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https://www.investopedia.com/ask/answers/09/forex-currency-converter.asp | 1,516,144,404,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00467.warc.gz | 904,856,931 | 47,648 | A:
All currencies are quoted in pairs - one country's currency against another country's currency. A currency converter is used by traders to check the current exchange rates between two chosen currencies. You'll find that most currency traders will use price charts to determine the direction of any given pair. A chart shows the price (or exchange rate) of the currency pair, which plotted on the y axis, over the time period, which is plotted on the x axis. A chart can be constructed for any time frame - from months, weeks, and days to hours and minutes; it can provide the trader with a historical perspective on the range of exchange rates over a period of time.
Chartists believe certain repeatable patterns allow them to gain an edge in determining the future movement of rates. Thus, when a currency trader wants to buy or sell a currency, he or she will use a chart for guidance to determine the likely currency rates in the future. Chart patterns, areas of support or resistance, and the consequent trading range within which a currency may fluctuate all factor into speculation on future rates. Once trends on future rates are decided upon, a trader will turn to a currency converter to determine the current rate of exchange.
A trader's use of a currency converter is similar to the need for any visitor to another country to physically exchange the currency of his or her home country for that of the host country. He or she will have to refer to a currency converter to obtain the current rate of exchange. When the trader finally converts the currency from his or her local currency into that of the country being visited, he or she will have to pay whatever rate is being charged by the local bank.
(For more on this topic, see Forces Behind Exchange Rates.)
This question was answered by Selwyn Gishen.
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The current ratio is a liquidity ratio that measures a company's ability to pay short-term and long-term obligations. | 1,080 | 5,365 | {"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-2018-05 | latest | en | 0.943551 |
http://openstudy.com/updates/55c787afe4b0c7f4a9796c3b | 1,496,092,163,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612553.95/warc/CC-MAIN-20170529203855-20170529223855-00448.warc.gz | 333,930,761 | 9,060 | • anonymous
Which inequality models this problem? The length of a rectangle is four times its width. If the perimeter is at most 130 centimeters, what is the greatest possible value for the width? A. 2w + 2 • (4w) < 130 B. 2w + 2 • (4w) > 130 C. 2w + 2 • (4w) ≤ 130 D. 2w + 2 • (4w) ≥ 130
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Looking for something else?
Not the answer you are looking for? Search for more explanations. | 350 | 1,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-22 | longest | en | 0.292335 |
https://math.stackexchange.com/questions/2820110/a-variation-on-dirichlets-theorem-on-arithmetic-progressions | 1,558,250,064,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254253.31/warc/CC-MAIN-20190519061520-20190519083520-00470.warc.gz | 556,791,366 | 34,634 | # A variation on Dirichlet's theorem on arithmetic progressions
Dirichlet's theorem on arithmetic progressions says that if $a$ and $b$ are coprime, then $\{a+bL\}_{L \in \mathbb{N}}$ contains infinitely many prime numbers.
I wonder if the following claim is true:
If $a$ and $b$ are coprime, then $\{a+b^L\}_{L \in \mathbb{N}}$ contains infinitely many prime numbers.
Notice that if my claim is true, then Dirichlet's theorem is true.
Thank you very much!
Edit: After receiving a few helpful comments, perhaps I should change my question to: Is there an additional condition that will make my claim true (a condition on $a$ and $b$)?
• No one knows if, for instance, there are infinitely many primes of the form $2^n+1$, though only a few are known. – lulu Jun 14 '18 at 23:22
• Note: you have to avoid silly counterexamples, like $1+3^n$ is always even. – lulu Jun 14 '18 at 23:23
• We don't even know if there are infinitely many Mersenne primes, i.e. those of the form $2^p-1$ (it's easy to show you need prime $p$ here), which is about the simplest possible. – Chappers Jun 14 '18 at 23:24
• @lulu, thanks for your comments. – user237522 Jun 14 '18 at 23:27
• @Chappers, thanks for your comment. Please, do you claim that if $L$ is not prime then $a+b^L$ is not prime? – user237522 Jun 14 '18 at 23:29
Let $a=4, b=n^4\,$ with $\,n \gt 1\,$, then $a+b^L$ is not a prime for any $L \ge 1\,$ by Sophie Germain's:
$$a+b^L=4 + n^{4L}=(n^{2L}+2+2n^L)(n^{2L}+2-2n^L)$$
• Nice, thank you. (I have added an edit to my question). – user237522 Jun 14 '18 at 23:38
• @user237522 Regarding the edit, an even weaker question is whether there exists any pair $\,a,b\,$ for which the proposition was proved. I don't know the answer to that. – dxiv Jun 14 '18 at 23:54
• ok.. thank you. Please, do you think that there is any hope to say something interesting about primes in $a^{L_1}+b^{L_2}$ (or $a^p+b^q$), where $L_1,L_2 \in \mathbb{N}$ ($p$ and $q$ are prime numbers)? Of course, $a$ and $b$ are assumed to be coprime. – user237522 Jun 14 '18 at 23:56
Here's an alternative view that doesn't try to answer your question but merely points out some observations you could make.
