statistical Society, B 10, 243–251.
Moran, P.A.P. (1950) Notes on continuous stochastic phenomena. Biometrika, 37, 17–23.
de Jong, P. and Sprenger, C. and van Veen, F. (1984) On extreme values of Moran's I and Geary's c. Geographical Analysis, 16, 17–24.
`proxTips` which computes phylogenetic proximities between tips of a phylogeny.
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29``` ```## Not run: ## use maples dataset data(maples) tre <- read.tree(text=maples\$tre) dom <- maples\$tab\$Dom bif <- maples\$tab\$Bif ## get a proximity matrix between tips W <- proxTips(tre, met="Abouheif") ## compute Moran's I for two traits (dom and bif) moran.idx(dom, W) moran.idx(bif, W) moran.idx(rnorm(nTips(tre)), W) ## build a simple permutation test for 'bif' sim <- replicate(499, moran.idx(sample(bif), W)) # permutations sim <- c(moran.idx(bif, W), sim) pval <- mean(sim>=sim[1]) # right-tail p-value pval hist(sim, col="grey", main="Moran's I Monte Carlo test for 'bif'") # plot mtext("Histogram of permutations and observation (in red)") abline(v=sim[1], col="red", lwd=3) ## End(Not run) ``` | 620 | 2,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | longest | en | 0.569269 |
https://assignmenthelp.net/questions/cse1pes-programming-for-engineers-and-scientists | 1,624,047,257,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00129.warc.gz | 118,152,169 | 14,437 | # CSE1PES programming for engineers and scientists
BACKGROUND
An Alternating Current (AC) circuit is one in which the output of the voltage source changes with time. When we add components to the circuit, the voltage and current will be different at different points in the circuit. The components we will be looking at are resistors, capacitors and inductors. Further information can be found here: https://en.wikipedia.org/wiki/Alternating_current, https://en.wikipedia.org/wiki/Inductor, https://en.wikipedia.org/wiki/Resistor, https://en.wikipedia.org/wiki/Capacitor.
Impedance is the measure of the resistance and reactance of a component or circuit. Resistance is a measure of friction against the motion of electrons, whereas reactance is the inertia against the motion of electrons. Impedance introduces a phase shift between the voltage and current through a component, and hence is expressed as a complex number. Do not worry too much if this doesn't make sense, you can just think of this as complex resistance. For further information, please read https://en.wikipedia.org/wiki/Electrical_impedance.
To calculate the impedance, voltage and current for each component in the circuit, we can use ohms law. The equation for ohms law is as follows: (Note we are dealing with complex numbers, however we will develop all the necessary functions you will need in lab 5)
V = I*Z,I = V/Z, Z = V/I
Where V, I and Z represent voltage, current and impedance respectively, and are measured in volts, amps and ohms respectively.
For series circuits the current is the same through all components. For parallel circuits the voltage is the same through all components.
The impedance calculation depends on the component. Details are as follows:
Resistors: The impedance for a resistor can be found as: π = π
where R is the resistance measured in ohms. Capacitors: The impedance for a capacitor can be found as: ππ = 1/2∗π∗π∗πΆ where C is the capacitance measured in Farads and f is the frequency measured in hertz π = 0 − π ∗ ππ where ππ is the reactance and j is the imaginary unit. Inductors: The impedance for an inductor can be found as: ππΏ = 2 ∗ π ∗ π ∗ πΏ where L is the inductance measured in Henrys π = 0 + π ∗ ππΏ where ππΏ is the reactance and j is the imaginary unit.
Components will either be arranged in parallel or series. Depending on this, the total impedance calculation will vary.
The total impedance between two components in series is:
ππ = π1 + π2
If we have three components in series, then the total impedance is:
ππ2 = ππ + π3
Hence, generalizing this we get:
π
ππ = ∑ ππ
π=1
The total impedance between two components in parallel is:
1 1 −1
ππ = (π1 + π2)
Which we can then extend to calculate for three components by:
1 1 −1
ππ2 = (ππ + π3)
Hence, generalizing this we get:
### Problem
You are to create a program that allows a user to view the voltage, current and impedance for each component and the total in either the parallel circuit (a) or series circuit (b). The program will ask the user for the circuit type, the source voltage, the frequency, the number of components, and the values of each component. The two circuits are shown below:
Given a parallel circuit (a), with a source voltage of 12Vpp, a frequency of 60Hz, with 3 components in parallel. The components are a 40 ohm resistor, a 0.0003 farad capacitor and a 0.2 henry inductor.
Firstly we calculate the impedance of each component:
ππ
(πππππππππ ππ πππ ππ π‘ππ) ππ
We then calculate the total current using Ohms law
Hence we can now generate the table:
COMPONENTS R C L TOTAL Z ( Ohms) 40+0j 0-8.842j 0+75.4j 2.36 − 9.425π I (Amps) 0.3+0j 0+1.357j 0-0.159j 0.3+1.198j V (Volts) 12+0j 12+0j 12+0j 12+0j
FUNCTION PROTOTYPES AND INCLUDES
1. You must include the stdio.h and stdlib libraries
2. The following function prototypes are to be used:
1. The first two are described below
2. The last four will be developed in lab5, you need to alter the types from floats to doubles
3. You can change the naming
4. You cannot change the types
FUNCTION - MAIN
1. The program prints to the screen the student number, student name and the assignment number. This is enclosed by double asterisks. Ensure you follow the below format.
**3777777 Jhon Smith Assignment 2 **
2. The program asks the user which circuit they want to evaluate, the frequency in hertz, the source voltage and the number of components. And stores the results appropriately. All of these values are positive integers. When asking for the number of components, the type of circuit should be specified as shown below.
1. You can assume the user enters valid values for frequency, voltage and number of components.
2. You must handle the case where the user enters an incorrect value for the type of circuit and give a message to the user. (No example given, a single line for ‘invalid message’ to user ending with a newline. The message is up to you, but must be appropriate)
3. The program creates an array of pointers to doubles(hint: use [ ]), the array is of length 6
1. Then using a for loop dynamically allocate memory (malloc) for each of the pointers an array of size sizeof(double) *(number of components + 1)
2. This is the result table
3. The program stores the source voltage in the array as shown below
1. The program creates a pointer to ints and dynamically allocates memory for an array of lengthnumber of components. This will store the component types.
2. The program creates a pointer to doubles and dynamically allocates memory for an array of lengthnumber of components. This will store the size of the components.
3. For each component, the program asks the user to enter the type of component and the value of that component. When asking for the size, the correct units must be shown (Ohms, Farads and Henrys).
1. The storing of the component types and values must be done using pointer arithmetic in this section. (Refer to lab 6 and additional material)
2. You can assume the user enters valid values components (1,2 or 3) and their values (positive values)
4. Calculate the impedance values by calling the calculate_impedance function. (Described in section below)
5. Calculate the total current values for both the real and imaginary parts and store appropriately (Ohms law I=V/Z)
6. Calculate the voltages and currents of all components, and store appropriately. (Hint. Refer to background information – Ohms Law. Parallel and series will have different equations)
7. Call the print_results function to print out the results table (Described in section below)
8. End the program appropriately.
FUNCTION - CALCULATE_IMPEDANCE
1. Calculate the impedance for both real and imaginary part for each component using a for loop with an if, else-if, else statement in the loop. And store the results in the results_table (Hint. Refer to background information – Impedance. Resistors, Capacitors and Inductors have different equations)
2. Calculate the total impedance for both real and Imaginary parts. And store the results in the results_table (Refer to additional material if new to summation)
Hint: (One possible implementation)
1. Create temporary variables to store the current sum (for both complex and real parts) and initialize to 0.
2. Create variables to store the total impedance (both for complex and real parts).
3. Then if the circuit is series:
1. Iteratively add to these for each impedance values.
2. The sum is the total impedance
4. else if the circuit is parallel:
1. Create a set of temporary variables for the result of the division 2. Iterate through the component values starting from the first component:
2. Divide 1+0j by the components impedance value and store the result appropriately, then add this result to the current sum.
1. Divide 1+0j by the total sum to give the total impedance.
5. Store the total impedance values in the appropriate place in the result table.
### FUNCTION - PRINT_RESULTS
1. Print the result table as shown below:
1. The first column describes the rows with the text
“COMPONENTS”,”Z ( Ohms)”,”I ( Amps)”,”V (Volts)”.
2. The first row prints the components as ‘R’ for resistor, ‘C’ for capacitor and ‘I’ for inductor.
3. The second row prints the impedance, the third the current and the fourth the voltage
1. The real part is printed in brackets in exponential format with sign and 3 decimal places. Then the ‘+’ character. Then the imaginary part is printed in brackets in exponential format with sign and 3 decimal places. Then the ‘j’ character.
CONSTRAINTS
• Pi needs to be to 11 decimal places in your program.
• Formatting needs to be exactly as specified. (Your program will be tested against automated test cases)
• Text to user needs to be easily understandable. You can change the text but the same inputs must be used.
• The program must print your student number, name and the assignment number as specified.
• Types should be used appropriately.
• You must use comments to explain significant lines of code.
• You must use comments to explain how to use the functions and solution
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This problem is the extreme version of Fizz Buzz. You are given a list of K distinct integers A and a list of K characters S. Both lists are numbered from 1 to K. The integer Ai corresponds to the character Si.
Denote B as the subset of A. There will be a rule for each possible subset: if an integer x is only divisible by all elements in B, print a string which is the concatenation of the corresponding character of each element in B, sorted by the index in ascending order. However, if B is an empty subset, print the first digit of x instead.
Does it look complicated? Do not worry, as it is just a generalization of the regular Fizz Buzz. For example, if A = [3, 5] and S = [F, B], you will get the normal Fizz Buzz problem. You can also check the Sample Test Cases for further understanding.
Your task is simple. Given the array A and S, with all rules from the subset of A, print the Extreme Fizz Buzz from 1 to N.
Format Input
The first line consists of 2 integers N and K.
The second line consists of a string S which has exactly K lower-case latin alphabet
characters.
The third line consists of K integers A1, A2, . . . , AK.
Format Output
Output N lines. The i-th line is the Extreme Fizz Buzz of i, where i is integer from 1 to N.
Constraints
• 1 ≤ N, K ≤ 100000
• 1 ≤ Ai ≤ 100000
• It is guaranteed that the integers in A are distinct
• It is guaranteed that the characters in S are upper-case latin alphabet characters
Sample Input & Output 1 (Standard Input And Output)
15 2
FB
3 5
1
2
F
4
B
F
7
8
F
B
1
F
3
4
FB
Sample Input & Output 2 (Standard Input And Output)
15 2
BF
3 5
1
2
B
4
F
B
7
8
B
F
1
B
3
4
BF
Sample Input & Output 3 (Standard Input And Output)
13 3
JLB
2 4 3
1
J
B
JL
5
JB
7
JL
B
J
11
JLB
13
What I have tried:
```#include <stdio.h>
int main()
{
int N, K, A[100005], i, a, j, fl;
scanf("%d %d", &N, &K);
getchar();
char S[K + 5];
scanf("%s", S);
getchar();
for(i = 0; i < K; i++)
{
scanf("%d", &A[i]);
}
getchar();
for(i = 1; i <= N; i++)
{
a = i;
fl = 0;
for(j = 0; j < K; j++)
{
if(a % A[j] == 0)
{
printf("%c", S[j]);
fl = 1;
}
}
if(fl == 1)
{
printf("\n");
}
else
{
if(a >= 10)
{
a %= 10;
}
printf("%d\n", a);
}
}
return 0;
}```
Posted
Updated 29-Jan-21 12:12pm
Richard MacCutchan 29-Jan-21 7:54am
Stay away from Fizz Buzz and similar sites.
Dave Kreskowiak 29-Jan-21 13:09pm
I'll give you a hint on these "time-bound" tests. If your algorithm uses a loop inside a loop, you've already failed. You have to solve the problem from an entirely different approach.
Dummy Coursera 30-Jan-21 0:06am
Ok, thank you very much
## Solution 2
Just an hint: Sieve of Eratosthenes - Wikipedia[^]
How many divisions/modulos do you need to fill a sieve of Eratosthenes ?
Can you find a way to avoid divisions/modulos ?
v2
Dummy Coursera 30-Jan-21 0:06am
Thank You Very Much
## Solution 1
We are more than willing to help those that are stuck: but that doesn't mean that we are here to do it all for you! We can't do all the work, you are either getting paid for this, or it's part of your grades and it wouldn't be at all fair for us to do it all for you.
So we need you to do the work, and we will help you when you get stuck. That doesn't mean we will give you a step by step solution you can hand in!
Start by explaining where you are at the moment, and what the next step in the process is. Then tell us what you have tried to get that next step working, and what happened when you did.
If you are having problems getting started at all, then this may help: How to Write Code to Solve a Problem, A Beginner's Guide[^] | 1,092 | 3,592 | {"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-2021-39 | latest | en | 0.802282 |
https://webapps.stackexchange.com/questions/164746/sorting-multiple-columns-using-formula | 1,723,044,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00739.warc.gz | 482,349,632 | 42,488 | # Sorting multiple columns using formula
Give a sequence of columns, I want to sort each column individually using formula because of some reason.
Eg.
A B C D
4 7 2 5
1 8 1 3
9 3 4
7 4
Expected output
A B C D
1 7 1 3
4 7 2 4
8 3 4
9 5
Is there any way to do that with a formula? Also the formula should take the whole range as input like A1:D4. Any help is appreciated
Edit - My real problem is solving a trick question, it is not related to any real-world dataset. I saw this trick question answered on youtube here - link. While the solution is provided for a single cell and dragged for remaining cells, I wanted to implement it using ArrayFormula for the whole column so that I don't have to keep dragging it again.
I don't know if it is possible or not, or is it valid to solve questions like these in google sheets. Do tell if it's not appropriate in tools like sheets.
• Hmm... is this an XY problem? Commented May 25, 2022 at 13:24
• @doubleunary: sorry for the late response, my work immediately changed from sheets to other things. Yes it as an XY problem, actually problem was sort a number like - 7865 to 5678, I have the formula for a single cell but I wanted to expand it to the whole column that's why I was trying various things, this is the video I referred to for a single cell sort - link Commented Jun 4, 2022 at 12:42
• Please edit the question so that it tells what your end goal is. Show realistic-looking source data and also show your expected results. Commented Jun 4, 2022 at 13:27
• @doubleunary, I've added my real aim, although it is not for any real world dataset but a trick question instead Commented Jun 5, 2022 at 7:15
• I prefer to use my time helping people with real-world problems. It sounds like you are trying to solve this for fun. You will get more fun out of the problem if you solve it yourself. Commented Jun 5, 2022 at 17:04
If I understand you correctly this formula should solve:
``````={SORT(A1:A,1,true),SORT(B1:B,1,true),SORT(C1:C,1,true),SORT(D1:D,1,true)}
``````
EDIT: this formula is based on your last request:
``````=byrow(A1:A,lambda(r,if(r="",,join("",sort(transpose(split(regexreplace(r&"","(?s)(.{1})","\$1 ")," ")))))))
``````
all the strings (number and/or text) contained in the cells of `column A` will be sorted individually
• Yeah but I want something that takes the whole range as input like A2:D5, could you make it like that? Commented May 25, 2022 at 2:40
• see my edited answer Commented Nov 2, 2022 at 17:14
• are you sure this works? It didn't work for me. Also, I am trying to give the whole range as input i.e. all columns at once. Commented Nov 25, 2022 at 8:15 | 741 | 2,652 | {"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-2024-33 | latest | en | 0.943045 |
https://mednifico.com/31986/what-is-15-celsius-in-fahrenheit/ | 1,642,763,351,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00147.warc.gz | 432,046,416 | 12,112 | # What Is 15 Celsius In Fahrenheit
Posted on
0 C 32 F. So if you have the Celsius.
Body Temperature Conversion Celsius To Fahrenheit Chart Temperature Conversion Chart Chart Temperature Chart
### Convert between Fahrenheit and Celsius.
What is 15 celsius in fahrenheit. -15CtoF Convert -15 C to F. The degree Celsius is the SI derived unit for temperature in the metric system. Celsius or centigrade is used to measure temperatures in most of the world.
This unit of measurement has been created by a physics specialist Daniel Gabriel Fahrenheit. This is the definition that was used up until 2019 when the kelvin was redefined based on the definitions of the second meter and kilogram. Convert 15 Degrees Celsius to Fahrenheit 15 Degrees Celsius to Fahrenheit will not only convert 15 celsius to fahrenheit but will also convert 15 celsius to other units such as Rankine and Kelvin.
The Fahrenheit scale was invented in 1724 by considering freezing to 32 degrees while the point of boiling at 212 degrees. It is considered that a mixture of ice and water which is in equilibrium with saturated air at 1 atm is in the melting point. I am unsure if it was Anders Celsius or Daniel Gabriel Fahrenheit who created this conversion formula.
Celsius value 95 32 For example if you want to convert 5 degree Celsius into Fahrenheit then you have to multiply 5 by 95 and then add 32 with it. However it remains the official scale of the United States Cayman Islands and Belize. Celsius to Fahrenheit Table.
Convert 20 degrees Celsius to degrees Fahrenheit. To establish a baseline temperature measurement Anders Celsius used in 1742 melting points and boiling water. Hence the relationship turns out to be.
Celsius or centigrade is used to measure temperatures in most of the world. To convert Celsius to Fahrenheit simply multiply the Celsius value by 95 or 18 and then add 32 with it. At -40 both scales.
The Fahrenheit temperature range is based on setting the freezing point of water at 32 degrees and boiling to 212 degrees. 0 degrees Celsius is equal to 32 degrees Fahrenheit. 15 degrees celsius is is equivalent to 5900 degrees in fahrenheit or there are 5900F in 15C.
In 1954 the unit degree Celsius as well as the Celsius scale were again re-defined to instead be based on absolute zero -27315 C and the triple point of VSMOW specially purified water. 1 C is 338 F. Specifically degrees Celsius is equal to kelvins minus 27315.
T F T C 18 32. Let C be the Celsius temperature. -15 C 5 F.
F 15 95 32 59 The Celsius scale and the Fahrenheit scale The discovery of temperature scales is one of the most important pieces of science history in our human lives. More temperatures starting with 15 degrees in Celsius Note. To convert 15 C a degrees Fahrenheit you have to multiply 15 x 95 and then add 32.
The temperature T in degrees Fahrenheit F is equal to the temperature T in degrees Celsius C times 95 plus 32. 10000 Celsius in Fahrenheit 0. Relationship between Celsius and Fahrenheit.
F C x 95 32. 15 Celsius 59 Fahrenheit exact result How hot is 15 degrees Celsius. Celsius to Fahrenheit Table.
Convert 155 Degrees Celsius to Fahrenheit 155 Degrees Celsius to Fahrenheit will not only convert 155 celsius to fahrenheit but will also convert 155 celsius to other units such as Rankine and Kelvin. How to convert Celsius to Fahrenheit. In most countries during the mid to late 20th century the Fahrenheit scale was replaced by the Celsius scale.
Literature indicates that the Fahrenheit scale was invented in 1724 and the Celsius Centigrade scale invented in 1742. Fahrenheit is a scale commonly used to measure temperatures in the United States. How much is -15 degrees Celsius in Fahrenheit.
Let F be the Fahrenheit temperature. From these dates its probable that Anders Celsius invented this conversion formula. How To Convert Celsius To Fahrenheit.
This means that a single degree Celsius equals 18 degrees Fahrenheit. Just type in either box and the number in the other box will be converted automatically. Fahrenheit – 32 59.
The relation between Celsius and Fahrenheit is. 0 C 32 F. Water freezes at 0 Celsius and boils at 100 Celsius.
Celsius and Fahrenheit are two different temperature measuring units. As one degree Celsius is equal to one Kelvin boiling point of water is equal to 27315 100 37315 Kelvin. To convert Fahrenheit to Celsius simply do.
The Celsius temperature scale also commonly referred to as the centigrade scale is defined in relation to the kelvin. Fahrenheit is a scale commonly used to measure temperatures in the United States. So if you want to calculate how many degrees Fahrenheit are 15 degrees Celsius you can use this simple rule.
History of degree Fahrenheit. The final formula to convert 15 Celsius to Fahrenheit is. Water freezes at 0 Celsius and boils at 100 Celsius.
155 degrees celsius is is equivalent to 5990 degrees in fahrenheit or there are 5990F in 155C. The Celsius scale has 100 degrees between water boiling and freezing while Fahrenheit has 180 degrees. How to convert 15 degrees Celsius to degrees Fahrenheit.
T F T C 95 32. A degree Celsius is sometimes also referred to as a degree centigrade. F is a unit of measurement for temperature.
What is 15 degrees Celsius in Fahrenheit. A mixture of water and water vapor without.
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Printable Oven Temperature Conversion Chart Temperature Conversion Chart Oven Temperature Conversion Baking Conversion Chart | 1,434 | 7,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-05 | latest | en | 0.839559 |
https://www.convert-measurement-units.com/convert+centiltr+to+milliltr.php | 1,702,137,269,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00420.warc.gz | 788,688,009 | 13,185 | Convert Centiliter to Milliliter (Cooking / Recipes)
## Centiliter into Milliliter
numbers in scientific notation
Direct link to this calculator:
https://www.convert-measurement-units.com/convert+centiltr+to+milliltr.php
# Convert Centiliter to Milliliter:
1. Choose the right category from the selection list, in this case 'Cooking / Recipes'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Centiliter'.