$\Bbb{Z}$ under the open set basis $U(a,b) = a + b\Bbb{N}$ forms a topological ring.
Fixing $a$ you can form a topology at least with $U(b) = a + b^{\Bbb{N}}$ since if $a + b^{\Bbb{N}} \cap a + c^{\Bbb{N}} \neq \varnothing$ say $x \in$ the intersection, then $a + b^n = x = a + c^m$ or $b^n = c^m$. There is a some work involved but you need to conclude that there is another basic open set $a + d^{\Bbb{N}}$ containing $x$ contained in the intersection.
• Thank you, interesting. Is it possible to prove Dirichlet's theorem based on this idea? – user237522 Jun 26 '18 at 10:42
• @user237522 not sure. Idk how Dirichlet's is proved. But in both cases you have a topology on $\Bbb{Z}$. It's only a basic observation I'm afraid. Note, that your case $a + b^{\Bbb{N}}$ may not be a topological ring basis, you'd have to prove that it is, but it's likely not because of addition not being continuous. However, even though the ring operations aren't continuous, other things might be such as certain maps reminiscent of our basis set formula. – BananaCats Category Theory App Jul 7 '18 at 15:08
• Thank you very much for the clarification/explanation. – user237522 Jul 8 '18 at 3:43 | 1,073 | 3,360 | {"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} | 3.421875 | 3 | CC-MAIN-2019-22 | latest | en | 0.914848 |
https://afteracademy.com/problems/k-pairs-with-smallest-sums | 1,606,485,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193221.49/warc/CC-MAIN-20201127131802-20201127161802-00660.warc.gz | 191,533,230 | 14,701 | You are given two integer arrays `arr1[]` and `arr2[]` sorted in ascending order and an integer `k`. Suppose a pair `(a, b)` which consists of one element from the first array and one element from the second array. Write a program to find the k pairs `(a1, b1)`, `(a2, b2)...(ak, bk)` with the smallest sums.
Example 1
``````Input: arr1[] = [2,8,12], arr2[] = [4,6,8], k = 3
Output: [[2,4],[2,6],[2,8]]
Explanation: The first 3 pairs are returned from the sequence: [2,4],[2,6],[2,8],[8,4],[8,6],[12,8],[8,8],[12,6],[12,8]``````
Example 2
``````Input: arr1[] = [3,3,4], arr2[] = [3,4,5], k = 2
Output: [3,3],[3,3]
Explanation: The first 2 pairs are returned from the sequence: [3,3],[3,3],[3,5],[4,3],[3,4],[4,4],[3,5],[3,5],[4,5]``````
Example 3
``````Input: arr1[] = [5,6], arr2[] = [7], k = 3
Output: [5,7],[6,7]
Explanation: All possible pairs are returned from the sequence: [5,7],[6,7]`````` | 356 | 903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-50 | latest | en | 0.815905 |
https://www.physicsforums.com/threads/object-sliding-down-inclined-plane.34158/ | 1,519,589,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816912.94/warc/CC-MAIN-20180225190023-20180225210023-00603.warc.gz | 929,703,669 | 14,453 | # Object sliding down inclined plane
1. Jul 7, 2004
### akatsafa
An object slides down an inclined plane of angle 30 degrees and of incline length 4m. If the initial speed of the object is 5m/s directed down the incline, what is the speed at the bottom? Neglect friction.
I used Kinitial+Uinitial=Kfinal+Ufinal. I got 2gh+5^2=v^2, but when I solve for v, I get 10.17m/s. What am I doing wrong?
2. Jul 7, 2004
### Staff: Mentor
What did you use for h? h must be the vertical distance the object moves, not the distance along the incline.
3. Jul 7, 2004
### akatsafa
Should I use 4sin30 as my height?
4. Jul 7, 2004
### Staff: Mentor
Yes. That's the height. | 211 | 668 | {"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} | 3.296875 | 3 | CC-MAIN-2018-09 | longest | en | 0.858263 |
https://oeis.org/A229428 | 1,720,946,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00489.warc.gz | 378,676,095 | 4,502 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A229428 T(n,k) = Number of n X k 0..2 arrays with horizontal differences mod 3 never 1, vertical differences mod 3 never -1, rows lexicographically nondecreasing, and columns lexicographically nonincreasing. 9
3, 5, 5, 8, 12, 8, 12, 27, 27, 12, 17, 55, 83, 55, 17, 23, 102, 222, 222, 102, 23, 30, 175, 524, 754, 524, 175, 30, 38, 282, 1116, 2204, 2204, 1116, 282, 38, 47, 432, 2187, 5700, 7816, 5700, 2187, 432, 47, 57, 635, 4005, 13345, 24126, 24126, 13345, 4005, 635, 57 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Table starts ..3...5....8....12.....17.....23......30.......38.......47........57........68 ..5..12...27....55....102....175.....282......432......635.......902......1245 ..8..27...83...222....524...1116....2187.....4005.....6936.....11465.....18219 .12..55..222...754...2204...5700...13345....28794....58053....110550....200533 .17.102..524..2204...7816..24126...66503...166972...387738....842802...1731129 .23.175.1116..5700..24126..87648..281016...812352..2152643...5297329..12231874 .30.282.2187.13345..66503.281016.1037193..3420692.10260128..28379127..73192023 .38.432.4005.28794.166972.812352.3420692.12768612.43042290.132960319.380811699 LINKS R. H. Hardin, Table of n, a(n) for n = 1..480 FORMULA Empirical for column k: k=1: a(n) = (1/2)*n^2 + (1/2)*n + 2 k=2: a(n) = (1/24)*n^4 + (5/12)*n^3 + (11/24)*n^2 + (25/12)*n + 2, A229422 k=3: [polynomial of degree 6], A229423 k=4: [polynomial of degree 8], A229424 k=5: [polynomial of degree 10], A229425 k=6: [polynomial of degree 12], A229426 k=7: [polynomial of degree 14] EXAMPLE Some solutions for n=4 k=4 ..2..2..1..0....1..1..0..0....1..1..0..0....2..1..1..1....1..0..0..0 ..2..2..1..1....2..1..0..0....2..1..0..0....2..1..1..1....1..1..1..1 ..2..2..1..1....2..1..1..1....2..1..1..1....2..2..2..1....2..2..1..1 ..2..2..2..1....2..1..1..1....2..2..1..1....2..2..2..2....2..2..2..1 CROSSREFS Column 1 is A022856(n+4). Main diagonal is A229421. Sequence in context: A063285 A316938 A112507 * A348374 A029639 A087349 Adjacent sequences: A229425 A229426 A229427 * A229429 A229430 A229431 KEYWORD nonn,tabl AUTHOR R. H. Hardin, Sep 22 2013 STATUS approved
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Last modified July 14 03:51 EDT 2024. Contains 374291 sequences. (Running on oeis4.) | 1,059 | 2,649 | {"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 | 0.637287 |
https://tolstoy.newcastle.edu.au/R/e2/help/06/10/4054.html | 1,586,177,830,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371624083.66/warc/CC-MAIN-20200406102322-20200406132822-00227.warc.gz | 728,512,144 | 4,253 | # Re: [R] Multivariate regression
From: Andris Jankevics <andza_at_osi.lv>
Date: Mon 30 Oct 2006 - 09:01:16 GMT
Andris Jankevics
On Sestdiena, 28. Oktobris 2006 06:04, Ritwik Sinha wrote:
> You can use gee (
> http://finzi.psych.upenn.edu/R/library/geepack/html/00Index.html) or maybe
> the function gls in nlme.
>
> Ritwik.
>
> On 10/27/06, Ravi Varadhan <rvaradhan@jhmi.edu> wrote:
> > Hi,
> >
> >
> >
> > Suppose I have a multivariate response Y (n x k) obtained at a set of
> > predictors X (n x p). I would like to perform a linear regression taking
> > into consideration the covariance structure of Y within each unit - this
> > would be represented by a specified matrix V (k x k), assumed to be the
> > same
> > across units. How do I use "lm" to do this?