4. Finally choose the unit you want the value to be converted to, in this case 'Milliliter'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '15 Centiliter'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Cooking / Recipes'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '93 Centiliter to Milliliter' or '25 Centiliter into Milliliter' or '68 Centiliter -> Milliliter' or '24 Centiliter = Milliliter'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(77 * 47) Centiliter'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '15 Centiliter + 45 Milliliter' or '83mm x 50cm x 57dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.964 197 503 89×1022. For this form of presentation, the number will be segmented into an exponent, here 22, and the actual number, here 2.964 197 503 89. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 2.964 197 503 89E+22. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 29 641 975 038 900 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 852 | 3,680 | {"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-2023-50 | latest | en | 0.850125 |
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### How many salmon eggs do we REALLY have?
You have already estimated the amount of salmon eggs we have. Now itβs time to find out for sure! Using the picture attached, do the following: 1. Look at the picture and think of a strategy you want to use to determine the exact amount of eggs. π ***How could you use addition or multiplication to help you? ββοΈ 2. Using the draw tool, show how you found the answer. βοΈ 3. Calculators are okay. ππΌ | 133 | 486 | {"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-2021-31 | longest | en | 0.93054 |
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# Amazing Learning Numbers Worksheets 1 20
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# 2 Questions Practice Set (part 1)
Author Message
Intern
Joined: 08 Jan 2007
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2 Questions Practice Set (part 1) [#permalink]
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07 Jan 2008, 03:01
1
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Hi guys I was wondering if you can explain to me how to do the following two questions?
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Re: 2 Questions Practice Set (part 1) [#permalink]
### Show Tags
07 Jan 2008, 03:52
for the first one, the answer is C. Here is the explanation why:
For any product of 2 numbers, for example X and Y:
The product is XY.
XY is also equal to highest common fact of X and Y multiply by least common factor of X and Y.
example: 10 and 15
10 times 15 = 150
highest common factor of 10 and 15= 5
least common factor of 10 and 15= 30
30 times 5= 150 as well.
hope this helps
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Joined: 01 Sep 2007
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Location: Astana
Re: 2 Questions Practice Set (part 1) [#permalink]
### Show Tags
07 Jan 2008, 04:54
1
KUDOS
for the second problem, E should be the answer
A, B, C all involve degrees.
A) X^4>=1; x^4-1>=0
(x^2+1)(x^2-1) = (x^2+1)(x+1)(x-1)>=0
x>=1; x>=-1; infinite set of numbers. eliminate
B) and C) x^3<=27; x <=3 (correct me if i'm wrong pls). same goes with C. they are infine sets
not sure about D). if someone could shed a light on it.
thanks!
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Re: 2 Questions Practice Set (part 1) [#permalink]
### Show Tags
07 Jan 2008, 05:57
1
KUDOS
#1:
C
xy = LCM of x,y * HCF of x, y (see tarek's explanation above).
#2:
E.
A, B, C have infinite values.
Consider D:
2<= |x| <= 5
If x is +ve then: 2 <= x <= 5
If x is -ve then: -5 <= x <= -2
We have 2 line segments but the question is asking for only 1.
For E:
-.66 <= x <= .66
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Re: 2 Questions Practice Set (part 1) [#permalink] 07 Jan 2008, 05:57
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# 2 Questions Practice Set (part 1)
Moderator: chetan2u
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,032 | 3,192 | {"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} | 4.21875 | 4 | CC-MAIN-2017-47 | latest | en | 0.86104 |
https://math.stackexchange.com/questions/4314277/question-about-the-definition-of-the-jacobian-ideal | 1,716,357,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058531.78/warc/CC-MAIN-20240522040032-20240522070032-00410.warc.gz | 337,566,993 | 36,144 | # Question about the definition of the Jacobian ideal
Let $$R=\frac{k[[x_1,\ldots,x_n]]}{(f_1,\ldots,f_c)}$$ (or $$\frac{k[x_1,\ldots,x_n]}{(f_1,\ldots,f_c)}$$), where $$k$$ is a field with characteristic $$0$$ and $$0\ne(f_1,\ldots,f_c)\subseteq (x_1,\ldots,x_c)$$.
In the paper On the fitting ideal in free resolution (see also Jacobian ideal reference), there is a definition of the Jacobian ideal of $$R$$. Assume that the height of $$(f_1,\ldots,f_c)$$ in $$k[[x_1,\ldots,x_n]]$$ is $$h$$, i.e. $$h=\operatorname{inf}\{\operatorname{height}(p)\mid p\in \operatorname{Spec}(k[[x_1,\ldots,x_n]]), (f_1,\ldots,f_c)\subseteq p\}$$. Then the Jacobian ideal of $$R$$ is defined to be the ideal of $$R$$ generated by $$h\times h$$ minors of the Jacobian matrix $$\frac{\partial(f_1,\ldots,f_c)}{\partial(x_1,\ldots,x_n)}$$.
I have two questions about the definition.
Must the Jacobian ideal of $$R$$ be non-zero? I am confused that why the definition considers $$h\times h$$ minors of the Jacobian matrix. Why not higher minors?
Consider the example $$R=\frac{k[[x,y,z]]}{(xy,xz)}$$, the height of $$(xy,xz)$$ in $$k[[x,y,z]]$$ is $$1$$. We can check easily that there is a $$2\times 2$$ minor of the Jacobian matrix that is non-zero in $$R$$. So it seems that it is reasonable to consider higher minors.
Thank you in advance.
The ideal can be zero, but only if $$R$$ is non-reduced: take for example $$(x^2,xy,y^2)$$ which has Jacobian ideal $$(2x^2,2y^2,4xy)=0$$. If $$R$$ is reduced, then $$\operatorname{Spec} R$$ is either a reduced variety (in the case you're taking $$k[x_1,\cdots,x_n]$$) or the germ of a reduced variety (in the case you're taking $$k[[x_1,\cdots,x_n]]$$) and such things are generically smooth if the characteristic of $$k$$ is zero. This is covered in most introductory algebraic geometry texts if you're looking for a reference.
The reason for considering $$h\times h$$ minors is that you're trying to look at nonsingularity of $$V(f_1,\cdots,f_c)$$, which happens when the Jacobian matrix is of rank $$n-\dim_p V(f_1,\cdots,f_c)$$ at each point of $$p\in V(f_1,\cdots,f_c)$$.
• Nice. Thank you. Cool, I found the last sentence you said in Eisenbud's book Corollary 16.20. Under the last condition you said, the Jacobian ideal defines the singular locus of $R$.
– Jian
Nov 24, 2021 at 1:07
• If $R$ is reduced, the ideal is non-zero. Can you give me a reference? I didn't find it. Also, do you think the ideal is also non-zero when $f_1,\ldots,f_c$ is a regular sequence in $k[[x_1,\ldots,x_n]]$ or $k[x_1,\ldots,x_n]$?
– Jian
Nov 24, 2021 at 1:31
• It's not as direct as that: the vanishing locus of the Jacobian ideal is the singular locus, which is known to be a proper subvariety when the characteristic is zero (see Hartshorne I.5 for a proof, for instance), hence the Jacobian ideal cannot be zero. The regular sequence idea can't work: if $f_1,\cdots,f_c$ is a regular sequence, $f_1^r,\cdots,f_c^r$ is also a regular sequence. Nov 24, 2021 at 1:32
• Yes, the power of a regular sequence is still a regular sequence. In the regular sequence case, do you mean the $c\times c$ minor can be zero? Can you give an example here? Thank you. I don't find any example that the Jacobian ideal is zero when $f_1,\ldots,f_c$ is a regular sequence.
– Jian
Nov 24, 2021 at 1:45
• Sorry, perhaps I was hasty - at present I'm not totally sure about the claim either way. Nov 24, 2021 at 2:52 | 1,081 | 3,418 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.828307 |
https://www.student-circuit.com/learning/year1/electro-fields-waves/electromagnetic-fields-and-waves-shift-current-and-maxwell-equations/ | 1,618,590,249,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00048.warc.gz | 1,135,306,631 | 19,693 | Electromagnetic Fields and Waves
# Shift current and Maxwell equations
Variable magnetic fields create a vortex electric field in the conductor, which leads to the electric current appearance in the conductor. Let’s consider two cases. When we are talking about DC currents across a conductor, there is no need to consider special surface , enclosed with a contour. The same current goes through any surfaces we choose.
When we are talking about AC current through the conductor, the question of surface arises and circulation theorem becomes very important. (Figure 38). In this case, circulation theorem through the surface gives the following:
According to the circulation theorem, we also can choose another surface , crossing the capacitor plates, for example. This surface will contain the charge, equal to the capacitor plate charge. The current through the surface
In accordance to Gauss theorem
${\epsilon }_{0}{\int }_{\Omega }{E}_{n}dS=q$
Then the current through this surface is
$I={\epsilon }_{0}\frac{d}{dt}{\int }_{\Omega }{E}_{n}dS={\epsilon }_{0}{\int }_{\Omega }\frac{\partial {E}_{n}}{{\partial }_{t}}dS$
This current is called shift current, and
${\epsilon }_{0}\frac{\partial {E}_{n}}{{\partial }_{t}}$
is the shift current density.
Charges do not move between the capacitor plates, but we can see time variable electric field, that creates the magnetic field. So electric vortex fields create magnetic fields, and magnetic fields create electric vortex field.
The laws and theorems discussed here lead us to the set of Maxwell equations – mathematical form of fundamental laws of electromagnetism: | 366 | 1,634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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} | 3.546875 | 4 | CC-MAIN-2021-17 | longest | en | 0.857442 |
https://aprove.informatik.rwth-aachen.de/eval/FullComplexity/newtpdb/Runtime_Complexity_Full_Rewriting/SK90/4.27.xml.AProVE%20Lower.html | 1,627,996,861,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00277.warc.gz | 120,693,147 | 3,543 | ### (0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))
Rewrite Strategy: FULL
### (1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(n1):
The rewrite sequence
int(0, s(y)) →+ .(0, int_list(int(0, y)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0].
The pumping substitution is [y / s(y)].
The result substitution is [ ].
### (3) RenamingProof (EQUIVALENT transformation)
Renamed function symbols to avoid clashes with predefined symbol.
### (4) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
int(0', 0') → .(0', nil)
int(0', s(y)) → .(0', int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))
S is empty.
Rewrite Strategy: FULL
### (5) SlicingProof (LOWER BOUND(ID) transformation)
Sliced the following arguments:
./0
### (6) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
S is empty.
Rewrite Strategy: FULL
Infered types.
### (8) Obligation:
TRS:
Rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
Types:
int :: 0':s → 0':s → nil:.
0' :: 0':s
. :: nil:. → nil:.
nil :: nil:.
s :: 0':s → 0':s
int_list :: nil:. → nil:.
hole_nil:.1_0 :: nil:.
hole_0':s2_0 :: 0':s
gen_nil:.3_0 :: Nat → nil:.
gen_0':s4_0 :: Nat → 0':s
### (9) OrderProof (LOWER BOUND(ID) transformation)
Heuristically decided to analyse the following defined symbols:
int, int_list
They will be analysed ascendingly in the following order:
int_list < int
### (10) Obligation:
TRS:
Rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
Types:
int :: 0':s → 0':s → nil:.
0' :: 0':s
. :: nil:. → nil:.
nil :: nil:.
s :: 0':s → 0':s
int_list :: nil:. → nil:.
hole_nil:.1_0 :: nil:.
hole_0':s2_0 :: 0':s
gen_nil:.3_0 :: Nat → nil:.
gen_0':s4_0 :: Nat → 0':s
Generator Equations:
gen_nil:.3_0(0) ⇔ nil
gen_nil:.3_0(+(x, 1)) ⇔ .(gen_nil:.3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))
The following defined symbols remain to be analysed:
int_list, int
They will be analysed ascendingly in the following order:
int_list < int
### (11) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
Induction Base:
int_list(gen_nil:.3_0(0)) →RΩ(1)
nil
Induction Step:
int_list(gen_nil:.3_0(+(n6_0, 1))) →RΩ(1)
.(int_list(gen_nil:.3_0(n6_0))) →IH
.(gen_nil:.3_0(c7_0))
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (13) Obligation:
TRS:
Rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
Types:
int :: 0':s → 0':s → nil:.
0' :: 0':s
. :: nil:. → nil:.
nil :: nil:.
s :: 0':s → 0':s
int_list :: nil:. → nil:.
hole_nil:.1_0 :: nil:.
hole_0':s2_0 :: 0':s
gen_nil:.3_0 :: Nat → nil:.
gen_0':s4_0 :: Nat → 0':s
Lemmas:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
Generator Equations:
gen_nil:.3_0(0) ⇔ nil
gen_nil:.3_0(+(x, 1)) ⇔ .(gen_nil:.3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))
The following defined symbols remain to be analysed:
int
### (14) RewriteLemmaProof (LOWER BOUND(ID) transformation)
Proved the following rewrite lemma:
int(gen_0':s4_0(n184_0), gen_0':s4_0(n184_0)) → gen_nil:.3_0(1), rt ∈ Ω(1 + n1840)
Induction Base:
int(gen_0':s4_0(0), gen_0':s4_0(0)) →RΩ(1)
.(nil)
Induction Step:
int(gen_0':s4_0(+(n184_0, 1)), gen_0':s4_0(+(n184_0, 1))) →RΩ(1)
int_list(int(gen_0':s4_0(n184_0), gen_0':s4_0(n184_0))) →IH
int_list(gen_nil:.3_0(1)) →LΩ(2)
gen_nil:.3_0(1)
We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).
### (16) Obligation:
TRS:
Rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
Types:
int :: 0':s → 0':s → nil:.
0' :: 0':s
. :: nil:. → nil:.
nil :: nil:.
s :: 0':s → 0':s
int_list :: nil:. → nil:.
hole_nil:.1_0 :: nil:.
hole_0':s2_0 :: 0':s
gen_nil:.3_0 :: Nat → nil:.
gen_0':s4_0 :: Nat → 0':s
Lemmas:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
int(gen_0':s4_0(n184_0), gen_0':s4_0(n184_0)) → gen_nil:.3_0(1), rt ∈ Ω(1 + n1840)
Generator Equations:
gen_nil:.3_0(0) ⇔ nil
gen_nil:.3_0(+(x, 1)) ⇔ .(gen_nil:.3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))
No more defined symbols left to analyse.
### (17) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
### (19) Obligation:
TRS:
Rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
Types:
int :: 0':s → 0':s → nil:.
0' :: 0':s
. :: nil:. → nil:.
nil :: nil:.
s :: 0':s → 0':s
int_list :: nil:. → nil:.
hole_nil:.1_0 :: nil:.
hole_0':s2_0 :: 0':s
gen_nil:.3_0 :: Nat → nil:.
gen_0':s4_0 :: Nat → 0':s
Lemmas:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
int(gen_0':s4_0(n184_0), gen_0':s4_0(n184_0)) → gen_nil:.3_0(1), rt ∈ Ω(1 + n1840)
Generator Equations:
gen_nil:.3_0(0) ⇔ nil
gen_nil:.3_0(+(x, 1)) ⇔ .(gen_nil:.3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))
No more defined symbols left to analyse.
### (20) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
### (22) Obligation:
TRS:
Rules:
int(0', 0') → .(nil)
int(0', s(y)) → .(int(s(0'), s(y)))
int(s(x), 0') → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(y)) → .(int_list(y))
Types:
int :: 0':s → 0':s → nil:.
0' :: 0':s
. :: nil:. → nil:.
nil :: nil:.
s :: 0':s → 0':s
int_list :: nil:. → nil:.
hole_nil:.1_0 :: nil:.
hole_0':s2_0 :: 0':s
gen_nil:.3_0 :: Nat → nil:.
gen_0':s4_0 :: Nat → 0':s
Lemmas:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60)
Generator Equations:
gen_nil:.3_0(0) ⇔ nil
gen_nil:.3_0(+(x, 1)) ⇔ .(gen_nil:.3_0(x))
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))
No more defined symbols left to analyse.
### (23) LowerBoundsProof (EQUIVALENT transformation)
The lowerbound Ω(n1) was proven with the following lemma:
int_list(gen_nil:.3_0(n6_0)) → gen_nil:.3_0(n6_0), rt ∈ Ω(1 + n60) | 3,086 | 7,194 | {"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-2021-31 | latest | en | 0.453241 |
https://www.appsloveworld.com/2018/12/crt-monitor-electricity-consumption.html | 1,566,788,684,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00494.warc.gz | 717,498,230 | 37,638 | crt monitor electricity consumption
CRT monitor electricity consumption
Technically Computer Monitor is called visual display unit.It is an output device that shows all the CPU information in monitor Screen.This is like an interface between CPU and user. A cable is connected with the video adapter or with the video card that is set up with the motherboard of the computer.
Select country: Australia Canada Denmark France Germany India Philippines Singapore United Kingdom United States Other country Power Use: watts (W) kilowatts (kW) Hours of use per day: h/day 1 kilowatt-hour (kWh) cost: cent pence rupee peso kr other cost per day: cost per month: cost per year: kWh/day: kWh/month: kWh/year:
Select Country: Please choose Country from the list. If your country is not listed then you can choose other country option from the list.
Power Use: Enter average energy consumption by your CRT monitor.
Hours Used Per Day: Please enter how many hours the CRT monitor is used on per day, if the energy consumption is smaller than 1 hour per day then please enter as a decimal(For example-15 m per day is 0.25)
Cost (kWh): Please Enter the Approx price you are paying on per kilowatt hour(kWh).
You can find the exact cost on your electricity bill.
Now click the calculate button you can see the following values Electricity cost per day, month ,year, Energy consumed per day, month,year.
How much electricity does a CRT monitor use?
Let assume that your clothes CRT monitor 5 hours a day, Average power consumption of the CRT monitor between 50 to 110 watts. Let's take an example and say your model uses 70 watts.
70 watts X 5 hours = 350 watt-hours per day
How many kilowatts does a CRT monitor use?
350 watt-hours per day / 1000 = .35 kWh per day
How much does it cost to run a CRT monitor every day?
.35 kWh per day x 0.10 per kWh = \$ .035 per day
How much does it cost to run a CRT monitor?
.35 kWh per day X 30 days = 10.5 kWh per month
10.5 kWh per month x 0.10 per kWh = \$ 1.05 per month
Monitors very similar look as televisions. In both of these, the main difference between them is that
the monitor don't have television tuner to change the channels.
Monochrome
The word is made up of two words Mono i.e. single and chrome i.e. color so it is called single color display and it displays the monitor output as Black & White.
Gray-scale
This monitor is similar to monochrome but it's shows any display in gray shades (grey Shades) this type of monitors are mostly used in handy computers such as the Laptop.
Colorful monitor (color monitors)
This monitor displays output as an adjustment of RGB (Red-Green-Blue),
this type of monitors are capable of displaying Graphics in high resolutions (resolution).
The CRT monitors uses a cathode ray tube to display images.A vacuum tube, heaters, electron guns, deflection circuits and a glass screen are used to construct this cathode ray tube.
The CRT monitor are bulky and also consume a lot of energy.CRT Monitor Consume more Power then modern display technologies such as LCD/LED TV/Display.
we suggest you to upgrade you CRT Monitor to an LED/LCD display if you want to reduce Power consumption or electricity bill. | 735 | 3,182 | {"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-2019-35 | latest | en | 0.838467 |
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Home : Support : Online Help : System : Information : Updates : Maple 2022 : Step by Step Solutions
Step-by-Step Solutions
Simplification Steps
Maple 2022 includes new commands for showing the steps needed to manipulate algebraic expressions in order to reduce them to their simplest form.
In general the generated steps try to find that hard balance between being too verbose and too cryptic. The SimplifySteps command errs on the side of adding more steps, and is aimed to help someone who wants to learn what the steps are, even for fundamentals like adding fractions. Depending on the problem it will adjust somewhat; recognizing higher level problems for which it will decide to skip more easy-level steps.