> >
> >
> >
> > One approach that I was thinking of is as follows:
> >
> >
> >
> > Flatten Y to a vector, say, Yvec (n*k x 1). Create Xvec (n*k, p*k) such
> > that it is made up of block matrices Bij (k x k), where Bij is a diagonal
> > matrix with X_ij as the diagonal (i = 1,.n, and j = 1,.,p). Now I can
> > use "lm" in a univariate mode to regress Yvec against Xvec, with
> > covariance matrix Vvec (n*k x n*k). Vvec is a block-diagonal matrix with
> > blocks of V along the diagonal. This seems like a valid approach, but I
> > still don't know how to specify the covariance structure to do weighted
> > least squares.
> >
> >
> >
> > Any help is appreciated.
> >
> >
> >
> > Best,
> >
> > Ravi.
> >
> >
> >
> >
> > -------------------------------------------------------------------------
> >--- -------
> >
> > Ravi Varadhan, Ph.D.
> >
> > Assistant Professor, The Center on Aging and Health
> >
> > Division of Geriatric Medicine and Gerontology
> >
> > Johns Hopkins University
> >
> > Ph: (410) 502-2619
> >
> > Fax: (410) 614-9625
> >
> > Email: rvaradhan@jhmi.edu
> >
> > Webpage:
> >
> >
> >
> >
> > -------------------------------------------------------------------------
> >--- --------
> >
> >
> >
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Mon Oct 30 20:14:42 2006
Archive maintained by Robert King, hosted by the discipline of statistics at the University of Newcastle, Australia.
Archive generated by hypermail 2.1.8, at Mon 30 Oct 2006 - 09:30:14 GMT.
Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help. Please read the posting guide before posting to the list. | 843 | 2,931 | {"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-2020-16 | latest | en | 0.753404 |
https://www.physicsforums.com/threads/generating-circular-polarization.325509/ | 1,611,306,367,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00770.warc.gz | 956,531,270 | 15,560 | # Generating Circular Polarization
Statement:
Consider two dipole antennas, oriented 90degrees apart [imagine the x-y plane, let "a" be the dipole oriented along the x-axis, and the "b" be the dipole oriented along the y-axis]. If "a" dipole radiates $$cos(\omega t)$$ and "b" dipole radiates $$sin(\omega t)$$, the field radiated by the two antennas will be circularly polarized:
$$\vec{E}(z, t) = E_{0}[cos(\omega t - \beta z)\hat{x} + sin(\omega t - \beta z)\hat{y}]$$ (#1)
Side note: Very often, helical antennas are used to generate a circularly-polarized (CP) wave. The isolation between a left-handed CP wave and a right-handed CP wave can be significant. Also, a CP wave will change handedness upon reflection.
My thoughts:
I understand that $$E_0$$ is the magnitude of the sinusoid- and in this case it is circular thus both $$\hat{x}, \hat{y}$$ have the same amplitudes respectively. And since both $$sin(\omega t), cos(\omega t)$$ are perpendicular to one another, if one has a phase shift, the other will have the same phase shift $$\beta$$.
Relevant questions:
Is my thoughts above reasonable? What I would really like to know is why the electric field is a function of z also. What is the variable z, and how does it influence the electric field?
Also, can someone explain to me what is meant by
Also, a CP wave will change handedness upon reflection?
Thanks,
Jeffrey
Related Engineering and Comp Sci Homework Help News on Phys.org
Redbelly98
Staff Emeritus
Homework Helper
My thoughts:
I understand that $$E_0$$ is the magnitude of the sinusoid- and in this case it is circular thus both $$\hat{x}, \hat{y}$$ have the same amplitudes respectively. And since both $$sin(\omega t), cos(\omega t)$$ are perpendicular to one another, if one has a phase shift, the other will have the same phase shift $$\beta$$.
Relevant questions:
Is my thoughts above reasonable? What I would really like to know is why the electric field is a function of z also. What is the variable z, and how does it influence the electric field?
Your thoughts look pretty reasonable, but β is not a phase shift. z is the distance away from the antenna in the direction of the E-M wave's propagation, and β=2π/λ is related to the wavelength λ.
Those equations represent a wave travelling in the +z direction. At any fixed time t, the electric field direction makes a rotating, helical pattern as one moves along the z direction.
I'm not familiar with helical antennas, so I can't comment on them.
Also, can someone explain to me what is meant by
Also, a CP wave will change handedness upon reflection.
The wave changes from right-handed to left-handed (or vice versa) CP if it is reflected.
Your thoughts look pretty reasonable, but β is not a phase shift. z is the distance away from the antenna in the direction of the E-M wave's propagation, and β=2π/λ is related to the wavelength λ.
Shouldn't z be $$\hat{z}$$ then? So if it becomes the unit vector (since that is the direction of propagation), then the electric field cannot be a function of z, and thus should it be a function of $$\beta$$ instead- along with $$t$$?