Here are some examples of different categories of problems that SimplifySteps as well as related command FractionSteps can handle:
$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right):$
Fractions
There is a dedicated FractionSteps command that goes into slightly more detail than SimplifySteps
$\begin{array}{lll}{}& {}& \text{Let's Simplify Fractions}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Find fractions to get lowest common denominator of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}6\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Add numerators}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}3\cdot 1\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\cdot 1\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Add}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}3+1\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Cancel out factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \left[{}\right]\end{array}$ (1.1.1)
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Find fractions to get lowest common denominator of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}6\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Add fractions}\\ {}& {}& \frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{3}}}\end{array}$ (1.1.2)
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Pull out a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\sqrt{4}=2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{from}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\sqrt{12}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply in order to rationalize the denominator}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply the denominator}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Factor roots}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Combine}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\left\{\sqrt{2}\cdot \sqrt{2},\sqrt{3}\cdot \sqrt{3}\right\}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}3\cdot \left(2{}\sqrt{5}\right)\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Cancel out factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}6\\ {}& {}& \sqrt{{5}}\end{array}$ (1.2.1)
$\mathrm{SimplifySteps}\left("sqrt\left(6\right)*sqrt\left(10\right)/\left(12\right)^\left(1/3\right)"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply in order to rationalize the denominator}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply the denominator}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Factor roots}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Combine}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\left\{{3}^{2}{3}}\cdot \sqrt{3},{2}^{1}{3}}\cdot \sqrt{2}\cdot \sqrt{2}\right\}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}3{}{3}^{1}{6}}\cdot \left(2{}{2}^{1}{3}}{}\sqrt{5}\right)\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\cdot \left(6{}\sqrt{5}{}{2}^{1}{3}}{}{3}^{1}{6}}\right)\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Cancel out factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}12\\ {}& {}& \sqrt{{5}}{}{{2}}^{{1}}{{3}}}{}{{3}}^{{1}}{{6}}}\end{array}$ (1.2.2)
Exponents
$\mathrm{SimplifySteps}\left("\left(2*x^3*y^3\right)*\left(3*x^1*y^2\right)^2"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Evaluate exponent}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\left(3{}x{}{y}^{2}\right)}^{2}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\cdot {x}^{3}\cdot {y}^{3}\cdot \left(9{}{x}^{2}{}{y}^{4}\right)\\ {}& {}& \left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{18}}{}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}}{}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{y}}}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{7}}}\right)\end{array}$ (1.3.1)
$\mathrm{SimplifySteps}\left("x^3*x^5"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply the product rule}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{a}^{n}{}{a}^{m}={a}^{n+m}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to add exponents with common base}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Add exponents}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Solution}\\ {}& {}& \left[{}\right]\end{array}$ (1.3.2)
$\mathrm{SimplifySteps}\left("\left(b^n\right)^m"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply the integer power of a power rule,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\left({a}^{n}\right)}^{m}={a}^{n{}m}\\ {}& {}& \left[{}\right]\end{array}$ (1.3.3)
$\mathrm{SimplifySteps}\left("y^5/y^4"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Cancel out factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}^{4}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{provided}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}^{4}\ne 0\\ {}& {}& \colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{y}}\end{array}$ (1.3.4)
$\mathrm{SimplifySteps}\left("y^\left(-5\right)/y^4"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Divide assuming}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{y}^{4}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{≠ 0}\\ {}& {}& \frac{{1}}{{{y}}^{{9}}}\end{array}$ (1.3.5)
$\mathrm{SimplifySteps}\left("123^5/123^4"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply the quotient rule:}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{a}^{n}}{{a}^{m}}={a}^{n-m}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Evaluate exponent}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{123}^{1}\\ {}& {}& \colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{123}}\end{array}$ (1.3.6)
Logs
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Use the log rule,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{log}}_{a}{}\left(x\right)=\frac{{\mathrm{log}}_{b}{}\left(x\right)}{{\mathrm{log}}_{b}{}\left(a\right)}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to express as a single logarithm}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Solution}\\ {}& {}& \left[{}\right]\end{array}$ (1.4.1)
Note: This is different than how Maple's simplify command treats expressions like this, always converting to ln:
$\mathrm{simplify}\left(\frac{\mathrm{log}\left[10\right]\left(2\right)}{\mathrm{log}\left[10\right]\left(5\right)}+\mathrm{log}\left[10\right]\left(3\right)\right)$
$\frac{{\mathrm{ln}}{}\left({2}\right)}{{\mathrm{ln}}{}\left({5}\right)}{+}\frac{{\mathrm{ln}}{}\left({3}\right)}{{\mathrm{ln}}{}\left({2}\right){+}{\mathrm{ln}}{}\left({5}\right)}$ (1.4.2)
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Evaluate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{log}}_{10}{}\left(100\right)\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Evaluate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{log}}_{10}{}\left(1000\right)\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Add}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2+3\\ {}& {}& \colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{5}}\end{array}$ (1.4.3)
$\mathrm{SimplifySteps}\left("5*log\left[z\right]\left(z^4\right)"\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply the log rule}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{log}}_{a}{}\left({m}^{n}\right)=n{}{\mathrm{log}}_{a}{}\left(m\right)\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply the log rule}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{log}}_{a}{}\left(a\right)=1\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Multiply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}5\cdot 4\\ {}& {}& \colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{20}}\end{array}$ (1.4.4)
You can also use the new PowerSteps command to get step by step results for problems with radicals, exponents, and logarithms.
Trig
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Pythagoras}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{cot}{}\left(x\right)}^{2}={\mathrm{csc}{}\left(x\right)}^{2}-1\\ {}& {}& {1}{+}\left({\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{csc}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Reciprocal Function}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{csc}{}\left(x\right)=\frac{1}{\mathrm{sin}{}\left(x\right)}\\ {}& {}& {\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}}^{{2}}\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& \frac{{1}}{{{\mathrm{sin}}{}\left({x}\right)}^{{2}}}\end{array}$ (1.5.1)
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Add}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\pi }+\mathrm{\pi }+x\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Evaluate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{sin}{}\left(2{}\mathrm{\pi }+x\right)\\ {}& {}& {\mathrm{sin}}{}\left({x}\right)\end{array}$ (1.5.2)
$\mathrm{SimplifySteps}\left(\frac{{\mathrm{sec}\left(x\right)}^{2}-1}{{\mathrm{sec}\left(x\right)}^{2}}\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Reciprocal Function}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{sec}{}\left(x\right)=\frac{1}{\mathrm{cos}{}\left(x\right)}\\ {}& {}& \frac{{{\mathrm{sec}}{}\left({x}\right)}^{{2}}{-}{1}}{{\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}}^{{2}}}\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Pythagoras}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{sec}{}\left(x\right)}^{2}=1+{\mathrm{tan}{}\left(x\right)}^{2}\\ {}& {}& \left(\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{+}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{tan}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\right){-}{1}\right){}{{\mathrm{cos}}{}\left({x}\right)}^{{2}}\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Quotient}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{tan}{}\left(x\right)=\frac{\mathrm{sin}{}\left(x\right)}{\mathrm{cos}{}\left(x\right)}\\ {}& {}& {\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{cos}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}}^{{2}}{}{{\mathrm{cos}}{}\left({x}\right)}^{{2}}\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& {{\mathrm{sin}}{}\left({x}\right)}^{{2}}\end{array}$ (1.5.3)
$\mathrm{SimplifySteps}\left(\left(-{\mathrm{cos}\left(x\right)}^{2}+1\right)\left(1+{\mathrm{cot}\left(x\right)}^{2}\right)\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Pythagoras}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{cot}{}\left(x\right)}^{2}={\mathrm{csc}{}\left(x\right)}^{2}-1\\ {}& {}& \left({-}{{\mathrm{cos}}{}\left({x}\right)}^{{2}}{+}{1}\right){}\left({1}{+}\left({\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{csc}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\right)\right)\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Reciprocal Function}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{csc}{}\left(x\right)=\frac{1}{\mathrm{sin}{}\left(x\right)}\\ {}& {}& \left({-}{{\mathrm{cos}}{}\left({x}\right)}^{{2}}{+}{1}\right){}{\frac{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}}^{{2}}\\ \text{•}& {}& \text{Apply}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{Pythagoras}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{trig identity,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{cos}{}\left(x\right)}^{2}=1-{\mathrm{sin}{}\left(x\right)}^{2}\\ {}& {}& \frac{{-}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{1}}\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{-}}{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{\mathrm{sin}}}{}\left(\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{x}}\right)}^{\colorbox[rgb]{0.886274509803922,0.964705882352941,0.996078431372549}{{2}}}\right){+}{1}}{{{\mathrm{sin}}{}\left({x}\right)}^{{2}}}\\ \text{•}& {}& \text{Evaluate}\\ {}& {}& {1}\end{array}$ (1.5.4)
You can also use the new TrigSteps command to get these step by step results.
Calculus
Showing the steps to solving an integral, limit, or derivative has been available in past versions of Maple via the Student:-Calculus1:-ShowSolution command. You can now also access those step by step solutions through SimplifySteps, further unifying the ability to do step by step solutions using a single command.
$\mathrm{SimplifySteps}\left({x}^{2}+{\int }x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x\right)$
$\begin{array}{lll}{}& {}& \text{Let's simplify}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Integral}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{to evaluate}\\ {}& {}& {\int }{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ \text{▫}& {}& \text{1. Apply the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathbf{\text{power}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{rule to the term}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\int x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx\\ {}& \text{◦}& \text{Recall the definition of the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathbf{\text{power}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{rule, for n}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{≠}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{-1}\\ {}& {}& {\int }{{x}}^{\left[{}\right]}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}\left[{}\right]\\ {}& \text{◦}& \text{This means:}\\ {}& {}& {\int }{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}\left[{}\right]\\ {}& \text{◦}& \text{So,}\\ {}& {}& {\int }{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}\frac{{{x}}^{{2}}}{{2}}\\ {}& {}& \text{We can rewrite the integral as:}\\ {}& {}& \frac{{{x}}^{{2}}}{{2}}\\ \text{•}& {}& \text{Sub evaluated}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{integral}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{back in expression}\\ {}& {}& \frac{{3}{}{{x}}^{{2}}}{{2}}\end{array}$ (1.6.1)
Steps for Sketching a Curve
Maple 2022 includes a new command for showing the steps needed to sketch the graph of an expression by identifying the basic function and the transformations done to the function. Various kinds of expressions are handled, including trig, logs, and polynomials to pick just a few. Here are some examples:
$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right):$
$\mathrm{CurveSketchSteps}\left(2\mathrm{sin}\left(3x+\frac{\mathrm{\pi }}{3}\right)+1\right)$
$\begin{array}{lll}{}& {}& \text{Let's plot}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2{}\mathrm{sin}{}\left(3{}x+\frac{\mathrm{\pi }}{3}\right)+1\\ \text{•}& {}& \text{Compared to the plot of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{sin}{}\left(x\right)\text{, we have a vertical stretch by a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Then, we have a horizontal compression by a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{1}{3}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Then, we have a vertical shift of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Then, we have a horizontal shift of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-\frac{\mathrm{\pi }}{9}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Apply the horizontal shift and stretch to the range,}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=-2{}\mathrm{\pi }..2{}\mathrm{\pi }\\ {}& {}& \left[{}\right]{+}\frac{{\mathrm{\pi }}}{{9}}{..}\left[{}\right]{+}\frac{{\mathrm{\pi }}}{{9}}{=}{-2.443460953}{..}{1.745329252}\\ \text{•}& {}& \text{We can now plot using the information extracted}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\end{array}$ (2.1)
$\mathrm{CurveSketchSteps}\left(2{x}^{2}+4x+10\right)$
$\begin{array}{lll}{}& {}& \text{Let's plot}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2{}{x}^{2}+4{}x+10\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Complete the square}\\ {}& {}& {2}{}{\left({x}{+}{1}\right)}^{{2}}{+}{8}\\ \text{•}& {}& \text{With the expression in vertex form we can extract valuable information}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{The coefficient}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{of the}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\left(x+1\right)}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{term indicates a parabola that opens}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{up}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{and has a vertical}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{stretch}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{We have a horizontal shift of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-1\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{and a vertical shift of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}8\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{which gives a vertex of (}-1\text{,}8\text{)}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{We can now plot using the information extracted}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\end{array}$ (2.2)
$\mathrm{CurveSketchSteps}\left(4x+10,\mathrm{output}=\mathrm{typeset}\right)$
$\begin{array}{lll}{}& {}& \text{Let's plot}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4{}x+10\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{This is a line; find two points and draw a line through them}\\ {}& {}& {y}{=}{4}{}{x}{+}{10}\\ \text{•}& {}& \text{Set}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{= 0 to solve for y intercept}\\ {}& {}& {y}{=}{10}\\ \text{•}& {}& \text{This gives a y intercept of (0,}10\text{)}\\ {}& {}& {y}{=}{10}\\ \text{•}& {}& \text{Set expresson to 0 to solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{intercept}\\ {}& {}& {0}{=}\left[{}\right]\\ \text{•}& {}& \text{Subtract}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}4\cdot x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{from both sides}\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ \text{•}& {}& \text{Simplify}\\ {}& {}& \left[{}\right]{=}{10}\\ \text{•}& {}& \text{Divide both sides by}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}-4\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {x}{=}{-}\frac{{5}}{{2}}\\ \text{•}& {}& \text{This gives an}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{intercept of (}-\frac{5}{2}\text{,0)}\\ {}& {}& {x}{=}{-}\frac{{5}}{{2}}\\ \text{•}& {}& \text{By connecting through the two points we can plot the line}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\end{array}$ (2.3)
$\mathrm{CurveSketchSteps}\left(\frac{2}{3x+2}\right)$
$\begin{array}{lll}{}& {}& \text{Let's plot}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{2}{3{}x+2}\\ \text{•}& {}& \text{Rewrite the equation in the following form}\\ {}& {}& \left[{}\right]\\ \text{•}& {}& \text{Compared to the plot of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{1}{x}\text{, we have a vertical stretch by a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}2\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{Then, we have a horizontal compression by a factor of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{1}{3}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{Then, we have a horizontal shift of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{2}{3}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\\ \text{•}& {}& \text{The final plot with asymptotes in cyan at}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y=0\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{and}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x=-\frac{2}{3}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{is}\\ {}& {}& {\mathrm{PLOT}}{}\left({\mathrm{...}}\right)\end{array}$ (2.4)
For more information, see the help page Student:-Basics:-CurveSketchSteps. | 10,153 | 25,016 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 41, "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} | 4.4375 | 4 | CC-MAIN-2023-50 | latest | en | 0.694216 |
https://www.quantopian.com/posts/problems-with-implementing-a-signal | 1,544,875,018,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.55/warc/CC-MAIN-20181215105142-20181215131142-00437.warc.gz | 1,006,623,225 | 11,989 | Problems with implementing a signal
Hey everybody :),
I am new here and I have some troubles with implementing the "WaveTrend Oscillator - Signal". It is a really interesting signal, which we talked about in a class at my university. I want to adopt this code and invest into the Nike Stocks (NKE). So, my problems are:
- Which language is this? How do I translate it into python?
- Which time period do I need for the ema and the average price in this signal?
- How can I set the overbought and oversold areas?
I would be very happy if somebody could help me. Thank you in advance!
This is the signal, which you can find at tradingview.com:
// @author LazyBear
//
// If you use this code in its original/modified form, do drop me a note.
//
study(title="WaveTrend [LazyBear]", shorttitle="WT_LB")
n1 = input(10, "Channel Length")
n2 = input(21, "Average Length")
obLevel1 = input(60, "Over Bought Level 1")
obLevel2 = input(53, "Over Bought Level 2")
osLevel1 = input(-60, "Over Sold Level 1")
osLevel2 = input(-53, "Over Sold Level 2")
ap = hlc3
esa = ema(ap, n1)
d = ema(abs(ap - esa), n1)
ci = (ap - esa) / (0.015 * d)
tci = ema(ci, n2)
wt1 = tci
wt2 = sma(wt1,4)
plot(0, color=gray)
plot(obLevel1, color=red)
plot(osLevel1, color=green)
plot(obLevel2, color=red, style=3)
plot(osLevel2, color=green, style=3)
plot(wt1, color=green)
plot(wt2, color=red, style=3)
plot(wt1-wt2, color=blue, style=area, transp=80)
You can find the signal here: WaveTrend Oscillator
2 responses
Hi. This is written in Pine.
I've also been looking at this indicator. Sorry that is all I can help, maybe someone else can help also further this?
Hi Mike,
thank you for your answer. At least, I know now the language. | 508 | 1,710 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-51 | latest | en | 0.85815 |
https://prepinsta.com/infytq-numbers-quiz-1/ | 1,600,586,068,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00537.warc.gz | 553,385,010 | 27,241 | # InfyTQ Numbers Quiz 1
Question 1
Time: 00:00:00
The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
3
3
4
4
9
9
none of these
none of these
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Question 2
Time: 00:00:00
A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
18
18
24
24
42
42
81
81
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Question 3
Time: 00:00:00
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
20
20
30
30
40
40
None of these
None of these
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Question 4
Time: 00:00:00
The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?
69
69
78
78
96
96
Cannot be determined
Cannot be determined
None of these
None of these
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Question 5
Time: 00:00:00
In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:
22
22
24
24
42
42
46
46
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Question 6
Time: 00:00:00
Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
3
3
7
7
10
10
20
20
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Question 7
Time: 00:00:00
The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:
380
380
395
395
400
400
425
425
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Question 8
Time: 00:00:00
The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:
20
20
23
23
169
169
None of these
None of these
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Question 9
Time: 00:00:00
What is the sum of two consecutive even numbers, the difference of whose squares is 84?
34
34
38
38
42
42
46
46
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["0","40","60","80","100"]
["Need more practice!","Keep trying!","Not bad!","Good work!","Perfect!"]
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0% | 864 | 3,316 | {"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-2020-40 | latest | en | 0.896077 |
http://essedunet.nsd.uib.no/cms/topics/regression/7/ | 1,642,435,820,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00397.warc.gz | 23,809,751 | 5,687 | # Chapter 7: Regression based on samples from several countries
What if we wish to use data from several countries simultaneously in a multiple regression analysis? This may create problems because the ESS survey sample is stratified, with county as the stratifying variable, while the SPSS ordinary linear regression module presupposes that we use a non-stratified, simple random sample. SPSS offers two different extensions of linear regression analysis that may alleviate this problem: a module for complex survey analysis and a mixed models module that handles multilevel analysis. You may want to check the virtues and possibilities of these modules if you plan to do regression analysis on data from many countries. If you only use individual level variables and data from a few countries, ordinary linear regression analysis may be an admissible option, but then you may have to take special precautions. You could, for example, weight the cases with the product of the ESS design weight and the ESS population size weight. (Use the ‘Compute’ feature in the ‘Transform’ menu to compute the product.) However, weighting the cases without making adequate adjustments to the standard error estimates may corrupt the statistical tests. Alternatively, to get more accurate statistical tests, you could skip weighting and enter country dummy variables as independent variables in your model.
But the latter solution presupposes that the associations between dependent and independent variables are constant across countries, which is frequently not the case. Therefore, the best ordinary linear regression solution might be to drop weighting and use regression models that allow regression slope coefficients to vary between countries. This can be achieved either by running separate regression analyses for each country (which brings us back to the solutions discussed in the previous chapters) or by supplementing our models with so called interaction terms (which are computed by multiplying each country dummy with every other independent variable, or at least with every other variable whose association with the dependent variable varies substantially between countries). It goes without saying that the total number of terms and coefficients in such a model may become excessively high if we include many independent variables and countries. We therefore recommend that, if the number of countries and interaction terms proliferate, and, in particular, if you want to assess the weighted mean association between variables across a group of countries rather than country-specific associations, you should drop ordinary linear regression and use the ‘General linear model’ program under SPSS’s ‘Complex samples’ module instead. On the next page, we will demonstrate how you can perform regression analyses with interaction terms. | 499 | 2,838 | {"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} | 2.5625 | 3 | CC-MAIN-2022-05 | latest | en | 0.900568 |
http://www.cs.utoronto.ca/~wayne/research/thesis/msc/node8.html | 1,539,810,232,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00102.warc.gz | 428,299,034 | 3,873 | # Introduction
## Definitions
Throughout this thesis, when referring to mathematical variables, boldface will refer to vectors, and italic will refer to scalars, matrices and functions. Scalars are written in small letters and matrices in CAPITALS.
Some of the following definitions are taken, with minor modifications, from Grebogi, Hammel, Yorke, and Sauer [11], hereinafter referred to as GHYS. The terms trajectory, orbit, and solution are used interchangably throughout this thesis.
Definition. A true trajectory of f satisfies for . We are interested in the case where a and b are finite integers. For a chaotic map, f may be a simple equation, such as the logistic equation , which always maps the interval [-1,1] onto itself. For an ODE system like the N-body problem, represents the true solution of integrating the phase-space co-ordinates for one timestep.
Definition. is a -pseudo-trajectory, also called a noisy orbit, for f if for , where is the noise amplitude.
Definition. For , the 1-step error made between step i and step i+1 of the pseudo-trajectory is the vector . Thus, a true trajectory is one whose 1-step errors are identically zero.
Definition of shadowing. The true trajectory -shadows on if for . The two stages, of proving that a shadow exists, are refinement and containment, defined below.
Definition. A shadow step is an interval, that can be larger than the internal timestep of a numerical integrator, across which a 1-step error is computed.
Definition. The pseudo-trajectory has a glitch at point if for some relevant there exists a true trajectory that -shadows for , but no true trajectory that -shadows it for , for .
The first group of chaotic systems for which it was proven that shadow orbits exist was hyperbolic systems [2, 7]. In a 2-dimensional hyperbolic system, there are two special directions called the unstable (or expanding) and the stable (or contracting) directions, which are generally not orthogonal. Small perturbations along the stable direction decay exponentially in forward time, while small perturbations in the unstable direction grow exponentially in forward time. The two directions reverse roles in backwards time. A ``trajectory'' for such a system can be imagined as a point moving in a 2D plane, evolving through a third dimension, representing time.
For such a system it was shown that, if the angle between the stable and unstable directions is uniformly bounded away from 0, then a noisy trajectory can be shadowed for all time. For non-hyperbolic systems, it appears that shadows may exist only for finite time. The most important question in this regard is, how long can a noisy orbit be shadowed? If the time is at least as long as most typical simulations of chaotic non-hyperbolic systems, then simulations have great validity; if the shadowing time turns out to be too short, then a less stringent error measure must be resorted to, such as one listed previously.
Definition. Refinement is an iterative process that perturbs each point of a noisy orbit in an attempt to produce a nearby orbit with less noise. A refinement iteration is successful if the trajectory before the iteration has noise and the trajectory after the iteration has noise , and , for some reasonable . Otherwise the iteration is unsuccessful.
Definition. Containment is a rigorous method to prove the existence of a shadow orbit. See GHYS for details. Although this is a good area for further work, containment is beyond the scope of this thesis.
The refinement algorithm that concerns us in this thesis is the one first presented in two dimensions by GHYS, and generalized to handle arbitrary Hamiltonian systems by Quinlan and Tremaine [29], hereinafter referred to as QT.
QT make the distinction between dynamical noise and observational noise. Observational noise does not effect the future evolution of the system. Laboratory measurements of a macroscopic system are usually of this type; another example is computer output that prints fewer digits than are represented internally. In contrast, dynamical noise does effect the future evolution of the system. The noise introduced by numerical solution of a system of ODE's is dynamical. QT studied some existing noise-reduction algorithms in an attempt to refine noisy trajectories of chaotic systems, but none worked as well as that presented by GHYS. QT postulates that this may be because the other noise reduction algorithms were designed to reduce observational noise, whereas the GHYS procedure was designed to reduce dynamical noise in a chaotic system. | 933 | 4,584 | {"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} | 2.984375 | 3 | CC-MAIN-2018-43 | latest | en | 0.903982 |
https://lancek4.com/2014/11/page/2/ | 1,548,326,574,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584520525.90/warc/CC-MAIN-20190124100934-20190124122934-00148.warc.gz | 574,351,397 | 8,476 | Uncategorized
## The 4th Dimension
If we were exclusively 3 dimensional beings, then we would not be able to experience a 3 dimensional world. Why? Below is the reason:
Construction:
1. Circle is a geometrical construct that needs two dimensions for its construction.
2. The construction of a circle is possible only in a 2-D space.
3. Two dimensional space is a necessary and sufficient condition for the construction of a circle.
4. When we say that a certain geometrical or topological construct is N dimensional we mean that an N dimensional space is the necessary and sufficient condition for its construction.
Perception:
1. Circle as a 2-D construct can be perceived fully only from above, from above the 2-D space in which it is constructed.
2. To fully perceive a circle one needs to transcend the circle’s space.
3. One has to be transcendent to the circle in order to perceive the circle, or else the circle itself…
View original post 323 more words | 208 | 967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-04 | latest | en | 0.93464 |
https://chem.libretexts.org/Courses/University_of_Kansas/KU%3A_CHEM_110_GOB_Chemistry_(Sharpe_Elles)/06%3A_Introduction_to_Chemical_Reactions/6.1%3A_The_Law_of_Conservation_of_Matter | 1,674,892,306,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00596.warc.gz | 180,071,274 | 30,025 | # 6.1: The Law of Conservation of Matter
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Skills to Develop
• Correctly define a law as it pertains to science.
• State the law of conservation of matter.
In science, a law is a general statement that explains a large number of observations. Before being accepted, a law must be verified many times under many conditions. Laws are therefore considered the highest form of scientific knowledge and are generally thought to be inviolable. Scientific laws form the core of scientific knowledge. One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved.
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter.
## Law of Conservation of Mass
According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.
$\overbrace{\underbrace{\ce{HgO (s)}}_{\text{100 g}}}^{\text{Mercuric oxide}} \rightarrow \underbrace{\overbrace{\ce{Hg (l) }}^{\text{Mercury}}}_{\text{92.6 g}} + \underbrace{\overbrace{\ce{O2 (g)}}^{\text{Oxygen}}}_{\text{7.4 g}}$
The law of conservation of mass is also known as the "law of indestructibility of matter."
Example $$\PageIndex{1}$$
If heating 10 grams of $$\ce{CaCO3}$$ produces 4.4 g of $$\ce{CO2}$$ and 5.6 g of $$\ce{CaO}$$, show that these observations are in agreement with the law of conservation of mass.
A sample of calcium carbonate (CaCO3). Image used with permission (Public Domain; Walkerma).
SOLUTION
• Mass of the reactants: $$10 \,g$$
• Mass of the products: $$4.4 \,g+ 5.6\, g = 10\, g$$.
Because the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.
What does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and type of atoms after the chemical change as were present before the chemical change.
Before looking at explicit examples of the law of conservation of matter, we need to examine the method chemists use to represent chemical changes.
Exercise $$\PageIndex{1}$$
1. What is the law of conservation of matter?
2. How does the law of conservation of matter apply to chemistry?
The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
## Summary
The amount of matter in a closed system is conserved.
## Exercises
1. Express the law of conservation of matter in your own words.
2. Explain why the concept of conservation of matter is considered a scientific law. | 1,199 | 4,622 | {"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} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.677406 |
https://www.studypool.com/discuss/1175471/i-don-t-know-i-don-t-know-if-the-statistics-is-done-on-this-site-however-here-is-my-proble | 1,477,603,298,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721392.72/warc/CC-MAIN-20161020183841-00441-ip-10-171-6-4.ec2.internal.warc.gz | 992,651,354 | 750,902 | # I don’t know I don’t know if the statistics is done on this site however here is my proble
Sep 18th, 2015
FratBro23
Category:
Statistics
Price: \$5 USD
Question description
Listed below are the errors between the predicted temperature and the actual temperature of a certain city. Find the mean and median for each of the two samples. Do the means and medians indicate that the temperature predicted one day in advance are more accurate than those predicted five days in advance, as we might expect?
(actual high)-(predicted high one day earlier) 0,0,0,3,2,2,0,0,-2,0
(actual high)-(predicted high 5 days earlier)
(Top Tutor) Daniel C.
(997)
School: Boston College
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## Review from our student for this Answer
SoccerBoss
Sep 18th, 2015
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https://gr.spek.gr/9ax4l/excel-array-constant-with-cell-reference-7dcd10 | 1,623,639,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00031.warc.gz | 288,548,104 | 8,571 | Want to master Microsoft Excel and take your work-from-home job prospects to the next level? [Calendar Year].&[2014] 2. An Array Constant is one of those Excel features that goes virtually unnoticed. In the last section, you saw how we can easily read from a group of cells to an array and vice versa. In the example shown, the LARGE function returns 3 values, which are "caught" by the SUM function, which returns the sum as a final result: The curly braces surrounding an array constant signify "constant" to Excel's formula parsing engine. When you enter an array formula, you most often use a range of cells in your worksheet, but you don't have to. You could also enter the formula as =SUM(A1*1,B1*2,C1*3,D1*4,E1*5). You can use the Paste Special function to multiply a range of cells by a number as follows:. To enter values in a single column, such as 3 cells in column C, you: Enter an equal sign and your constant. To work correctly, many array... A multi-cell array formula is an array formula that returns multiple results to more than one cell at the same time. There are two ways you can add the \$ sign to a cell reference in Excel. Select cell D9, then press F2 to switch to edit mode. This is a guide to Vlookup Table Array. Shortcut to add \$ Sign to Cell References. The problem I have is that all of my searching only results in information on how to go the other way (i.e. Enter an equal sign and your constant. Where A1 = [2015]. The term "array" comes from programming, but you'll hear it often in the context of more advanced Excel formulas. In this video, we'll answer the question "what is an array? The acronym "CSE" stands for "Control + Shift + Enter". Now that you're familiar with array constants, here's a working example. Another example of array constants inside the COUNTIFs function. There are two formats of the function, which are the Array Format (which is the most basic format), and the Range Format of the function. This notation ensures that when the formula is dragged across, it will reference Row 1 in the current column at all times. In your array formula, type an opening brace, the values you want, and a closing brace. [Calendar Weeks]. Array constants are often used in array formulas to create or manipulate several values at once, rather than a single value. Get the formula text for the formula we created in step 2 and store that in the InputText field. You just need to add the \$ symbol to the cell reference, or you can do the following steps: #1 select the cell reference in the formula box or double click on the cell reference that you want to make it as constant. [Calendar Weeks]. Enter the rest of your formula and press Ctrl+Shift+Enter. provide the column and row and ‘find’ the value). You may have selected a range of cells that doesn't match the number of elements in your constant. If you don't select enough cells, Excel omits the values that don't have a corresponding cell. For example, the LARGE function can be used with an array constant like this to get the top 3 values in B3:B11: In many cases, formulas that use array constants do not require Ctrl+Shift+Enter, even though they are in fact array formulas. Press Ctrl+Shift+Enter, and the constant looks like this: In geek terms, this is a one-dimensional horizontal constant. ", array constants inside the COUNTIFs function, I've watched three of your videos this morning and while I would consider myself somewhat advanced, I thought the tutorials were helpful and extremely well done. In a mixed cell … CHOOSE(H5, Table1, Table2) The CHOOSE function returns a value from a list using a given position or index. Fortunately, the header row of a data table always stays in the same place, so you can lock it like a normal cell reference in Excel. The CUBEMEMBER function used an array of values for retrieving data: =CUBEMEMBER(“AW BI Cube”,{“[Date]. To return a reference to specified cells. Entire formula (the array constant goes where the {#####} is): {=SUM ( ( (1-References!M1:M12)* ( {#####}* (G3:G14+F3:F14-0.11)))+ ( (References!M1:M12)* ('Chart Data'!A12:A23* (G3:G14+F3:F14-0.11)))+ ( (H2:H13*X3:X14)+ (H3:H14*Y3:Y14)+ (I2:I13* (V3:V14-X3:X14))+ … In any blank cell, enter (or copy and paste) this formula, and then press Ctrl+Shift+Enter: What happened? In the example, the numeric index is provided in column H. When you provide an array constant to an Excel function as an argument, you will often receive more than one result in an array. Well, most of the Excel users don’t even know about the existence of R1C1 reference style. Create an array constant from existing values. ts \$ Symbol in excel is used to lock a specific cell or rows or columns in a worksheet, the shortcut to lock down a reference in excel is by pressing ALT + F4, this feature is used while working in formulas when we do not want the reference to be changed when we copy or drag down the formula to other cell references, this feature will keep the reference same for all the further calculations. As a result, you can't include references or functions in an array constant, since doing so would make an array constant variable (i.e. Inside braces: { } formula above, that ’ s exactly we! Spilled array formula should still be = C8: E10 co-founder Kasper Langmann Microsoft! Another function that handles arrays natively selected by the row reference or both Accounts.... Of more advanced Excel formulas \$ a \$ 2 format of cell reference when creating formulas mathematical on... Curly braces { }: =SUM ( A1: A100=C2 ) ) {! Values you just enter in the cell references to values braces ( { ) }, and excel array constant with cell reference. Langmann, Microsoft Office Specialist now that you 're familiar with array constants, here a! Inputvalidation cell need the data from the cells to an array we can the... To keep a cell that lies in a formula is entered with CSE, Excel automatically wraps formula. With my wife, Lisa '' Quarter2 '' ; '' Quarter 3 '' } it precede... Which Excel array constants, here 's an example: =SUM ( A1: A100=C2 ) ) omits... Row with commas, and charts across, it will reference row 1 in the constant with other one.. A1: E1 * { 1,2,3,4,5 } ) the choose function returns a reference to a cell reference styles is! Seen in C8 through E10 reference to a cell in Excel: absolute, Relative, the! R1C1 reference style reference row 1 in the constant is one of those Excel features goes! Excel index function returns a reference to an array, selected by the row and ‘ find ’ the of! Type those braces manually to cell from the array formula and convert that into its component.! Find ’ the value returned in the example, if your constant so it 's easier use. The Excel users don ’ t accep… Let me try to explain the changes with a basic. Function will only accept a cell address that contains a dollar sign ( )! That goes virtually unnoticed dates into the worksheet, and then reference these cells as holidays the... To use it and found it more convenient than A1 answer the ``! ] with & A1 wife, Lisa or by themselves examples of formulas, it reference. ] with & A1 often used in array formulas to create or manipulate several at... Comes from programming, but you 'll hear it often in the current column at all times just! The choose function returns a reference to a cell constant in a formula into each cell where you a... Those braces manually want deal with these arrays using another function that handles arrays natively [ Accounts ] &... ) to the reference of that lookup value which related to array Table, always use the Paste Special to!, or a range here is a two-dimensional constant because it fills and... To appear when creating formulas '' } when creating formulas the current column at all times enough cells, omits! Excel provides a quick way to apply a mathematical operation on a range to help you faster... Into the worksheet, and a closing brace ) to the cells you ca n't create a three-dimensional,. To values functions, pivot tables, conditional formatting, and mixed references Explained constants do n't have a cell. Have selected a range of cells this is a one-dimensional vertical constant the formula bar inside:!, always use the reference does n't match the number of rows and columns you matches. Users don ’ t simply replace & [ 2014 ] with & A1 'm Dave,! As input a 3x3 array of cells \$ symbol automatically will add the \$ sign to cell... '' Quarter2 '' ; '' Quarter2 '' ; '' Quarter2 '' ; '' Quarter 3 ''.... Handles arrays natively reference these cells as holidays in the B2 cell cell or range of cells the data... Type an opening brace, the row reference or both Excel users ’... To reference a cell that lies in a specified row and column number indexes when a formula entered. In an Excel cell or array of cells that does n't match the of... Bar inside braces excel array constant with cell reference { } surrounding the values that do n't select enough cells, Excel omits values... That lookup value which related to array Table ca n't create a three-dimensional constant, a cell reference as.! Array Table, always use the Paste Special function to return the value of an in... That ’ s an array, selected by the row reference or both, and the is... So it 's easier to use again I have is that all of my searching only results information. And three rows, select that many columns and rows the second is.... Lookup value which related to array Table, always use the reference of that lookup which. Returned in the InputText cell in Excel is an array we can ’ the.
excel array constant with cell reference 2021 | 2,193 | 9,597 | {"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} | 2.921875 | 3 | CC-MAIN-2021-25 | latest | en | 0.903296 |
https://encyclopedia2.thefreedictionary.com/Foot-pound+force | 1,571,275,108,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00034.warc.gz | 488,209,739 | 11,051 | # foot-pound
(redirected from Foot-pound force)
Also found in: Dictionary, Thesaurus, Medical, Wikipedia.
## foot-pound,
abbr. ft-lb, unit of workwork,
in physics and mechanics, transfer of energy by a force acting to displace a body. Work is equal to the product of the force and the distance through which it produces movement.
or energyenergy,
in physics, the ability or capacity to do work or to produce change. Forms of energy include heat, light, sound, electricity, and chemical energy. Energy and work are measured in the same units—foot-pounds, joules, ergs, or some other, depending on the system of
in the customary English gravitational system; it is the work done or energy expended by a force of 1 pound acting through a distance of 1 foot. It is equal to 1.356 joulesjoule
, abbr. J, unit of work or energy in the mks system of units, which is based on the metric system; it is the work done or energy expended by a force of 1 newton acting through a distance of 1 meter. The joule is named for James P. Joule.
. The term foot-pound is also used to designate a unit of torquetorque,
in physics, that which tends to change the rate of rotation of a body; also called the moment of force. The torque produced by rotating parts of an electric motor or internal-combustion engine is often used as a measure of its ability to do useful work.
that is sometimes called the pound-foot to distinguish it from the energy unit. A force of 1 pound applied 1 foot from and perpendicular to the direction to an axis of rotation produces a 1 foot-pound (or pound-foot) torque at the axis.
## foot-pound
[′fu̇t ¦pau̇nd]
(mechanics)
Unit of energy or work in the English gravitational system, equal to the work done by 1 pound of force when the point at which the force is applied is displaced 1 foot in the direction of the force; equal to approximately 1.355818 joule. Abbreviated ft-lb; ft-lbf.
Unit of torque in the English gravitational system, equal to the torque produced by 1 pound of force acting at a perpendicular distance of 1 foot from an axis of rotation. Also known as pound-foot. Abbreviated lbf-ft.
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Open / Close | 518 | 2,152 | {"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} | 3.53125 | 4 | CC-MAIN-2019-43 | longest | en | 0.940348 |
http://www.notimeforflashcards.com/tag/math-activities | 1,408,814,512,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500826322.0/warc/CC-MAIN-20140820021346-00003-ip-10-180-136-8.ec2.internal.warc.gz | 527,493,278 | 54,391 | ## Shoot & Add { Nerf Gun Math For Kids }
This outdoor math for kids can be done with beanbags, baseballs, water balloons… or Nerf guns like we used. My mantra is to use what they love to make them love learning and that is what we did. Nerf guns are big news at our house right now because I just relented and let my kids have them. Using the darts are an outside only activity although they are loved less for their shooting ability and more for a pretend play prop. Either way they are the bright shiny new and novel item at our house so I used them to make learning fun after school. My goal with all the activities in the Learning After School series is to make the activities for school age kids that are educational without making them feel like homework. Here is what we did with this math for kids activity and yes they LOVED it!
Gather your materials. You will need some plastic cups, a permanent marker, something to use to write down the tally, a nerf blaster , and eye protection. If you don’t want to use a Nerf gun you could use a bean bag or ball.
Start by writing out numbers on the cups. We did 1-6 but you could number them anything you like. Set them up in a pyramid.
Knock those numbers down!!
Add all the numbers on the cups together and record on the chalkboard.
This activity was made for my 6 year old not my 3 year old but so when she shot them down I had her simply identify the numbers on the cups. I am planning on doing this again with her using a bean bag instead of a Nerf gun and the goal of hitting specific cups down. I will post it when I do.
Set the cups back up and go again. Add the tally up. I did the addition for my daughter but got her to write some. My son did his own writing. It’s not a ton of practice but it’s a little something in the midst of a fun activity so I will take it!
Like I said this was designed for my almost 7 year old not my 3 year old. She did well with it and had fun but I would never use this activity with a group of 3 year olds – nerf guns while safe to play with still hurt if you get one in the eye or at close range which is why they had safety glasses on. You know best what your kids are ready for but I just wanted to be clear that this is meant for our big kids !
## Jar Of Hearts – Valentine’s Day Math Printable
This is another part of our learning after school series. I try to engage my son in fun things after school and when I asked him what he wanted to work on he said math. He loves math and comes by it very very naturally. His dad has a bachelors degree in engineering with a minor in math and a MBA in finance. Yeah numbers are well loved in our house . Even I love math although I didn’t always. It wasn’t until I was teaching it to students that I found the fun in numbers.
1. Gather your materials. You will need a jar , some hearts ( any hearts will do – we used foam ones but paper, heart erasers, heart candies will all work), The printable found here and again below and a pencil with an eraser.
3. Lay out the sheet, pencil and hearts and let your little math whiz at it. Of course this is more than just a math lesson. My son is working on writing numbers in kindergarten right now and this is my tricky way of practicing without making him write them over and over again. If the 8 questions are too much do 2 or 3 and put it away for a day or two and complete the rest after. Learning isn’t a sprint it’s a marathon so go slow if needed! He estimated, counted, sorted and counted and wrote some more.
## Shape Princess Craft
After the holidays we packed up the kids and went to Disneyland. We had a ball and my already into princesses daughter is now officially obsessed. When I asked her what sort of craft she wanted to make she said “Princess!” so I grabbed a few simple materials and decided to focus the craft on shapes since we’ve been talking a lot about them lately.
1. Gather your materials. You will need 2 sheets of construction paper , scissors, glue , sequins, googly eyes and a marker or two.
2. Start by cutting out the shapes. I kept this so simple for my 2 year old because she likes to do it all herself and I want to encourage that. I cut out a circle for a head,2 rectangles for her arms, 3 small triangles for a crown, a large triangle for her gown and a square for her torso. Encourage older children to draw and cut out their own shapes.
3. Time to call your little princess, prince or court jester to explore. My daughter sorted through the shapes labeling each one and counting them then I grabbed the glue.
4. Glue! To give her a guide I drew the shape in glue on the paper. Older children could skip this. As you can see it’s still a big challenge . She was so excited when she found the correct shape each time.
5. We turned the princess upside down to glue on her crown.
6. Next up the sequins. I included these for a few reasons beyond their sparkle. They are small and challenging to pick up which makes them a really good fine motor skills building exercise. Also when I do something as structured as this activity is I try to make sure there is a completely free creative element to it. So I gave her the glue and let her at it.
7. She needed a little help with the crown jewels.
9. Add hair and a smile if you want.
10. Let dry. She was very excited to show her Belle doll the princess she made.
The Princess Knight by Cornelia Funke is a tale about a princess named Violet who was raised with her 3 older brothers by her father after her mother dies in childbirth. Her brothers are trained to be knights and she to learns to joust , ride horses and sword fighting. Her brothers ( like most) tease her and tell her that she’ll never be as strong and it’s a maid who tells her that she won’t be as strong but she can be smarter. That message stood way out for me and is why I think this is such a great book. When her father sets up a tournament for Violet’s hand in marriage she takes things into her own hands. She shows everyone how she is smarter than all the other knights and with hard work trains to win her own hand in marriage . I love this book not only as a great empowering one for girls but also to show boys that girls don’t have to fit a specific mold either.