Thanks,
Jeff
Redbelly98
Staff Emeritus
Homework Helper
No, it is z. The electric field is a function of z and t.
No, it is z. The electric field is a function of z and t.
Ok I almost get it now. I am going to start a new thread, this one is getting long.
THanks,
JL
Last edited: | 855 | 3,407 | {"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} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.878214 |
http://primes.utm.edu/curios/cpage/11633.html | 1,534,852,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218122.85/warc/CC-MAIN-20180821112537-20180821132537-00224.warc.gz | 346,799,964 | 2,571 | 19 (another Prime Pages' Curiosity)
Curios: Curios Search: Participate: Single Curio View: (Seek other curios for this number) Here is the mathematical code to determine if a number is divisible by 19. Multiply everything in front of the last digit by nine and subtract the last digit. If you get zero, nineteen, or a multiple of nineteen, the number will be divisible by 19. For example 57. Five times nine minus seven equals 38. Since 19 divides evenly into 38, it divides evenly into 57. Next example 114. Eleven times nine minus four equals 95. Since 19 divides evenly into 95, it divides evenly into 114. [Young] Submitted: 2005-04-03 23:12:38; Last Modified: 2008-01-30 11:28:00. Prime Curios! © 2000-2018 (all rights reserved) privacy statement | 209 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-34 | latest | en | 0.738446 |
https://fr.mathworks.com/matlabcentral/cody/problems/1164-sum-the-digits-of-a-number/solutions/309571 | 1,596,849,876,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00062.warc.gz | 316,613,738 | 16,515 | Cody
# Problem 1164. Sum the Digits of a Number
Solution 309571
Submitted on 23 Aug 2013 by Ashish
• Size: 10
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
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### Test Suite
Test Status Code Input and Output
1 Pass
%% x = '123456789'; y_correct = 9; assert(isequal(digit_sum(x),y_correct))
2 Pass
%% x = '13579'; y_correct = 7; assert(isequal(digit_sum(x),y_correct))
3 Pass
%% x = '1036654257757615301164620529930689045676735109259113932133140605724504628985272966102896725849035075'; y_correct = 5; assert(isequal(digit_sum(x),y_correct))
4 Pass
%% x = '5851147873501164141085965889086954824958752606678975950184825606304112110625645414882256429011165097708998751310932346085834016381957924478113053129649177515212802040810341932020576007951832700665777265307367115487700079617116367572798033657320723526417122504117269467461912747320644603761100467516110111332287512097531691230649461317836258532443574410236994277771642081168571956087153534120969197542720767643838785694086392663104173875192923061073636098783655224289050890906758861210169349969736226546755550793938442137760897037722646218791104180057313259613054984813997639176837835953637446938790362276560342782718153854834909165636800962412231318093037756803017785098259784452756314377610539928858957504653988358962604698474998342789551842878266142728834686534787064418323355335697481001330501689595534408048368891285568524496673551564873437746977135402808065251650010486580915150789952155706519549648556325841434843312042241472703020112115992435204109497067652723884369953849057131345052221998713'; y_correct = 3; assert(isequal(digit_sum(x),y_correct)) | 577 | 1,783 | {"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-2020-34 | latest | en | 0.388029 |
https://leetcode.ca/2024-04-29-3109-Find-the-Index-of-Permutation/ | 1,725,717,556,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00816.warc.gz | 338,183,170 | 9,282 | # 3109. Find the Index of Permutation 🔒
## Description
Given an array perm of length n which is a permutation of [1, 2, ..., n], return the index of perm in the lexicographically sorted array of all of the permutations of [1, 2, ..., n].
Since the answer may be very large, return it modulo 109 + 7.
Example 1:
Input: perm = [1,2]
Output: 0
Explanation:
There are only two permutations in the following order:
[1,2], [2,1]
And [1,2] is at index 0.
Example 2:
Input: perm = [3,1,2]
Output: 4
Explanation:
There are only six permutations in the following order:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]
And [3,1,2] is at index 4.
Constraints:
• 1 <= n == perm.length <= 105
• perm is a permutation of [1, 2, ..., n].
## Solutions
### Solution 1: Binary Indexed Tree
According to the problem requirements, we need to find out how many permutations are lexicographically smaller than the given permutation.
We consider how to calculate the number of permutations that are lexicographically smaller than the given permutation. There are two situations:
• The first element of the permutation is less than $perm[0]$, there are $(perm[0] - 1) \times (n-1)!$ permutations.
• The first element of the permutation is equal to $perm[0]$, we need to continue to consider the second element, and so on.
• The sum of all situations is the answer.
We can use a binary indexed tree to maintain the number of elements that are smaller than the current element in the traversed elements. For the $i$-th element of the given permutation, the number of remaining elements that are smaller than it is $perm[i] - 1 - tree.query(perm[i])$, and the number of permutation types is $(perm[i] - 1 - tree.query(perm[i])) \times (n-i-1)!$, which is added to the answer. Then we update the binary indexed tree and add the current element to the binary indexed tree. Continue to traverse the next element until all elements are traversed.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the permutation.
• class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
this.c = new int[n + 1];
}
public void update(int x, int delta) {
for (; x <= n; x += x & -x) {
c[x] += delta;
}
}
public int query(int x) {
int s = 0;
for (; x > 0; x -= x & -x) {
s += c[x];
}
return s;
}
}
class Solution {
public int getPermutationIndex(int[] perm) {
final int mod = (int) 1e9 + 7;
long ans = 0;
int n = perm.length;
BinaryIndexedTree tree = new BinaryIndexedTree(n + 1);
long[] f = new long[n];
f[0] = 1;
for (int i = 1; i < n; ++i) {
f[i] = f[i - 1] * i % mod;
}
for (int i = 0; i < n; ++i) {
int cnt = perm[i] - 1 - tree.query(perm[i]);
ans = (ans + cnt * f[n - i - 1] % mod) % mod;
tree.update(perm[i], 1);
}
return (int) ans;
}
}
• class BinaryIndexedTree {
private:
int n;
vector<int> c;
public:
BinaryIndexedTree(int n)
: n(n)
, c(n + 1) {}
void update(int x, int delta) {
for (; x <= n; x += x & -x) {
c[x] += delta;
}
}
int query(int x) {
int s = 0;
for (; x > 0; x -= x & -x) {
s += c[x];
}
return s;
}
};
class Solution {
public:
int getPermutationIndex(vector<int>& perm) {
const int mod = 1e9 + 7;
using ll = long long;
ll ans = 0;
int n = perm.size();
BinaryIndexedTree tree(n + 1);
ll f[n];
f[0] = 1;
for (int i = 1; i < n; ++i) {
f[i] = f[i - 1] * i % mod;
}
for (int i = 0; i < n; ++i) {
int cnt = perm[i] - 1 - tree.query(perm[i]);
ans += cnt * f[n - i - 1] % mod;
tree.update(perm[i], 1);
}
return ans % mod;
}
};
• class BinaryIndexedTree:
__slots__ = "n", "c"
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, delta: int) -> None:
while x <= self.n:
self.c[x] += delta
x += x & -x
def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= x & -x
return s
class Solution:
def getPermutationIndex(self, perm: List[int]) -> int:
mod = 10**9 + 7
ans, n = 0, len(perm)
tree = BinaryIndexedTree(n + 1)
f = [1] * n
for i in range(1, n):
f[i] = f[i - 1] * i % mod
for i, x in enumerate(perm):
cnt = x - 1 - tree.query(x)
ans += cnt * f[n - i - 1] % mod
tree.update(x, 1)
return ans % mod
• type BinaryIndexedTree struct {
n int
c []int
}
func NewBinaryIndexedTree(n int) *BinaryIndexedTree {
return &BinaryIndexedTree{n: n, c: make([]int, n+1)}
}
func (bit *BinaryIndexedTree) update(x, delta int) {
for ; x <= bit.n; x += x & -x {
bit.c[x] += delta
}
}
func (bit *BinaryIndexedTree) query(x int) int {
s := 0
for ; x > 0; x -= x & -x {
s += bit.c[x]
}
return s
}
func getPermutationIndex(perm []int) (ans int) {
const mod int = 1e9 + 7
n := len(perm)
tree := NewBinaryIndexedTree(n + 1)
f := make([]int, n)
f[0] = 1
for i := 1; i < n; i++ {
f[i] = f[i-1] * i % mod
}
for i, x := range perm {
cnt := x - 1 - tree.query(x)
ans += cnt * f[n-1-i] % mod
tree.update(x, 1)
}
return ans % mod
}
• class BinaryIndexedTree {
private n: number;
private c: number[];
constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
}
update(x: number, delta: number): void {
for (; x <= this.n; x += x & -x) {
this.c[x] += delta;
}
}
query(x: number): number {
let s = 0;
for (; x > 0; x -= x & -x) {
s += this.c[x];
}
return s;
}
}
function getPermutationIndex(perm: number[]): number {
const mod = 1e9 + 7;
const n = perm.length;
const tree = new BinaryIndexedTree(n + 1);
let ans = 0;
const f: number[] = Array(n).fill(1);
for (let i = 1; i < n; ++i) {
f[i] = (f[i - 1] * i) % mod;
}
for (let i = 0; i < n; ++i) {
const cnt = perm[i] - 1 - tree.query(perm[i]);
ans = (ans + cnt * f[n - i - 1]) % mod;
tree.update(perm[i], 1);
}
return ans % mod;
} | 1,895 | 5,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-38 | latest | en | 0.85891 |
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Q: If P is false, and Q is false, the truth-value of "P ?Q" is false. true. Cannot be determined. All of the above.