The Storytelling Princess (Picture Puffin Books) by Rafe Martin is a cute story about an independent princess who saves herself after a shipwreck and even though she ends up with the prince her parents had arranged for her to marry she does it on her own terms. Her bibliophile prince isn’t interested in an arranged marriage either and ends up agreeing to marry if his father can find someone to tell him a story that he doesn’t know the ending to. The princess saves herself from the sea and dressed as a sailor she ends up vying for the King’s prize by telling the prince her own story. After she reveals herself and in the end the betrothed fall in love . I love that we don’t loose the happily ever after but that the prince and princess are unique and strong .
The Paper Bag Princess by Robert Munsch has enchanted me for years. I bought it while volunteering as a leader in training at a day camp when I was a teen. It’s followed me to many schools, children I babysat and finally my own son. I can’t remember one child ever not liking it. It’s a story of Elizabeth a princess who outwits a dragon to rescue her prince. I love that the author has switched the typical damsel in distress and has the princess as the heroine. Some parents have expressed concern about Elizabeth calling the prince a “Bum” in the end of the book, personally I love it. I have always used it to explain why she was so angry, and as a reminder why calling names hurt. That said I think she is totally justified !
Not All Princesses Dress in Pink by Jane Yolen .This book is not so much about being different and facing adversity but about being yourself even if the world has decided you should fit perfectly into the mold it’s given you. When we think of Princesses we think pink, sparkly and dry clean only! These princesses can’t be pigeonholed, they do what’s in their heart not what’s expected of them just because they are princesses. The princesses have all different interests, all different looks and I love that there are some with glasses too. Strong girls being themselves isn’t too different but for a book about princesses it is and it’s refreshing to read.
## Hot Chocolate Math
My son loves math and adding marshmallows into the mix pretty much made this activity one of the highlights of our snow storm fun last week. You don’t have to add the contact paper but by adding it becomes reusable and I was able to quickly make the activity more challenging when my original problems were too simple. If marshmallows are not something you want to use as a manipulative try buttons, packing peanuts or cotton balls.
1. Gather your materials. You will need 3 full sheets of construction paper ( backing, cups and strips of hot chocolate), a washable marker, double stick tape, a bunch of mini marshmallows, a white crayon for the steam, scissors, a tray or cookie sheet to keep it all together and a barely damp baby wipe for erasing the marker. I like dollar store cookie sheets because they keep the manipulatives in one place and I can attach the contact paper on top .
2. Start by drawing a mug and cutting it out. I then used it as a template for 4 more.
3. Tape the mugs down , add a strip of brown to suggest hot chocolate and add some stem with a white crayon.
4. Cover with contact paper. Don’t you hate it when you make a ridge in the contact paper , the stuff is too costly to just redo it too. I dream of one day having a laminator….
5. Using a washable marker write numbers or basic equations. I started with the ones above but as you will see had to change it to offer more of a challenge. That’s the beauty of the write and wipe surface.
7. I quickly changed 3+1 into 3-1 and the 5 into a 5+3 to test the waters.
8. Magic of manipulatives- not a problem. Finding just the right level of challenge is a bit of trail and error. My son didn’t think he could do 9-1 but he did. That is the perfect zone when they feel unsure of success but are willing to take a chance and take on the challenge. That is when new connections are being made. This stuff excited me beyond words.
9. This is such a great activity because you can wipe it clean and customize it to your needs 2 months from now or right away for another learner.
## Counting Around The House – Math Activity
This activity was inspired by my Halloween candy counting over at my other blog . Seeing how resistant my son is to writing tells me one thing, that he needs to practice lots but it needs to be within activities he loves. This is a math activity with gross motor, and some writing on a vertical surface which is great for beginning writers because it forces the correct wrist position and strengthens the correct muscles in the hand and arm.
1. Gather your materials. You will need some paper ( I get this craft paper in the mailing supply section of Walmart , way cheaper than real butcher paper), markers and something to put the paper on the wall with. I use painter’s tape and I use a lot so my toddler doesn’t pick it off.
2. Draw a house .
3. Decide what things you want to count and write them on. I made the windows and doors spots to write the numbers in. | 2,548 | 11,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-35 | longest | en | 0.972685 |
https://adlmag.net/is-rough-lumber-cheaper-2/ | 1,686,332,808,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00799.warc.gz | 103,308,403 | 25,647 | Basically, rough sawn lumbers are thicker as well rougher in surface than the finished woods. If you are on a tight budget, opting for an air dried lumber makes a lot of sense, as it is cheaper than its kiln dried counterparts.
Also, How thick is a 1×6?
Nominal vs. Actual Measurements of Dimension Lumber
Nominal Size Actual Size
1 x 6 3/4 x 5 1/2 inches (19 x 140 mm)
1 x 8 3/4 x 7 1/4 inches (19 x 184 mm)
1 x 10 3/4 x 9 1/4 inches (19 x 235 mm)
1 x 12 3/4 x 11 1/4 inches (19 x 286 mm)
In this way, How long is a 1×6?
Usually, there is 1/2 inch difference in measurements over 2 inches and 1/4 inch in measurements less than 2 inches.
Finished Lumber Sizes.
NOMINAL SIZE ACTUAL SIZE
1×6 3/4″ x 5 1/2″
1×8 3/4″ x 7 1/4″
1×10 3/4″ x 9 1/4″
1×12 3/4″ x 11 1/4″
Contents
## How many pixels is a 4×4 photo?
For example, you have an image that has a width and height of 400 x 400 pixels, has a physical size of 4×4 inches, and has a resolution of 100 pixels/inch.
## How long is a 1×6 board?
A 1×6, for example, actually measures 3/4″ x 5 1/2″. In this case, the planing is necessary for a smoother surface, so that the board could be used for interior applications. The length of the boards is generally exactly what you order, though we usually only stock pieces of even lengths.
## How thick is a 2×3?
Nominal vs. Actual Measurements of Dimension Lumber
Nominal Size Actual Size
1 x 12 3/4 x 11 1/4 inches (19 x 286 mm)
2 x 2 1 1/2 x 1 1/2 inches (38 x 38 mm)
2 x 3 1 1/2 x 2 1/2 inches (38 x 64 mm)
2 x 4 1 1/2 x 3 1/2 inches (38 x 89 mm)
## What is the actual size of a 5×5 post?
Softwood lumber is generally used for framing walls and floors. In the United States, softwood lumber is governed by the National Institute of Standards American Softwood Lumber Standard (PS 20).
North American Standard 6-by Lumber.
Nominal Size (inches) Actual Size (inches) Actual Size (mm)
6 x 6 5 1/2 x 5 1/2 140 x 140
## What type of 2×4 is used for framing?
The real dimensions of light structural lumber and timber for framing homes:
Dimensional Lumber: Nominal Size vs. Actual Size
Two-by-four or 2 x 4 1 1/2 inches x 3 1/2 inches
Two-by-six or 2 x 6 1 1/2 inches x 5 1/2 inches
Two-by-eight or 2 x 8 1 1/2 inches x 7 1/4 inches
Two-by-ten or 2 x 10 1 1/2 inches x 9 1/4 inches
## How do you read dimensions?
For example, a window that is 24 inches wide by 30 inches tall would be written as 24″ X 30″. In the manufacturing industry, this standard window size is referred to as a 2030 or 2 feet by 3 feet. In a rectangular swimming pool, the dimension might read 16′ X 30′ X 9′ or 16 feet wide by 30 feet long and 9 feet deep.
## Is common board OK for outside?
For starters, the most common wood used outdoors in our area is treated yellow pine. This is primarily due to treated yellow pine being an economical option. The pressure treatment process should allow the chemicals to reach all the way through the yellow pine board to protect even when the boards are cut.
## What is actual length of 2x4x8?
Finished Lumber Sizes
NOMINAL SIZE ACTUAL SIZE
2×3 1 1/2″ x 2 1/2″
2×4 1 1/2″ x 3 1/2″
2×6 1 1/2″ x 5 1/2″
2×8 1 1/2″ x 7 1/4″
## Why is lumber not actual size?
Maybe you’ve noticed that lumber sizes are often misleading. The “nominal” cross-section dimensions of a piece of lumber, such as 2 X 4 or 1 X 6, are always somewhat larger than the actual, or dressed, dimensions. The reason is that dressed lumber has been surfaced or planed smooth on four sides (called S4S).
## How much do wooden boards cost?
But I would buy these boards again. In Stock. Only 16 left in stock – order soon. Smooth nice wooden panels were awesome for mounting 12×12 photos that I took.
Top Selected Products and Reviews.
List Price: \$49.99
You Save: \$7.12 (14%)
## What comes first length or width or height?
It needs to be written Length X Width X Height. That is standard for measurements. It makes no difference in the order you have them listed.
## Why is wood not actual size?
When referring to soft wood that is milled and used for building, we call it by the nominal measurement and not the actual measurement. A 2×4 has been called a “2 by 4” since, well… forever. Ultimately, the reason reason wood varies from the advertised size is due to the shrinking that occurs during the drying process.
## How much does a 1×6 cost?
Lumber Prices: Estimating the Cost of Dimensional Lumber
Board Per ft* 6′
1×6 \$0.88 \$5.37
1×8 \$0.75 \$5.98
1×10 \$0.98 \$7.98
1×12 \$1.75 \$9.98
## How many board feet are in a 2×4?
Back in the olden days of cutting trees into framing lumber, a 2×4 did actually measure more or less two inches by four inches. About the same time, dried lumber became popular for its dimensional stability and resistance to mold.
## How many board feet are in a 2×4?
Two-by-fours are actually 1.5-by-3.5s. The same is true for most other wood cuts, which are all systematically smaller than their names would suggest. Here’s why. And moist wood—called “green lumber”—is prone to bending and warping.
## How thick is a 1×3?
Lumber Dimensions
Nominal Actual Actual – Metric
1″ x 2″ 3/4″ x 1-1/2″ 19 x 38 mm
1″ x 3″ 3/4″ x 2-1/2″ 19 x 64 mm
1″ x 4″ 3/4″ x 3-1/2″ 19 x 89 mm
1″ x 5″ 3/4″ x 4-1/2″ 19 x 114 mm
## How much weight can a 2×4 hold?
Here you can operate by assumptions that there is no wind or it is not as strong. If there is no strong wind, a 2×4 measuring at least 8 feet can support at least 1,000 pounds vertically. Such calculations would hold if, for instance, the load is square.
## What size do 2×4 come in?
A piece of surfaced (sanded smooth) 2×4 lumber actually measures 1½ inches thick and 3½ inches wide. In rough-cut condition, a 2×4 is slightly less than 2 inches thick and approximately 4 inches wide.
## What is poplar wood used for?
Common Uses: Seldom used for its appearance, (except in the case of Rainbow Poplar), Poplar is a utility wood in nearly every sense. It’s used for pallets, crates, upholstered furniture frames, paper (pulpwood), and plywood.
## How do you size a tattoo?
Pine Common Board-458503 – The Home Depot.
## What type of wood is common wood?
Hickory is the hardest common wood that is commercially available, followed by pecan, hard maple and white oak. Softwood includes cedar, pine, spruce and fir wood.
## How many boards do I need?
A single board foot is one square foot of wood that is one inch thick. You can measure board feet by multiplying length x width x thickness in inches, and then dividing by 144. If ordering a large amount by board feet, just multiply the total board feet needed by the price per board foot for the total cost.
## How many boards do I need?
An 8′-2×4 has 5.3 board feet. A 16′ 2×4 will have 10.6 board feet and a 14′-2×4 will have 9.3 board feet. multiply the height in inches times the width in inches times the length in feet and divide by 12. It’s easy to remember.
## What is dressed wood?
Well, dressed wood is the name given to a specific type of timber after cutting and treatment. Prior to becoming dressed wood, timber is first cut and harvested in its rawest form. Basically, trees are logged and then sent to a processing plant for further treatment.
## How many boards do I need to cover a wall?
Boards are most commonly sold in 12′ lengths, though 16′ and 20′ lengths can often be ordered. Divide the linear footage by 12′ to find the number of boards needed to cover the wall. For example, installing 21 courses of siding on a 20′ wide wall will require 420 linear feet of siding, or 35 boards.
## What causes the difference between nominal and actual dimensions?
Nominal measurements traditionally refer to the size of the board when it was first rough cut, before it was dried and planed (smoothed). Actual measurements are the actual final size of the lumber after it has been dried and planed. If the final product has been planed on all four sides, it will be denoted as S4S.
## What is the size of deck boards?
Length: When possible, buy decking that can run the full length of your deck. All lumberyards and home centers carry 16-ft. deck boards, but many also stock or can order 20- and 24-ft. boards, though they may cost more.
## What type of wood is Home Depot common board?
Boards are most commonly sold in 12′ lengths, though 16′ and 20′ lengths can often be ordered. Divide the linear footage by 12′ to find the number of boards needed to cover the wall. For example, installing 21 courses of siding on a 20′ wide wall will require 420 linear feet of siding, or 35 boards.
## What does 2×4 mean truck?
2X4 is just 2 wheel drive. Either its front wheel drive, like a civic. Or rear wheel drive like an NSX.
## How big is a 4×6 photo?
Pine Common Board-458503 – The Home Depot. | 2,543 | 8,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-23 | longest | en | 0.860847 |
https://kr.mathworks.com/matlabcentral/answers/119857-beginner-matlab-project-function-trouble-instructions-say-create-a-sub-function-that-takes-as-i | 1,669,972,608,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00464.warc.gz | 388,816,569 | 33,763 | # Beginner Matlab project... function trouble. Instructions say "Create a sub-function that takes as input the x and y values of a closed polygon and calculates x and y values for a new polygon that has vertices at the midpoints of the input polygon. "
조회 수: 3(최근 30일)
cassie 2014년 3월 2일
편집: cassie 2014년 4월 19일
Here is my code so far, I have a LOT more to do but am stuck on this step. Help is appreciated!
function project_2 %to test subfunctions
clc;
[xinitial,yinitial]=generatepoly(10);%call subfunction from Step2 to generate polygon using given values
[xclose,yclose]=closepoints(xinitial,yinitial); %call subfunction from Step3 to close polygon
plot_function(xclose,yclose,'bo-',4);
%(closepoints(xclose,yclose,'bo-',4))%plot closepoints function?
%[newx,newy]=iterationloop(xclose,yclose); %call subfunction from Step4
[midx,midy]=calcmidpoints(xclose,yclose); %call subfunction from Step5 to generate midpoints
end %end main function
function [xinitial,yinitial]=generatepoly(n) %generates given test values in a matrix
xinitial=[1 2 3 4 3 2 1 3]; %given test points
yinitial=[1 2 1 2 4 3 5 6]; %given test points
end %end Step2
function [xclose,yclose]=closepoints(xinitial,yinitial)
xclose=[xinitial(1:end),xinitial(1)];%connect ending x value to first x value to close poly
yclose=[yinitial(1:end),yinitial(1)];%connect ending y value to first y value to close poly
end %end Step3
function [newx,newy]=plot_function(xclose,yclose,linespec,I)%to plot closed polygon with specified line type a defined number of times
numberofpoints=length(xclose);%define variable to be the number of items in the list xclose
plot(xclose,yclose,linespec); %plot polygon with line type
txt = ['Number of points: ',num2str(numberofpoints)];%define variable txt to include number of points
title(txt)%display title including number of points
end %end Step4, iteration function
function [midx,midy]=calcmidpoints(normalx,normaly)
for n=1:length(xclose); %while n is one less than number of inputs
midx(n)=.5.*(normalx(n)+normalx(n+1));
midy(n)=.5.*(normaly(n)+normaly(n+1));
end %end for loop
end %end Step5
##### 댓글 수: 6표시숨기기 이전 댓글 수: 5
Image Analyst 2014년 3월 2일
편집: Image Analyst 2014년 3월 2일
I didn't see it at first when I told Firefox to highlight all occurrences of the word, but I see it now. Maybe because it's not indented like usual I didn't notice it. You can type control-a and then control-i in MATLAB to "fix up" or standardize the indenting. Walter monitors this forum pretty actively so I'm sure he'll see your question, if he hasn't already, but he doesn't answer every question. Look who answers the questions here and you'll see Walter's name all over the place as well as here http://www.mathworks.com/matlabcentral/answers/contributors
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### 채택된 답변
Will 2014년 4월 2일
length(xclose) should be length(normalx) and this will work correctly. Too late, I know, because this referenced an assignment that I made :)
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
cassie 2014년 4월 19일
편집: cassie 2014년 4월 19일
Yes a little late Professor Schle... I mean "Will" :) Feel free to chime in on my most recent question though.. I want to use an image (a car) as a slider button for our last project.
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### 추가 답변(1개)
Image Analyst 2014년 3월 2일
Evidently it's not getting into your for loop and defining midx and midy. So what is the value of xclose? It seems your functions are not nested so actually I'm not even sure how it got past the length(xclose) without a "variable not defined error".
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Image Analyst 2014년 3월 2일
You can have a whole bunch of functions in a single m-file, as long as the m-file starts with a function and not a script. If the functions are not nested, then each function has only it's own internal variables and whatever was passed in. If a function is nested, contained within another, then it can see what ever variables are in the parent function when the nested/contained function is called.
Put a breakpoint in that function and when it stops there, see what the value of xclose is. Then step line by line and see if it ever gets inside the if. It's always best to define some default values so if something happens it will return something, even if it's just null. For example
function [midx,midy] = calcmidpoints(normalx,normaly)
midx = 0;
midy = 0;
whos xclose
class(xclose)
for n=1:length(xclose); %while n is one less than number of inputs
midx(n)=.5.*(normalx(n)+normalx(n+1));
midy(n)=.5.*(normaly(n)+normaly(n+1));
end % end for loop
end % of calcmidpoints()
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Translated by | 1,382 | 4,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-49 | latest | en | 0.545967 |
https://www.winsteps.com/winman/categoryboundaries.htm | 1,516,304,684,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00608.warc.gz | 1,018,188,932 | 6,828 | Category boundaries and thresholds
# Category boundaries and thresholds
Conceptualizing rating scales and partial-credit response structures for communication can be challenging. Rasch measurement provides several approaches. Choose the one that is most meaningful for you.
Look at this edited excerpt of Table 3.2:
# ------------------------------------------------------------(Mean)---------(Median)---
Here are three ways of conceptualizing and communicating the transition, threshold, boundary between category 1 and category 2:
(1) Rasch-half-point thresholds (Zone). Someone at the boundary between "1" and "2" would have an expected rating of 1.5, or 1000 persons at the boundary between "1" and "2" would have an average rating of 1.5. This boundary is the "zone" = "Expected Measure at 1+0.5 or 2 -0.5" which is -.03 logits, the Rasch-half-point threshold. To illustrate this, use the model item characteristic curve. The expected score ogive / model ICC (Table 21.2 - second on list in Graphs menu). The CAT+.25, CAT-0.5, AT CAT, and CAT-.25 columns in the ISFILE= plot points on this ogive. The expected score ogive relates most directly to the estimation of the Rasch parameters. Since it is only one line, it is also convenient for summarizing performance at any point on the latent variable by one number. Crucial points are the points on the variable corresponding to the lower category value + 0.5, i..e, more than the higher adjacent category value - 0.5. These Rasch-half-point thresholds are "average score thresholds" or "Rasch-ICC thresholds".
(2) Rasch-Thurstone thresholds (50% Cumulative Probability). Someone at the boundary between "1" and "2" would have a 50% chance of being rated 1 or below, and a 50% chance of being rated 2 or above. This is the Rasch-Thurstone threshold of -.07 logits. To illustrate this, use the cumulative probability curves. The cumulative probability curves (Table 21.3 - and third on list in Graphs menu). The 50%PRB columns in the ISFILE= are the crucial points on these curves. and are the Rasch-Thurstone thresholds, useful for identifying whether a person is most likely to respond below, at or above a certain category.
(3) Rasch-Andrich thresholds. Someone at the boundary between "1" and "2" would have an equal chance of being rated 1 or 2. This is the Rasch--Andrich Threshold of -.15 logits. To illustrate this, use the category probability curves. The probability curves (Table 21.1 - and top of list in Graphs menu). The Andrich Threshold in the ISFILE= gives the point of equal probability between adjacent categories. The points of highest probability of intermediate categories are given by the AT CAT values. These probability curves relate most directly to the Rasch parameter values, also called Rasch-Andrich thresholds. They are at the intersection of adjacent probability curves, and indicate when the probability of being observed in the higher category starts to exceed that of being observed in the adjacent lower one. This considers the categories two at a time, but can lead to misinference if there is Rasch-Andrich threshold disordering.
d) Empirical average measures. For any particular sample, there is the average ability of the people who scored in any particular category of any particular item. This is the "Average Measure" reported in Table 3.2. This is entirely sample-dependent. It is not reported in ISFILE=. In the Table, the empirical average measure of the persons responding in category 1 is .47 logits, and in category 2 is 1.07 logits. Half-way between is .77 logits.
Help for Winsteps Rasch Measurement Software: www.winsteps.com. Author: John Michael Linacre
Forum Rasch Measurement Forum to discuss any Rasch-related topic
Rasch Publications
Rasch Measurement Transactions (free, online) Rasch Measurement research papers (free, online) Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch Applying the Rasch Model 3rd. Ed., Bond & Fox Best Test Design, Wright & Stone
Rating Scale Analysis, Wright & Masters Introduction to Rasch Measurement, E. Smith & R. Smith Introduction to Many-Facet Rasch Measurement, Thomas Eckes Invariant Measurement with Raters and Rating Scales: Rasch Models for Rater-Mediated Assessments, George Engelhard, Jr. & Stefanie Wind Statistical Analyses for Language Testers, Rita Green
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Journal of Applied Measurement Rasch models for measurement, David Andrich Constructing Measures, Mark Wilson Rasch Analysis in the Human Sciences, Boone, Stave, Yale
in Spanish: Análisis de Rasch para todos, Agustín Tristán Mediciones, Posicionamientos y Diagnósticos Competitivos, Juan Ramón Oreja Rodríguez
Winsteps Tutorials Facets Tutorials Rasch Discussion Groups
Coming Rasch-related Events
Jan. 5 - Feb. 2, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Jan. 10-16, 2018, Wed.-Tues. In-person workshop: Advanced Course in Rasch Measurement Theory and the application of RUMM2030, Perth, Australia (D. Andrich), Announcement
Jan. 17-19, 2018, Wed.-Fri. Rasch Conference: Seventh International Conference on Probabilistic Models for Measurement, Matilda Bay Club, Perth, Australia, Website
Jan. 22-24, 2018, Mon-Wed. In-person workshop: Rasch Measurement for Everybody en español (A. Tristan, Winsteps), San Luis Potosi, Mexico. www.ieia.com.mx
April 10-12, 2018, Tues.-Thurs. Rasch Conference: IOMW, New York, NY, www.iomw.org
April 13-17, 2018, Fri.-Tues. AERA, New York, NY, www.aera.net
May 22 - 24, 2018, Tues.-Thur. EALTA 2018 pre-conference workshop (Introduction to Rasch measurement using WINSTEPS and FACETS, Thomas Eckes & Frank Weiss-Motz), https://ealta2018.testdaf.de
May 25 - June 22, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
June 27 - 29, 2018, Wed.-Fri. Measurement at the Crossroads: History, philosophy and sociology of measurement, Paris, France., https://measurement2018.sciencesconf.org
June 29 - July 27, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Further Topics (E. Smith, Winsteps), www.statistics.com
July 25 - July 27, 2018, Wed.-Fri. Pacific-Rim Objective Measurement Symposium (PROMS), (Preconference workshops July 23-24, 2018) Fudan University, Shanghai, China "Applying Rasch Measurement in Language Assessment and across the Human Sciences" www.promsociety.org
Aug. 10 - Sept. 7, 2018, Fri.-Fri. On-line workshop: Many-Facet Rasch Measurement (E. Smith, Facets), www.statistics.com
Sept. 3 - 6, 2018, Mon.-Thurs. IMEKO World Congress, Belfast, Northern Ireland www.imeko2018.org
Oct. 12 - Nov. 9, 2018, Fri.-Fri. On-line workshop: Practical Rasch Measurement - Core Topics (E. Smith, Winsteps), www.statistics.com
Our current URL is www.winsteps.com | 1,722 | 6,897 | {"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} | 3.109375 | 3 | CC-MAIN-2018-05 | latest | en | 0.901144 |
https://www.riddles.com/index.php/problem-of-the-week?page=4 | 1,685,707,289,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00512.warc.gz | 1,053,575,764 | 25,061 | A riddle problem is a question or statement intentionally phrased so as to require ingenuity in ascertaining its answer or meaning, typically presented as a game. Below is a list of the past problem of the week riddles in descending chronological order.
Riddle: When can you add two to eleven and get one as the correct answer?
Answer: When you add two hours to eleven o'clock, you get one o'clock.
Riddle: If you have two coins which total 35 cents and one of the coins is not a dime, what are the two coins?
Answer: A quarter and a dime. One coin is not a dime, but the other one is.
Riddle: Three playing cards in a row. Can you name them with these clues? There is a two to the right of a king. A diamond will be found to the left of a spade. An ace is to the left of a heart. A heart is to the left of a spade. Now, identify all three cards.
Riddle: A car's odometer shows 72927 miles, a palindromic number. What are the minimum miles you would need to travel to form another?
Riddle: An item is made from lead blanks in a lathe shop. Each blank suffices for 1 item. Lead shavings accumulated for making 6 items can be melted and made into a blank. How many items can be made from 36 blanks?
Answer: From 36 blanks there are 36 items made. The lead shavings are enough to make 6 blanks. Which make 6 more items. But don't stop here. The new shavings are good for 1 more item. Total: 43.
Riddle: While on my way to St. Ives, I saw a man with 7 wives. Each wife had 7 sacks. Each sack had 7 cats. Each cat had 7 kittens. Kitten, cats, sacks, wives, How many were going to St. Ives? | 416 | 1,586 | {"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} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.965668 |
https://www.physicsforums.com/threads/centripetal-force-question-other-question.147747/ | 1,721,645,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00277.warc.gz | 804,744,081 | 15,761 | # Centripetal force question + other question
• demode
In summary, the conversation is discussing problems related to circular motion and vertical motion of objects. The first problem involves finding the mass of a dog sitting on a merry-go-round, given its angular speed and the force keeping it in circular motion. The second problem involves calculating the vertical displacement of an olympic runner during a hurdle jump, with an initial vertical speed given. The conversation also includes tips for solving the problems, such as re-writing equations and identifying relevant information.
demode
1. A dog sits 1.50 meters from the center of a merry-go-round with an angular speed of 1.20 rad/s. If the magnitude of the force that maintains the dog's circular motion is 40.0 N, what is the dog's mass?
2. Fc = (mv^2) / r
3. We were provided with the answer "18.5" but I do NOT understand how you can get this value using the aforementioned equation. I'll demonstrate what I did:
40 * 1.50 = (1.20^2)m
60 = 1.44m
m = 41.6 kg
can someone explain to me what I'm doing wrong?
1. An olympic runner leaps over a hurdle. If the runner's initial vertical speed is 2.2 m/s, how much will the runner's center of mass be raised during the jump?
2. NO IDEA
3. Don't even know WHERE to begin. Can someone help me out?
To start off, I'd assume that for the speed, you mean 1.20 m/s, right?
For your first problem, I'll suggest rewriting the equation. Re-write it to solve for what you're looking for. Then, try it again, and see what you come up with.
For the second problem, think about what you would need to know to solve the problem, then write them down. This could get you to think of how you could actually solve the problem.
It appears you were given an angular speed of 1.2rad/s. This is NOT the same as linear speed. You must find the equation that relates angular speed to linear speed (v), then plug in that value of v into your correct equation to solve for m.
In part 2, you need to know either the motion equation that relates velocity, acceleration, and displacement, OR the conservation of energy theorem that relates kinetic and potential energies. Are you familiar with either of these equations?
## 1. What is centripetal force?
Centripetal force is the force that acts on an object to keep it moving in a circular path. It is always directed towards the center of the circle.
## 2. How is centripetal force calculated?
Centripetal force can be calculated using the formula Fc = mv^2 / r, where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.
## 3. What is the difference between centripetal force and centrifugal force?
Centrifugal force is often confused with centripetal force, but they are actually two different forces. Centripetal force is the inward force that keeps an object moving in a circular path, while centrifugal force is the outward force that is experienced in a rotating reference frame.
## 4. What are some real-life examples of centripetal force?
Some real-life examples of centripetal force include a ball being swung around on a string, a car turning around a curve, and the Earth revolving around the sun.
## 5. How does centripetal force relate to Newton's laws of motion?
Centripetal force is related to Newton's first law of motion, also known as the law of inertia, which states that an object in motion will stay in motion unless acted upon by an external force. In this case, the centripetal force is the external force that is acting on the object to keep it moving in a circular path.
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1K | 1,008 | 4,181 | {"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} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.959426 |
http://www.jiskha.com/display.cgi?id=1196657409 | 1,498,436,705,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320595.24/warc/CC-MAIN-20170625235624-20170626015624-00106.warc.gz | 544,965,736 | 3,854 | # math calculus
posted by .
Find the are under the given curve
y=square root of x + 1
from x=0 to x=9
Please walk me through each step.
• math calculus -
Do you mean sqrt(x+1) or sqrt x + 1?
If the former, the indefinite integral is [(x+1)^3/2]/(3/2)
Evaluate it at x = 9 and x = 0 and take the difference.
Area = (2/3)*[(10)^(3/2) - 1] | 119 | 344 | {"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} | 3.328125 | 3 | CC-MAIN-2017-26 | latest | en | 0.863419 |
https://www.dsprelated.com/freebooks/filters/Numerical_Computation_Group_Delay.html | 1,558,684,199,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257553.63/warc/CC-MAIN-20190524064354-20190524090354-00501.warc.gz | 768,185,947 | 10,096 | ### Numerical Computation of Group Delay
The definition of group delay,
does not give an immediately useful recipe for computing group delay numerically. In this section, we describe the theory of operation behind the matlab function for group-delay computation given in §J.8.
A more useful form of the group delay arises from the logarithmic derivative of the frequency response. Expressing the frequency response in polar form as
yields the following logarithmic decomposition of magnitude and phase:
Thus, the real part of the logarithm of the frequency response equals the log amplitude response, while the imaginary part equals the phase response.
Since differentiation is linear, the logarithmic derivative becomes
where and denote the derivatives of and , respectively, with respect to . We may therefore express the group delay as
imim
Consider first the FIR case in which , with
(8.9)
In this case, the derivative is simply
where denotes ramped'', i.e., the th coefficient of the polynomial is , for . In matlab, we may compute Br from B via the following statement:
Br = B .* [0:M]; % Compute ramped B polynomial
The group delay of an FIR filter can now be written as
imimre
In matlab, the group delay, in samples, can be computed simply as
D = real(fft(Br) ./ fft(B))
where the fft, of course, approximates the Discrete Time Fourier Transform (DTFT). Such sampling of the frequency axis by this approximation is information-preserving whenever the number of samples (FFT length) exceeds the polynomial order . The ratio of sampled DTFTs, however, is undersampled, in general. In fact, we may have at some frequencies (zeros on the unit circle''). The grpdelay matlab utility in §J.8 watches out for division by zero, and simply sets the group delay to zero at such frequencies. Note that the true group delay approaches infinite magnitude as either a zero or pole approaches the unit circle.
Finally, when there are both poles and zeros, we have
where is given in Eq.(7.9), and
(8.10)
Straightforward differentiation yields
(8.11)
and this can be implemented analogous to the FIR case discussed above. However, a faster algorithm (usually) results from converting the IIR case to the FIR case:
(8.12)
where
may be called the flip-conjugate'' or Hermitian conjugate'' of the polynomial .8.4In matlab, the C polynomial is given by
C = conv(B,fliplr(conj(A)));
It is straightforward to show (Problem 11) that
The phase of the IIR filter is therefore
and the group delay computation thus reduces to the FIR case:
re
This method is implemented in §J.8.
Next Section:
Computing Reflection Coefficients to Check Filter Stability
Previous Section:
Vocoder Analysis | 598 | 2,704 | {"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} | 3.453125 | 3 | CC-MAIN-2019-22 | longest | en | 0.905066 |
https://www.cemetech.net/forum/viewtopic.php?t=16194&start=0&postdays=0&postorder=asc&highlight= | 1,675,030,904,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00670.warc.gz | 729,972,362 | 9,453 | Do you think it's possible? Should I attempt to make it?
Yes, its possible, but you shouldn't make it. 11% [ 1 ]
No, it's not possible, and you shouldn't make it. 0% [ 0 ]
Yes, it's possible, and you should make it. 88% [ 8 ]
No, it's not possible, but you should still try to make it. 0% [ 0 ]
okay so im a new user and i recently had a breakthrough with a program i made on my ti 84 plus ce
i don't want to discuss fully what it was, but i will explain one key concept that the program used- a matrix. using a 10 row by 26 column matrix based off of the max "output(" values, i was able to create a basic level data system
i need your opinion in the poll above
Sure, go for it. I assume this will be in TI-Basic?
epsilon5 wrote:
Sure, go for it. I assume this will be in TI-Basic?
It's in the title, lol
Pieman7373 wrote:
epsilon5 wrote:
Sure, go for it. I assume this will be in TI-Basic?
It's in the title, 0x5
I should have read that a little more carefully, I was focused on the actual question and that didn't really register for some reason. In an attempt to make a constructive comment, however, I think that the matrix thing should work pretty well (that's basically what I did for Ace Recon). The issue will be storing the level data.
twistedstriker1234 wrote:
okay so im a new user and i recently had a breakthrough with a program i made on my ti 84 plus ce
i don't want to discuss fully what it was, but i will explain one key concept that the program used- a matrix. using a 10 row by 26 column matrix based off of the max "output(" values, i was able to create a basic level data system
i need your opinion in the poll above
This sounds pretty cool. I once did something sort of similar. By using the 15 TI-Basic Color numbers [e.g. blue =1, red=2], I was able to create a matrix containing the color number which would represent each "tile" in the tilemap. Downside? Each "tile" was a solid color and each time the screen had to reload to move to a different part of the tilemap, it took too long to load, as it was written in TI-Basic.
I look forward to seeing your game and how you overcome TI-Basic's limitations in making a non-text based rpg!
wow! the results are overwhelming.
i was planning to begin after the poll closed but i think i should get started now!
thanks to everyone who gave feedback.
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You cannot vote in polls in this forum | 746 | 3,010 | {"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-2023-06 | latest | en | 0.977161 |
https://www.indiabix.com/civil-engineering/water-supply-engineering/095005 | 1,652,996,954,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00700.warc.gz | 913,837,400 | 6,126 | # Civil Engineering - Water Supply Engineering
### Exercise :: Water Supply Engineering - Section 4
21.
Dead storage of a reservoir of Q capacity generally provided for silt deposition during its life time, is generally kept
A. 1/8th Q B. 1/6th Q C. 1/5th Q D. 1/4th Q E. 1/3rd Q.
Explanation:
No answer description available for this question. Let us discuss.
22.
If the specific capacity of a well is 0.3183 x 10-3 per sec, the discharge from a well of 4 m diameter under a depression head of 4 m, is
A. 8 litres/sec B. 10 litres/sec C. 12 litres/sec D. 14 litres/sec E. 16 litres/sec.
Explanation:
No answer description available for this question. Let us discuss.
23.
The factor to be considered for the source of city water supply, is
A. quantity and quality of the available water B. elevation of the source of water C. general terrain intervening the area D. all the above.
Explanation:
No answer description available for this question. Let us discuss.
24.
Pick up the correct statement from the following :
A. Domestic use of water is 50% of total consumption B. Average consumption of commercial use of water is 25% of total consumption C. Consumption of water on public use is 10% of total consumption D. Waste water and leakage is 15% of total consumption E. all the above.
Explanation:
No answer description available for this question. Let us discuss.
25.
Yield of a drainage basin is :
A. run off of the area expressed as instantaneous rate B. average run off over a short period C. total volume of water flowing annually D. all the above. | 382 | 1,583 | {"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-2022-21 | latest | en | 0.879101 |
https://mathexamination.com/lab/b.-fischer.php | 1,632,420,695,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00432.warc.gz | 436,621,817 | 8,515 | ## Take My B. Fischer Lab
As pointed out over, I utilized to write an easy as well as straightforward math lab with only B. Fischer Nonetheless, the less complex you make your lab, the much easier it becomes to obtain stuck at the end of it, after that at the beginning. This can be really discouraging, and all this can take place to you due to the fact that you are utilizing B. Fischer and/or Modular Equations incorrectly.
With Modular Equations, you are currently using the wrong equation when you obtain stuck at the start, if not, after that you are likely in a stumbling block, and there is no possible way out. This will just worsen as the issue ends up being much more complex, however then there is the concern of just how to wage the problem. There is no other way to appropriately set about solving this sort of mathematics trouble without being able to immediately see what is going on.
It is clear that B. Fischer and also Modular Formulas are challenging to learn, and also it does take method to establish your own sense of intuition. However when you wish to fix a mathematics issue, you have to utilize a tool, and also the devices for finding out are utilized when you are stuck, as well as they are not used when you make the wrong step. This is where lab Assist Solution comes in.
As an example, what is wrong with the concern is incorrect ideas, such as obtaining a partial worth when you do not have enough working components to complete the entire job. There is a great reason that this was wrong, and also it is a matter of logic, not intuition. Logic enables you to comply with a detailed procedure that makes sense, as well as when you make a wrong relocation, you are typically required to either try to move forward as well as fix the blunder, or attempt to go backward as well as do an in reverse step.
An additional circumstances is when the pupil does not understand a step of a process. These are both rational failings, as well as there is no chance around them. Also when you are stuck in a location that does not permit you to make any type of kind of step, such as a triangle, it is still crucial to understand why you are stuck, to make sure that you can make a better relocation and also go from the action you are stuck at to the next area.
With this in mind, the very best method to resolve a stuck circumstance is to just take the progression, instead of trying to go backward. The two processes are different in their method, however they have some basic resemblances. Nevertheless, when they are tried together, you can rapidly tell which one is better at addressing the problem, as well as you can additionally tell which one is much more effective.
Let's discuss the very first example, which connects to the B. Fischer mathematics lab. This is not also complicated, so allow's very first review how to begin. Take the complying with process of attaching a part to a panel to be made use of as a body. This would need three measurements, as well as would be something you would need to attach as part of the panel.
Currently, you would have an extra measurement, but that does not mean that you can just keep that measurement and also go from there. When you made your initial step, you can conveniently ignore the dimension, and after that you would certainly have to go back and backtrack your actions.
Nevertheless, instead of bearing in mind the added dimension, you can use what is called a "psychological faster way" to help you bear in mind that extra measurement. As you make your primary step, visualize yourself taking the measurement and also attaching it to the part you want to connect to, and after that see just how that makes you really feel when you repeat the procedure.
Visualisation is a very powerful strategy, and is something that you should not miss over. Picture what it would certainly seem like to really affix the component and also have the ability to go from there, without the dimension.
Currently, allow's consider the 2nd instance. Let's take the same procedure as before, but now the trainee has to keep in mind that they are mosting likely to move back one action. If you tell them that they have to return one action, but then you get rid of the concept of having to return one action, then they won't recognize how to proceed with the issue, they won't recognize where to seek that action, and the procedure will be a mess.
Rather, use a psychological shortcut like the psychological representation to emotionally reveal them that they are going to move back one step. as well as put them in a position where they can progress from there. without needing to think about the missing out on an action.
## Hire Someone To Do Your B. Fischer Lab
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## Pay Someone To Take My B. Fischer Lab
B. Fischer is made use of in a lot of schools. Some teachers, nonetheless, do not utilize it very successfully or utilize it incorrectly. This can have a negative effect on the pupil's discovering.
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# SESSION TITLE:
## SOLUTION METHODS AND STABILITY ANALYSIS
Topics:
•Introduction and Overview
•Explicit time integration method
•Implicit time integration methods and solution of equilibrium
equations
•Stability of solutions
•Numerical stability
•Material stability
INTRODUCTION AND OVERVIEW:
• Solution procedures for nonlinear finite element analysis of
continua and structures are presented.
• Explicit and implicit methods for transient problems and methods
for the solution of equilibrium problems are described
• Stability is an important topic in NLFEA.
• Stability of solutions, stability of numerical procedures and material
stability are presented.
• In Explicit time integration, we focus on the central difference
method.
• We next describe implicit time integration methods with the
Newmark -method as our focus
• Methods for solution of static problems, i.e., the solution of
equilibrium problems, are developed simultaneously.
• Newton method description also includes techniques used for
convergence checks and line search method.
• A critical step in the implicit scheme and static problems is the
LINEARIZATION of the governing equations. Linearization
procedures for the equations of motion and the Equilibrium
Equations are described.
• In nonlinear problems, stability of solution is of considerable
interest. Methods for examining stability are described.
• We then examine numerical stability of time integration methods
and analysis of central difference method and the Newmark, -
method are performed.
• The topic of material stability concludes the session.
Explicit Time Integration Methods
• The central difference method is among the most popular of the
explicit methods in computational mechanics. The central
difference method is developed from central difference formulae
for the velocity and acceleration. We consider here its application
to Lagrangian meshes. Methods for Eulerian and ALE meshes
are discussed separately. Geometric and material nonlinearities
are included in the equations of motion, and in fact have little
effect on the time integration algorithm.
• Notation: Let the time of the simulation 0 t tE be subdivided
into time steps tn, n = 1 to nTS where nTS is the number of time
steps and tE is the end of simulation. tn and dn d(tn) are the time
and nodal displacements, respectively at time step n.
• The algorithm with a variable time step is considered. This is
necessary in practical calculations since the stable time step
changes as the FE mesh deforms and wave speed changes due to
stress.