A: For clarification, the following paragraph is taken from Wikipedia. Biconditional joins two simpler propositions, and we write "above example (sic)", which is read "P if and only if Q". [ It expresses that P and Q have the same truth-value, thus P if and only if Q is true in cases 1 and 4, and false otherwise. 1. P is true and Q is true 2. P is true and Q is false 3. P is false and Q is
true 4. P is false and Q is false Here is my answer... Therefore, if P is false and Q is false, the truth-value of P iff (if and only if) Q is true. "True". ]
Original conversation
User: If P is false, and Q is false, the truth-value of "P ?Q" is false. true. Cannot be determined. All of the above.
Weegy: For clarification, the following paragraph is taken from Wikipedia. Biconditional joins two simpler propositions, and we write "above example (sic)", which is read "P if and only if Q". [ It expresses that P and Q have the same truth-value, thus P if and only if Q is true in cases 1 and 4, and false otherwise. 1. P is true and Q is true 2. P is true and Q is false 3. P is false and Q is true 4. P is false and Q is false Here is my answer... Therefore, if P is false and Q is false, the truth-value of P iff (if and only if) Q is true. "True". ]
jher000|Points 7504|
User: "~ P v Q" is best read as Not P and Q It is not the case that P and it is not the case that Q It is not the case that P or Q It is not the case that P and Q
Weegy: Biconditional joins two simpler propositions, and we write "above example (sic)", which is read "P if and only if Q". [ It expresses that P and Q have the same truth-value, thus P if and only if Q is true in cases 1 and 4, and false otherwise. 1. [ P is true and Q is true 2. P is true and Q is false 3. P is false and Q is true 4. P is false and Q is false Here is my answer... Therefore, if P is false and Q is false, the truth-value of P iff (if and only if) Q is true. "True". ] ]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,147 | 3,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-22 | latest | en | 0.943863 |
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## How does a lightning ro a fuse, and a circuit breaker increase
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Shop a wide range of server racks and server cabinets for your business needs. Solar and Wind-Powered Energy Devices Property Tax Exemption. Sugar Land metropolitan area and Liberty County. | 983 | 4,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.92247 |
https://scienceblogs.com/dotphysics/2010/04/13/basics-the-electric-field | 1,701,164,595,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099281.67/warc/CC-MAIN-20231128083443-20231128113443-00180.warc.gz | 564,007,155 | 11,684 | # Basics: The electric field
Suppose I take a 1 kg ball and hold it near the surface of the Earth. What would be the gravitational force the Earth exerts on this ball?
And I could say "g" is:
The magnitude of this force would then be 9.8 Newtons. And, if I replaced the ball with a 10 kg ball, the force would be 98 Newtons. What does this have to do with the electric field? Well, you are probably already familiar with this idea of the gravitational force. Guess what? "g" is the gravitational field. Basically, it is the force per unit mass due to the Earth. This is only approximately constant. If I get very far from the surface of the Earth, it might be better to write the gravitational field as:
This is the universal gravity force divided by the mass of the object at the location of interest. So, you could say take a small "test mass" and determine the force on it. If you divide the force on the test mass by the value of the test mass, you get the gravitational field. (note - r-hat points away from the surface of the Earth and G is the gravitational constant.) In general, the gravitational field is:
Maybe you see where this is going or maybe you are thinking "hey, this is supposed to be about the electric field". Yes, it is supposed to be about the electric field. Suppose I put some electric charge somewhere and measure the electric force on it. In that case, I can say the electric field is:
What about a point charge? If I take some electric charge - q1 and I want to find the electric field a distance r away. I will put a small "test charge" there with charge qt. The force on that charge is:
If I divide this force by the value of the test charge, I get the electric field due to a point charge.