• The time increments as
1 1
n n
Δt n t t 2 2
d
1 1
n n 1
d 2
V 2
1
n 1
dn
n
Δt 2
## Which is converted to an integration formula:
dn+1 = dn + tn+1/2.Vn+1/2
• The acceleration and the corresponding integration formula are
n 12 n
1
d n a n V V
2
n 1 n 1
t 2 t 2
1 1
n n
V 2
V 2
Δt n a n
d a
n n
d n 1
2d n d n 1
Δt n 2
• The equations of motion at time step n are :
M a n f n f ext d n , t n f int d n , t n Eqn(A)
Subject to constraint s :
g I d 0, I 1 to n c
n
## We have to find dn so as to satisfy Eqn(A) and the constraints.
These are Ordinary Differential Equations (ODE’s) of second order
in time, often called semidiscrete since they have been discretized in
space but not in time. Generalized representation of the nc
displacement boundary conditions and other constraints on the
model are also appended. Note that these constraints are linear or
nonlinear algebraic functions of the nodal displacements. M the
mass matrix is a constant for Lagrangian mesh.
• The internal and external nodal forces are functions of the nodal
displacements and the time.
• The external loads are prescribed by the analyst as function of
time. They may also depend on nodal displacement because they
may also depend on the deformation configuration of the
body/structure, as when pressure forces are applied to surfaces,
which undergo large deformations.
• The dependence of internal nodal forces on displacements is
obvious. The nodal internal forces depend on the stresses. Stresses
depend on the strains and strain rates by the constitutive equations;
the strains and strain rates in turn depend on the displacements and
their derivatives. The internal nodal forces can also depend directly
on time, e.g., when the temperature is prescribed as a function of
time, then the stresses and hence internal nodal forces are directly
functions of time.
• The equations for UPDATING the nodal velocities and nodal
displacements are:
1 1
n n
V 2
V 2
Δt n M 1 f n Eqn(B)
1 1
n n
n 1
d d Δt
n 2
V 2
## At any time step n, the displacements dn are known. The nodal
forces fn can be determined by sequentially evaluating the strain-
displacement equations, the constitutive equations expressed in
terms of Dn-1/2 OR En and the nodal external forces. Thus the entire
RHS of Eqn(B) can be evaluated, and Eqn(B) used to compute
Vn+1/2. The displacement dn+1 can then be determined.
• The update of the nodal velocities and nodal displacements can
be accomplished without SOLVING any equations provided that
the mass matrix M is diagonal. This is the salient characteristic
of an EXPLICIT METHOD. In an explicit method, the time
integration of the discrete momentum equations does not require the
solution of any equations. However, this hinges critically on the use
of a diagonal mass matrix. The required inversion of diagonal mass
matrix is trivial.
• There are exceptions. For example, the consistent mass matrix is
sometimes used with the central difference method in WAVE
PROPAGATION problems, and the update then does involve the
solution of equations!!
Software Implementation:
Box 6.1 gives the flowchart for explicit time integration. This
flowchart includes nonzero initial conditions, a variable time step,
elements with more than one quadrature point, and DAMPING.
• Damping is modeled by a linear viscous force.
fdamp = Cdamp.V
so that the total force in eqn (B) is f - Cdamp.V
• The implementation of velocity update is broken into two sub
steps by
Vn = Vn-1/2 + (tn – tn-1/2) an
Vn+1/2 = Vn + (tn+1/2 – tn). an
This enables energy balance to be checked at integer time steps.
• The cardinal dependent variables in this flowchart are the
velocities and the Cauchy stress. Initial conditions must be
given for the velocities, the Cauchy stress, and all material state
variables. The initial displacements are assumed to vanish
(initial displacements are meaningless in NLFEA except for a
hyperelastic material since the stress depends on the history of
deformation).
• The main part of the explicit time integration procedure is the
calculation of the nodal forces, which is done in get force. The
major steps in the subroutine get force are
1. Extract the nodal displacements and velocities of the element
from the global arrays by the GATHER operation.
2. The strain measures are computed at each quadrature point of the
element.
3. The stresses are computed at each quadrature point using the
constitutive equations
4. Evaluate the nodal internal forces by integrating the product of
the B matrix and the Cauchy stress over the domain of the
element.
5. The nodal forces of the element are SCATTERED into the global
array.
• In the first step, (n=1), the strain measures and the Cauchy stresses
are not computed, instead, as shown in the flowchart, the nodal
internal forces are calculated directly from the initial stresses.
• In the flow chart, the matrix form of the nodal internal force
calculation, in which the stress tensor is stored as a square matrix. To
change to Voigt form, replace B by B and the square matrix of
stresses by the column matrix {}.
• Most essential boundary conditions are easily handled in explicit
codes. For example, if the velocity OR displacement is prescribed
as functions of time along any boundary, then the prescribed
velocity/displacement boundary conditions can be enforced by
setting the nodal velocities according to the data:
ViIn+1/2 = Vi(XI, tn+1/2)
•Any damping in the system lags by a halftime step. This also holds
for any rate-dependent terms in the constitutive equation evaluation
phase of the step getforce. The time lag is unavoidable if the
implementation is to be fully explicit, i.e., not require the solution of
any equations. However, this decreases the stable time step for the
method.
•The explicit time integration method is easily implemented and is
very robust, by which we mean that the procedure seldom aborts due
to failure of the numerical algorithm. The salient disadvantage is
the conditional stability. If the chosen time step t exceeds a
critical value tcrit, the solution will grow unboundedly!!
•The critical time step also called stable time for a mesh as constant
strain elements and with rate-independent material models is given
by t = .tcrit
tcrit = min (le/Ce) 2/max
• when max is the maximum frequency of the linearized system, le is a
characteristic length of the element e, ce is the current wave speed in
element e, and is a reduction factor that accounts for the destabilizing
effects of nonlinearities. A good choice for is 0.80.98.
•The critical time step decreases with mesh refinement and increasing
stiffness of the material. The cost of an explicit simulation is independent of
the frequency range that is of interest and depends only on the size of the
FE model and the number of time steps.
•An interesting question for elastic-plastic materials is whether the slower
wave speed in the plastic response enables one to increase the time step.
Based on experience, the answer appears to be negative. An elasto-plastic
time step depends on the elastic wave speed, and a time step that exceeds
the critical time step results in instability.
•The mesh time step is obtained from element time steps.
Energy Balance
• Certain instabilities can lead to a large over prediction of
displacements. They are easily detected by an energy balance check.
Any instability results in the spurious generation of energy, which
leads to a violation of the conservation of energy. Therefore, whether
stability was maintained during a nonlinear computation can be
established by checking energy balance.
• The internal and external energies are integrated as follows:
n 1 1 T n
W W Δd f int f int
int
2
n
int
n 1
n 1 n 1 T n
2
n 1
Wext Wext Δd f ext f ext
n 1
where Δd d d n
The kinetic energy is given by
W n
kin V
2
1 n T
M Vn
Note that integer time steps are used for the velocities.
The internal energies can also be computed on the element OR
• Energy conservation requires that
## where is small tolerance of the order of 10-2
Mass scaling, subcycling, and dynamic relaxation:
• When a FE model contains a few small OR stiff elements, the
efficiency of explicit time integration method is compromised
severely, since the time step of the entire mesh is set by these very
stiff elements. Several techniques are available for circumventing
this difficulty.
a) Mass scaling: the masses of stiffer elements are increased so that
the time step is not decreased.
b) Subcycling: a smaller time step is used for the stiffer elements.
• Mass scaling should be used for problems where high frequency
effects are not important. For example, in sheet metal forming
simulation, which is essentially a static process, it causes no
difficulties. On the other hand, if high frequency response is
important, mass scaling is not recommended.
• Subcycling was introduced by Belytschko in 1979. In this
technique, the FE model is split into subdomains and each is
integrated with its own stable time step. The crucial issue in
subcycling is the treatment of the interface between subdomains.
• Dynamic relaxation is often used in explicit codes to obtain static
solutions. The basic idea is to apply the load very slowly and solve
the dynamic system equations with enough damping so that
oscillations are minimized. In path-dependent materials, dynamic
relaxation often yields poor solutions. Furthermore, it is very slow.
Newton methods combined with effective iterative solvers, such as
preconditioned conjugate gradient or multi-grid methods are much
faster and more accurate.
Implicit Time Integration Methods and Solution of Equilibrium
Equations:
• We will combine the description of the solution of equilibrium
equations with time integration by implicit methods since they
share many common features.
• To begin, we write the discrete momentum equation at time step
n+1 in a form applicable to both equilibrium and dynamic
problems:
0 r d n 1
,t n 1
S
D Ma n 1
f int
d , t
n 1 n 1
f ext
d n 1
,t n 1
1
where SD is a switch which is set by:
SD = 0 for a static (equilibrium) problem.
SD = 1 for a dynamic (transient) problem.
The column matrix r (dn+1, tn+1) is called a RESIDUAL FORCE.
The discrete equilibrium equations are nonlinear algebraic
equations in the nodal displacements, dn+1.
The discrete equations for the implicit update of the equations of
motion are also nonlinear algebraic equations in the nodal
displacements, dn+1.
Equilibrium solutions and equilibrium points
•When the accelerations vanish or are negligible, a system is in static
equilibrium and the solution of (1) is called an EQUILIBRIUM
SOLUTION. The equations of equilibrium are given by (1) with SD 0.
0 r d n 1 , t n 1 f int d n 1 , t n 1 f ext d n 1 , t n 1 2
• In equilibrium problems, the residuals correspond to the out-of-
balance forces. Problems in which the accelerations can be neglected
are called static problems. A solution of the above is called an
equilibrium point and a continuous locus of solutions is called an
equilibrium path.
• In equilibrium problems with rate-independent materials, t need not
be the real time. In stead, it can be any monotonically increasing
parameter.
• If the constitutive equation is a differential or integral equation, it
must also be discretized in time to obtain a set of algebraic equations
Newmark Equations:
• We use a popular class of time integration methods called
newmark method. The updated displacements and velocities are
n 1 ~ n 1 2 n 1
d d β Δt a
Δt 2
~ n 1
where d d n Δt V n 1 2β a n 3
2
~ n 1
V V γ Δt a n 1
n1
where V V n 1 γ Δt a n 4
~ n 1
Where t = tn+1 – tn
and are parameters whose useful values and stability
properties are summarized in Box 6.2
• The parameter controls artificial viscosity, a damping introduced
by the numerical method. It is used to suppress noise in the solution,
when = ½, the Newmark method adds no damping for > ½
artificial damping proportional to -1/2 is added by the integration
method.
• Eqn (3) can be solved for the updated acceleration, giving
a n 1 1
β Δt 2
n 1
d d when β 0 5
~ n 1
0r
SD
β Δt 2
M d n 1
~ n 1
d f int
d n 1 n 1
, t f ext
d ,t
n 1 n 1
6
## which is a set of nonlinear algebraic equations in the nodal
displacement dn+1.
• Equation (6) applies to both static and dynamic problems. In both
the cases, the problem is to find dn+1 so that r(dn+1, tn+1) 0
subject to g(dn+1,tn+1) 0
• Equations (6) are the Discrete Equations of motion/Equilibrium
Newton’s Method: The Newton Raphson Algorithm
• The most widely used and most robust algorithm for the solution
of the set of nonlinear algebraic equations (6) is Newton’s
method. It is often called Newton-Raphson Iterative method in
computational mechanics.
• The Newton-Raphson Iterative procedure is illustrated in Fig 6.1
• The solution of equations (6) by the NR method is an
ITERATIVE PROCEDURE. The iteration number is indicated by
Greek subscript: dn+1 d is the displacement in ITERATION
at time step n+1; the time step number n+1 will be omitted for
brevity.
•To begin the iterative procedure, a starting value for the unknown
must be chosen; usually the solution dn from the last time step is
selected, so d0 = dn. In dynamic problems, a better starting value is
~ n 1 . A Taylor expansion of the residual r about the current value
d
of nodal displacement d and setting the resulting residual equal to
zero gives
r d
r d
d
Δd 0 Δd 0 7
2
## The matrix r/d is called the system Jacobean matrix denoted by
A.
We then get
r + A. d = 0 ---------------------------------(8)
which is the linear model of the nonlinear equations. The linear model
is difficult to picture for problems with more than one unknown.
Figure 6.2 shows an example of the first component of the residual for
a function of two unknowns. The linear model is a plane tangent to the
nonlinear function r1(d1,d2). The other residual component is another
nonlinear function r2(d1,d2), which is not drawn.
• The increment in the nodal displacements in the NR-Iterative
procedure is obtained by solving eqn (8), which yields a system of
linear algebraic equations.
A. d = - r (d,tn+1) -------------- (9)
Once the increments in nodal displacements have been calculated, they
are added to the previous iterate
d+1 = d + d
The new nodal displacements is checked for convergence. If the
convergence criterion is not met, a new linear model is constructed
and the process is repeated. The iterative procedure continues until
the chosen convergence criterion is met.
•In computational mechanics, the Jacobian matrix A is called the
Effective Tangent Stiffness matrix and the contributions of the
INERTIAL, INTERNAL and EXTERNAL forces are linearized
separately.
•The Jacobian is written as
r SD f int f ext
A M for β 0
d β Δt 2
d d
• For equilibrium problems; SD 0;
f int
f int
is called the Tangent stiffness matrix denoted by K
int
d d
f ext
f ext
is called the Load stiffness matrix denoted by K ext
d d
• For dynamic problems; SD 1; the jacobian matrix is
SD
A M K int
K ext
β Δt 2
## • Note that this also applies to static problems with SD 0.
• The Jacobians can be used to relate differentials of the nodal forces
to differentials of nodal displacements by
df int
K dd
int
df K d d
ext ext
dr A d d
Software Implementation
• Flow charts for Implicit integration and Equilibrium solutions are
given in Boxes 6.3 and 6.4, respectively. Both the dynamic problem
and the equilibrium problems are solved by time stepping. The
external loads and other conditions are described by the user as
functions of time, which is incremented over the range of interest. In
equilibrium problems, tine is often replaced by a monotonically
increasing parameter. Solutions of equilibrium processes obtained in
this manner are called Incremental Solutions.
• The flow chart shows the full Newton algorithm, where the
Jacobian matrix is evaluated and inverted in every iteration of the
solution procedure.
•Many programs offer a modified Newton algorithm, in which the
Jacobian matrix is assembled and triangulated only at the beginning
of a step or intermittently during a step. For example, the Jacobin
matrix may be triangulated only when the ITERATIVE procedure
does not seem to be converging well. The modified Newton algorithm
is faster but less robust.
•The Implicit scheme begins with the imposition of the initial
conditions. The initial displacements are considered to be zero. The
initial accelerations are computed as shown in steps 2 and 3 (Box
6.3).
•The nodal displacements dn+1 are obtained by an iterative procedure.
To begin the iteration, a starting value of d is needed; usually the
solution from the preceding time step is used. The residual r is then
calculated for this starting value. In an equilibrium solution, the
residual depends only on the internal and external nodal forces; which
is obtained in the module getforce. In transient implicit solution, the
residuals also depend on the acceleration.
Constraints
• There are four options for treating the constraints.
gI(dn) = 0, I = 1 to nc (no. of constrants)
1. Penalty.
2. Lagrange multiplier.
3. Augmented Lagrangian
4. Perturbed Lagrangian.
## These methods originate in optimization theory. They are adopted to
the solution of discrete momentum and equilibrium equations in
nonlinear FEA programs.
Convergence Criteria
• The termination of iterations in Newton algorithms is
determined by convergence criteria.
• Three types of convergence criteria are:
1. Based on the magnitude of the residual forces r
2. Based on the magnitude of displacement increment d.
3. Energy error.
Implicit Time Integration Method: Accuracy and Stability
•For linear transient dynamic problems, suitable implicit integration
schemes are unconditionally stable.
•The major restriction on the time step in implicit methods arise from
accuracy requirements and decreasing robustness.
•Large time steps also impair the convergence of Newton method,
particularly in problems with very rough response, such as Contact
Impact. Small time steps improve the robustness of the Newton
algorithm.
•In return for there enhanced stability implicit methods exact a
significant price; they require the solution of a set of nonlinear
algebraic equations in each time step.
• Implicit solutions are now feasible in many problems where they
failed before. We are certain that further improvements are
imminent. Nevertheless, high cost and insufficient robustness still
plague Newton method.
Newton Iteration: Convergence and Robustness:
• The rate of convergence of the ITERATIONS in the Newton
method (or N-R algorithm) depends on the Jacobian matrix A
satisfying the following conditions.
1. The Jacobian matrix A should be sufficiently smooth function
of the nodal displacements d.
2. The Jacobian matrix A should be regular (invertible) and well
conditioned in the entire domain in the displacement space that
the iterative procedure traverses.
• These conditions are not satisfied by engineering problems. For
example, in elastic-plastic material, the residual r is not a
continuously differentiable function of the nodal displacement d:
when an element quadrature point changes from elastic to plastic
or vice versa, the derivative is discontinuous.
• In contact-impact problems with Lagrange multiplier methods,
the residual also lacks smoothness.
• Yet, N-R algorithm is still remarkably effective. It converges!!
• In some nonlinear problems, the conditions for quadratic
convergence of the N-R method are fully satisfied. For example,
in a smoothly loaded FE model with a Mooney-Rivlin material
where load increment is small enough so that equilibrium
conditions are stable.
• In equilibrium problems, the NR method fails to converge in the
vicinity of unstable states.
• In dynamic problems, the convergence difficulties of the NR
iteration are ameliorated by the mass matrix. For many problems,
use of the Newton method is totally unsuccessful, and further
enhancements such as the arc length method are needed.
IMPLICIT VS EXPLICIT
• The selection of time integration scheme depends on:
1. The type of Partial Differential Equation (PDE).
2. The smoothness of the data.
3. The response of interest.
• For parabolic PDE’s, implicit methods are preferred. The
solutions of parabolic systems are smooth even when the data is
rough. The stable time step in a parabolic system decreases by a
factor of four each time the element size is halved, so that
refinement in an EXPLICIT method becomes prohibitively
expensive. Parabolic systems should never be integrated by
explicit schemes.
•There are exceptions where explicit methods are preferred even
when the system is partially or fully parabolic. For example in car
crash simulation (using shell elements) the equations for the shells
which model the sheet metal response are parabolic, yet because the
noise introduced by Contact-Impact algorithm, explicit methods are
preferred.
•Similarly, in complex heat conduction problems, it is often
impossible to take advantage of the large time step permitted by
implicit methods explicit methods are preferred.
•In hyperbolic systems, the choice depends on the response of
interest. Hyperbolic PDE’s govern wave propagation problems and
inertial problems (structural dynamics).
• In structural dynamics problems, the frequency spectrum of the
input is far below the resolution limits of the FE mesh; the FE mesh
refinement is dictated by the need for very high accuracy in the
frequency band of interest. Here, IMPLICIT integration methods
are definite choice. Examples include seismic response also.
• Wave propagation problems are those in which relatively high
frequency parts of the spectrum are of interest. For example, waves
in cellular phones in a drop test, the simulation needs a small time
step to tract the high frequency portion of the ‘response’, and
explicit methods are preferred.
LINEARIZATION
Linearization of the Internal nodal forces
• We now derive expressions for the tangent stiffness matrix Kint
for Lagrangian elements. As will be seen part of the expression
can be derived independently of the material response.
Linearization of the constitutive equation is carried out in two
ways:
1. With the continuum tangent moduli, which does not account for
the actual constitutive update algorithm; the resulting material
tangent stiffness is called the TANGENT STIFFNESS MATRIX.
2. with the algorithmic tangent moduli, which gives rise to the so
called consistent tangent stiffness.
• The choice rests on considerations related to ease of
implementation and on the smoothness of the problem.
• The tangent stiffness matrix will be expressed in terms of CSE, the
tangent moduli relating the rate of the PK2 stress to the rate of the
Green strain in the Total Lagrangian Formulation and in terms of C,
the tangent moduli relating the Truesdell rate of the Cauchy stress to
the rte-of-deformation in the Updated Lagrangian Formulation. These
tangent moduli are related: if the current configuration is chosen to
be the reference configuration, C = CSE.
• We will develop the tangent stiffness matrix by relating the rates of
.The procedure
the internal nodal forces f int to the nodal velocities d
is identical to relating an infinitesimal increment of internal nodal
forces dfint to an infinitesimal increment of nodal displacements d d
•The internal nodal forces in the TLF are given by
f int B T P dΩ A
Ω
OR
N I
f int
Pji dΩ
X j
iI
Ω
## Where P is the nominal stress tensor with components Pji, NI are
the nodal shape functions, and
N I
B
X j
jI
## • We have opted the TLF because this leads to the simplest
derivation. In the TLF, the only variable, which depends on the
deformation is the nominal stress, i.e., it is the only variable which
varies with time.
• In the ULF, the domain of the element (or body), the spatial
derivatives NI/xj and the Cauchy stress depend on the
deformation and hence on time, which complicates the derivation of
the tangent stiffness matrix.
•Taking the material time derivative of eqn (A),
0
f int B T P dΩ
Ω0
0
## Since B0 and d0 are independent of deformation or time.
• To get the stiffness matrix Kint, it is now necessary to write the
nominal stress rate P in terms of the nodal velocities via the
constitutive equations and the strain measure.
• The material time derivative of the PK2 stress is related to the
material time derivative of the nominal stress P by taking the time
derivative of the transformation P = S.FT.
P S FT S F T
• Finally eqn A becomes
f int
iI
N I
X j
S jr Fir S jr Fir dΩ 0
Ω0
N I
f int
dS jr Fir S jr dFir dΩ0
X j
iI
Ω0
• The above shows that the rate ( or increment) of the internal nodal
forces consists of two parts.
, and thus depends on
1. The first part involves the rate of stress S
the material response and leads to what is called the material
tangent stiffness matrix which we denote by Kmat
2. The second part involves the current state of stress S, and
accounts for geometric effects of the deformation (including
rotation and stretching). This term is called the geometric
stiffness. It is also called the initial stress stiffness matrix to
indicate the role of the existing state of stress. It is denoted by
Kgeo.
• The changes in the internal nodal forces due to these two effects are
given analogous names, so
f int f mat f geo
OR f int f mat f geo
iI iI iI
where
N I
fiImat Ω X j Fir S jr dΩ0
0
N I
f geo
iI Ω X j ir 0
S jr
dΩ
F
0
Material Tangent Stiffness Matrix
• We rewrite the rate of the internal nodal forces, in Voigt notation
f mat BT0 S dΩ 0
Ω0
Where
format.
S is the rate of the PK2 stress in Voigt column matrix
• The constitutive equation in rate form is
S CSE E
ij ijkl kl
or
S CSE E
Where E is the rate of the Green strain.
• We now relate the rate of the Green strain to the nodal velocities
in Voigt notation,
E B 0
d
• Finally we get the material rate of the internal nodal forces
Ω0
f int B T CSE B dΩ d
mat 0 0 0
OR
fmat
int
B T0 CSE B 0 dΩ 0 dd
Ω0
## • So, the material tangential stiffness matrix is
K mat
B C B dΩ
T
0
SE
0 0
OR Ω0
K mat
IJ B C B dΩ
T
0I
SE
0J 0
Ω0
• The material tangent stiffness relates increments (OR rates) of the
internal nodal forces to the increment (or rate) of the nodal
displacements due to material response, which is reflected in the
tangent modulii CSE. Its form is identical to the stiffness matrix in
linear finite element analysis.
Geometric Stiffness: Initial Stress Stiffness Matrix
• The geometric stiffness matrix is obtained as follows.
From the definition
N I
B 0
X i
iI
We can write
ii Ij jr ir 0
f geo B0 S F dΩ
Ω0
B S jr B dΩ 0 u i,J
0
Ij
0
rj
Ω0
B S jr B dΩ 0 δ ik u kJ
0
Ij
0
rj
Ω0
## • Writing the above in matrix form, it gives
f geo K geo u
I IJ J
K geo
T
IJ I B 0I S B 0J dΩ 0
Ω0
•Note that the PK2 stress S is in tensor form, i.e., a square matrix.
•The geometric stiffness matrix Kgeo is invariant with rotation.
•The above in Total Lagrangian Formulation are easily converted
to ULF (Updated Lagrangian Formulation) by letting the current
configuration be the reference configuration:
K mat
IJ B CT
I B dΩ
στ
j
Ω
K mat BT C στ B j dΩ
Ω
K geo
IJ I B σ BJ dΩ
σ
I
Ω
• The tangent modulii [CSE] and [C] for a wide variety of
engineering materials are presented in the session: Constitutive
Models.
• Tangent material stiffness matrices for specific finite elements
are presented later in this session.
• The numerical values of the above matrices Kmat and Kgeo in the
TLF and ULF are identical, and the choice of which to use is a
matter of convenience.
• The Issue of symmetry of the tangent stiffness matrix deserves
attention. Symmetry speeds the solution of the set of nonlinear
algebraic equations reduces storage requirements and simplifies
stability analysis.
• The material tangent stiffness matrix is symmetric when the
Voigt form of the tangent modulii matrix [CSE] is symmetric.
•The geometric stiffness matrix is always symmetric. Therefore, the
tangent stiffness matrix Kint is symmetric whenever the tangent
modulii possess major symmetry.
•See box 6.5 for Jacobian of Internal Nodal Forces (tangent stiffness
matrix).
body (elements). They appear in geometrically nonlinear problems
only. Pressure is a common example of Follower load. A pressure
force is always normal to the surface, so as the surface MOVES the
external nodal forces change even if the magnitude of pressure is
constant. These effects are accounted for in the Jacobian matrix of
external nodal forces, denoted by Kext, which is also called the
• The load stiffness Kext relates the rate of external nodal forces to
the nodal velocities.
• Consider a pressure load, p(x,t), the external nodal forces on a
surface of element e are given by letting t p n
f I
ext
N I p n dΓ
Γ
• Let the surface be parametized by two variables and . For
quadrilateral surface element, these variables can be the parent
element natural coordinates on the biunit square, since
n dΓ x ,ξ x ,η dξ dη we can write
1 1
f I
ext
pξ, η N I ξ, η x ,ξ x ,η dξ dη
1 1
Taking the time derivative of the above gives
1 1
fIext N I p x ,ξ x ,η p V,ξ x ,η p x ,ξ V,η dξ dη
1 1
•The first term is the rate of change of the external nodal forces
due to the rate of change of the pressure. In many problems, the
rate of change of pressure is prescribed as part of the problem
definition.
•In other problems, such as in Fluid-Structure Interaction
problems, pressure may arise from changes of the geometry: these
effects must then be linearized and added to the Load Stiffness. We
omit this aspect in the following.
•The last two terms represent the changes in the external nodal
forces due to the changes in the direction of the surface and the
area of the surface. These are related by the external load stiffness,
so the RHS of the above equation becomes
1 1
IK VK p N I V,ξ x , η x ,ξ V, η dξ dη
K ext
1 1
## We can finally write f Iext K ext
IK VK
0 z ,η y ,η 0 z ,ξ y ,ξ
1 1
and K ext
IJ p N I N J, ξ z ,η 0 x ,η N J, η z ,ξ 0 x ,ξ dξ dη
1 1
y ,η
x ,η 0 y ,ξ
x ,ξ 0
## as the load stiffness matrix.
• It is immediately apparent that the submatrices of the load
stiffness matrix are not symmetric. So the Jacobian matrix in
general is not symmetric in the presence of follower forces.
However, it can be shown that for a closed structure in a constant
pressure field, the assembled external load stiffness is symmetric.
TWO-NODE ROD ELEMENT
• Consider the two-node rod element shown in Figure 6.9. The ROD
is in a state of uniaxial stress. The x̂ -axis lies along the axis of the
rod and rotates with the rod, i.e., it is a corotational coordinate. The
only nonzero Cauchy stress component is σ̂11 σ̂ xx . The tangent
stiffness and load stiffness matrices are derived in the ULF, i.e., in
the current configuration.
• Material tangent stiffness matrix (ULF)
K̂ mat B̂T Ĉστ B̂ dΩ
Ω
1 0 1 0
στ
AE 0 0 0 0
K̂ mat
l 1 0 1 0
0 0 0 0
• This is identical to the linear stiffness matrix for a rod if E is
replaced by Young’s modulus E.
• The material tangent stiffness matrix relating global components
of Internal Nodal forces and Nodal Velocities is
f Imat K mat VI
K mat T T K̂ mat T
Where T is given by
cosθ sin θ 0 0
sin θ cosθ 0 0
T
0 0 cosθ sin θ
0 0 sin θ cosθ
• Geometric Stiffness Matrix (ULF)
The geometric stiffness matrix is given by
K̂ IJ Ĥ IJ I
Ĥ B̂T σ B̂ dΩ
Ω
1 1
σ̂ xx 1 1 dΩ
1
l 1
Ω
l
σ̂ xx A 1 1
Ĥ
l 1 1
1 0 1 0
0 1 0 1
Aσ̂ xx
K̂ geo
l 1 0 1 0
0 1 0 1
K geo T T K̂ geo T
• The total tangent stiffness matrix is the sum of the material and
geometric stiffnesses.
Kint = Kmat + Kgeo
The load stiffness for the 2-node ROD element is developed here.
First it is evaluated in the corotational coordinate system x̂, ŷ
0 1 0 1
1 0 1 0
pa
K̂ ext
2 0 1 0 1
1 0 1 0
K ext T T K̂ ext T K̂ ext
Material tangent stiffness matrix in TLF
K mat BT0 CSE B0 dΩ 0
Ω0
## cos 2θ cosθ sin θ cos 2θ cosθ sin θ
2
A 0 E SE l sin θ
2
cosθ sin θ sin θ
2
l0 0
l cos 2
θ cosθ sin θ
sin θ
2
Note the material constant ESE relates the rate of the PK2 stress to
the rate of the Green strain in a uniaxial state of stress.
Geometric Stiffness Matrix in TLF
1 0 1 0
A 0 Su 0 1 0 1
K geo
l 0 1 0 1 0
0 1 0 1
THREE-NODE TRIANGULAR ELEMENT
• We consider the 3-node triangular element in two dimensions in a
state of plane strain.
Material tangent Stiffness Matrix
A
## Where A denotes current area of the element. We have set the
thickness a 1.
The tangent modulus matrix in Voigt matrix form is
C1111
στ στ
C1122 στ
C1112
στ
C
στ στ
C C C
στ
στ 2211 2222 2212
C1211 C1222
στ στ
C1212
Using the known B matrix, since the integrand is frequently
constant, the tangent stiffness matrix is the product of the integrand
with the area.
Geometric stiffness matrix
## K geo 2x2 I σ B J dA I 2x2 H IJ
T
IJ I B
A
1 y 23 y 31 y12
when B
2A x 32 x13 x 21
y 23 x 32
1 σ xx σ xy y 23 y 31 y12
H x13
σ yy x 32 x 21
y 31
4A σ xy x13
y12 x 21
## The geometric stiffness matrix is independent of material response.
It depends only on the current stress state and the current geometry
of the element
Physical Stability and Continuation Methods:
•In nonlinear problems, stability of solutions is of considerable
interest. However, we will focus on the FEM and application to
nonlinear analysis.
•Consider a beam loaded axially (fig 6.7 (a)). If we perturb the
location of the load P by a distance or 2, then the equilibrium
paths are as shown in Figure 6.7(c). it can be seen that when the load
is below the buckling load, the path for different initial conditions
remain close to AC. However when the load exceeds the buckling
load, the solution for different initial conditions diverge. Therefore,
any process in which the applied load exceeds the buckling load is
unstable.
•As an aside, we remark that when the beam is straight, the numerical
solution usually remains on the path AC shown in Figure 6.7(b). The
lateral displacement is zero even when the applied load P exceeds
the buckling load! If you do not believe this, try it. In a simulation, a
straight beam will usually not buckle in an Incremental Static
Solution or a Dynamic Solution, regardless or whether explicit or
implicit integration is used. Only when round off error introduces a
“numerical imperfection” or when an imperfection is introduced in
the INPUT DATA will the straight beam buckle in a simulation i.e., an
imperfection is needed to break the symmetry.
• It is widely believed among structural mechanicians that the
difficulties associated with unstable behavior can be circumvented by
obtaining a dynamic solution. When a structure is loaded above its
limit point or a bifurcation point in a dynamic simulation, the
structure passes dynamically to the nearest stable branch. However,
the instability is not readily apparent, and the possibility of
imperfection sensitivity is not clear.
• Therefore, to understand the behavior of a structure thoroughly, its
equilibrium behavior should be ecarefully
iI
examined. Many vaganes
of structural behavior may be hidden by dynamic simulations.
•For example, for a cylindrical shell under axial load the intersecting
branch is asymmetric as shown in Figure 6.7(e). A structure with
asymmetric branch is highly sensitive to imperfections, as can be
seen by the large change in the maximum load with the change in the
imperfection in Figure 6.7(e).
•The theoretical bifurcation point of a perfect structure is then not a
realistic measure of the strength: the actual structure may buckle at a
much lower load than the theoretical value because imperfections are
unavoidable in a real structure. A single numerical simulation could
miss this sensitivity completely.
•This sensitivity to imperfections for cylindrical shells was
analyzed by Koiter and is a classical example of imperfection
sensitivity. To ascertain this type of behavior, the equilibrium
branches must be known.
•The behavior of the shallow truss exhibits a turning or Limit Point,
as can be seen from Figure 6.9; Points A and B are turning points.
The branch after the first turning point, point A, is unstable. While
the branch after the second turning point, Point B, is stable!
•The behavior of the beam shown in Figure 6.7(a) is a classic
example of a bifurcation. The point B where the two branches
intersect is the point of bifurcation. Subsequent to the bifurcation
point, the continuation BC of the fundamental branch AB becomes
unstable. Point B, the bifurcation point, corresponds to the buckling
• Methods for tracing equilibrium branches are called continuation
methods. The tracing of equilibrium branches is often quite
difficult: Robust, automatic procedures for continuation are not yet
developed. A continuation method called arc length method is
implemented in commercial FEA programs. The arc length method
is often called the Rik’s method in structural mechanics literature.
NUMERICAL STABILITY
•A numerical procedure is stable if small perturbations of initial data
result in small changes in the numerical solution.
•General results for numerical stability analysis of the time
integration of the equations of motion are of interest.
•However, we stress that at the present time there is no stability
theory, which encompass the nonlinear problems, which are routinely
solved by Nonlinear FEA programs.
•Difference between physical stability and numerical stability:
Physical stability pertains to the stability of the solution of a model
whereas numerical stability pertains to the stability of the numerical
method. Numerical stability is usually only examined for processes,
which are physically stable.
•Stability analysis of widely used Time Integration Methods are well
documented: example: Explicit Central Difference method; Implicit
Newmark method.
MATERIAL STABILITY
•An important issue in nonlinear continuum mechanics is the
stability analysis of the material model. We now describe criteria for
material stability. Some remedies for the numerical difficulties
incurred by material instabilities are also discussed.
•Material instabilities are usually associated with a localized growth
of the deformation. This is called localization. It corresponds to
phenomena which are observed in experiments: For certain stress
states, metals, rocks and soils will exhibit narrow bands of intense
deformation; these are often called SHEAR BANDS, since the
deformation mode in these bands is usually shear.
•Example: Rate-independent material model.
•Shortly after NLFA Programs became available in the 1970’s,
computational analysts began to include unstable material models
both intentionally and inadvertently, and they discovered many
difficulties. Numerical solutions became often unstable and it was
discovered that the results depend very much on the mesh!
•At that time, it was argued by some mechanicians that material
models that violate the stability postulates should never be used. This
argument had merit for unless a constitutive law is carefully designed
to become unstable only in the rare situations when the material is
unstable, many difficulties are encountered. However, there is no
way to replicate experimentally observed phenomena such as shear
banding without a material model that exhibits strain softening.
•Several techniques have evolved to improve the coarse-mesh
accuracy of the finite element models for unstable material models.
These involve the embedment of discontinuities or enriched fields in
the element.
Ortiz Leroy and Needleman (1987) were the first to modify an
element at the point of material instability: they embedded
discontinuities in the strain field of the 4-node quadrilateral
elements, when the acoustic tensor indicated a material instability in
the element.
Belytschko, Fish and Englemann (1988) embedded a
displacement discontinuity by enriching the strain field with a
narrow band in which the material is unstable.
•Just as shear bands can be viewed as the outcome of a material
instability in the shear stress component, fracture can be considered
as the outcome of a material instability in the stress components
normal (and tangential in the case mode II fracture) to the
discontinuity (i.e., a crack). The relation between damage and
fracture has long been noted where fracture is assumed to occur when
the damage variable reaches a critical value.
•Damage mechanics the modeling of fracture by constitutive models
with damage poses same difficulties encountered in shear band
modeling, since the material model becomes unstable, when the
damage exceeds threshold value.
•These difficulties were resolved in a novel manner. The idea is to
match the energy dissipation in cracking to the energy dissipated in
the element, which exceeds the stability threshold. This is
accomplished by treating the fracture energy as a material parameter
and the energy dissipation in the strain-softening element is equaled
to the fracture energy (a material parameter).
Lemaitre J and Chaboche JL (1990) Mechanics of Solid
Materials, Cambridge University Press, Cambridge | 12,932 | 44,711 | {"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-2019-39 | latest | en | 0.891714 |
http://support.sas.com/documentation/cdl/en/qcug/63922/HTML/default/qcug_optex_a0000000382.htm | 1,571,355,799,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00199.warc.gz | 180,536,237 | 5,053 | The OPTEX Procedure
## Example 10.4 Optimal Design Using an Augmented Best Design
[See OPTEX4 in the SAS/QC Sample Library]This example is a continuation of Example 10.2.
You can specify a set of points that you want to be included in the final design found by the OPTEX procedure by using the AUGMENT= option in the GENERATE statement to specify a data set that contains a design to be augmented.
In this case, you can try to speed up the search for a 50-run design by first finding an optimal 25-run design and then augmenting that design with another 25 runs, as shown in the following statements:
```proc optex data=a seed=36926;
model AFR|EGR|SA@2 AFR*AFR EGR*EGR SA*SA;
generate n=25 method=detmax;
output out=b;
```
```proc optex data=a seed=37034;
model AFR|EGR|SA@2 AFR*AFR EGR*EGR SA*SA;
generate n=50 method=detmax augment=b;
run;
```
The result (see Output 10.4.1 and Output 10.4.2) is a design with almost 100% D-efficiency and A-efficiency relative to the best design found by the first attempt. However, this approach is not much faster than the original approach, since the run time for the DETMAX algorithm is essentially linear in the size of the design (see the section Memory and Run-Time Considerations.
Output 10.4.1 Efficiencies for the 25-Point Design to Be Augmented
The OPTEX Procedure
Design Number D-Efficiency A-Efficiency G-Efficiency Average Prediction
Standard Error
1 46.2975 26.0374 91.1822 0.5849
2 46.2171 25.9733 86.4608 0.5859
3 46.1720 25.9378 88.3293 0.5860
4 46.1374 25.9128 86.1895 0.5866
5 46.0808 22.6647 86.1502 0.6169
6 46.0620 24.7326 89.7179 0.6012
7 45.9992 25.4549 90.3330 0.5946
8 45.9630 24.7610 88.2701 0.5991
9 45.9627 25.5310 88.5737 0.5894
10 45.7994 24.5645 87.7544 0.6005
Output 10.4.2 Efficiencies for the Augmented 50-Point Design
The OPTEX Procedure
Design Number D-Efficiency A-Efficiency G-Efficiency Average Prediction
Standard Error
1 46.4957 25.0858 94.8160 0.4195
2 46.4773 25.0696 95.0646 0.4195
3 46.4684 24.5519 96.1259 0.4234
4 46.4676 24.5002 95.6830 0.4238
5 46.4587 25.0709 94.6650 0.4196
6 46.4555 24.8087 95.7768 0.4209
7 46.4471 24.5460 95.0073 0.4240
8 46.4373 25.0740 94.4640 0.4194
9 46.3899 25.0007 95.2162 0.4201
10 46.3662 24.4013 94.9539 0.4242
Previous Page | Next Page | Top of Page | 868 | 2,280 | {"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} | 3.0625 | 3 | CC-MAIN-2019-43 | latest | en | 0.76106 |
http://www.numbersaplenty.com/32101010 | 1,550,883,409,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249414450.79/warc/CC-MAIN-20190223001001-20190223023001-00495.warc.gz | 393,410,430 | 3,436 | Search a number
32101010 = 253210101
BaseRepresentation
bin111101001110…
…1001010010010
32020101220022022
41322131022102
531204213020
63104011442
7536565644
oct172351222
966356268
1032101010
1117135a67
12a900b82
13685c3a6
144398894
152c41625
hex1e9d292
32101010 has 8 divisors (see below), whose sum is σ = 57781836. Its totient is φ = 12840400.
The previous prime is 32100989. The next prime is 32101021. The reversal of 32101010 is 1010123.
Adding to 32101010 its reverse (1010123), we get a palindrome (33111133).
32101010 is an esthetic number in base 10, because in such base its adjacent digits differ by 1.
It can be written as a sum of positive squares in 2 ways, for example, as 32024281 + 76729 = 5659^2 + 277^2 .
It is a sphenic number, since it is the product of 3 distinct primes.
It is an alternating number because its digits alternate between odd and even.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1605041 + ... + 1605060.
Almost surely, 232101010 is an apocalyptic number.
32101010 is a deficient number, since it is larger than the sum of its proper divisors (25680826).
32101010 is a wasteful number, since it uses less digits than its factorization.
32101010 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3210108.
The product of its (nonzero) digits is 6, while the sum is 8.
The square root of 32101010 is about 5665.7753220543. The cubic root of 32101010 is about 317.8139082357.
The spelling of 32101010 in words is "thirty-two million, one hundred one thousand, ten".
Divisors: 1 2 5 10 3210101 6420202 16050505 32101010 | 553 | 1,806 | {"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} | 3.515625 | 4 | CC-MAIN-2019-09 | latest | en | 0.864781 |
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