## Superposition
The electric field due to a point charge is useful because stuff is made up of things that look like point charges - (electron, proton). But what about other things? What about the electric field due to two point charges? The cool thing about the electric field is that it obeys the idea of superposition. This means that the electric field at any point due to two point charges is the vector some of the electric fields due to each individual charge. Here is a diagram.
It doesn't matter how many charges you have. You can keep adding up these electric fields due to the individual point charges. This is quite useful - as I will hopefully show later.
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In the last paragraph s/vector some/vector sum/
By Grep Agni (not verified) on 14 Apr 2010 #permalink
Say, if charge in coulomb is used for electricity when calculating force and field, what is used for a magnetic field? And is the magnetic field equation similar to the electric field equation? | 607 | 2,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-50 | latest | en | 0.947176 |
https://pay4essay.net/2021/09/17/understanding-an-income-statement-accounting-homework-help/ | 1,679,570,648,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00136.warc.gz | 517,913,517 | 17,851 | # Understanding an income statement | Accounting homework help
Please follow the directions below. The Dehew Health System is attached below.
Instructions
Assessment #1: Understanding an Income Statement
You are an intern in a health care organization business office. The calculations can be done for FY 2018 or 2019. Make sure to identify the year you used. Use the Dehew Health System income statement in the attachment, please develop a 4-6 page paper responding to the following questions regarding the income statement:
1. What is an income statement used for?
2. What are the key components of an income statement?
3. On your example of an income statement, is the organization making a profit or loss? How much is that profit or loss? Show the calculations. (Total Revenues – Total Expenses = Net Profit/Loss)
4. On your example of an income statement, what is the “total contribution margin”? Show the calculations. (Net Sales – Total Variable Costs=Total Contribution Margin).
5. What three financial indicators does a contribution margin help determine? How?
6. What is a break-even analysis? What is the break-even point for this organization? List expenses assigned as variable and fixed. Show the calculations. (Total revenues – Total VC – Total FC= Profit)
7. How does a break-even analysis vary between fee-for-service and capitated payments? Explain it using the income statement you chose. What are the similarities and differences applicable to the Income statements in fee-for-service and capitated payment options?
The paper should not be more than 6 pages and should follow APA format, spelling and grammar. Include a cover page and reference page (not included in the page count). Make sure to properly cite and reference all the used sources. The calculations need to be shown as formulas using figures. Follow the mathematical rules while calculating numbers.
Acknowledgment: The Income Statement for the Dehew Health System was developed and contributed to HMGT 322 by Professor Rob Parker.
Note:
Total Contribution Margin= Net Sales-Total Variable CostNet Sales in our case= Net PT Revenue + Premium Revenue (recognized insurance revenue) Then Total Variable Cost is comprised of Supply expenses, postage/courier services, repair/maintenance, travel/seminars, equipment rentals (not lease), and ed. supplies TCM is straightforward. Break-even point: Total Revenue-Total Variable Cost -Total Fixed Cost (the rest of expenses+ deductions from revenue) Technically, the break-even point will be Total Revenue-(Total Operating Expenses + Deductions from Gross PT Revenue)Total Revenue =Gross PT Revenue+ Premium Revenue+ Other Revenue The statement has Units of Service listed on the bottom. However, it is not possible to calculate Total Revenue by a regular formula of Units*Unit Price as the price is PT specific.
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 896 | 4,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-14 | longest | en | 0.888392 |
https://www.thespiritualmovements.com/meaning-of-the-number/faq-what-is-the-meaning-of-expression-number-in-numerology.html | 1,656,720,764,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103947269.55/warc/CC-MAIN-20220701220150-20220702010150-00583.warc.gz | 1,083,034,340 | 11,334 | ## What is expressive number?
The concept of expression number uses the technique of assignment of single-digit numbers (1 to 9) to the English alphabet from A to Z.
## What is expression number in astrology?
Your expression number reveals your innate potential. That’s why it is sometimes called the “destiny number.” It can provide clues to your talents and skills, as well as your challenges. A careful analysis of your name can point you toward your career, personal goals, and any other aspirations you may have.
## What is my expression destiny number?
To find your Destiny Number, calculate the root number of your full name (first, middle, last) by reducing each name to a single digit, and adding up the total.
## How do you calculate your soul number?
The soul number reflects your most inner thoughts, habits, thoughts, and baseline reactions. This shows what you truly desire, and what you value most in this life. It’s reflective of your inner spirit and can’t be altered no matter what. To calculate your soul number, list out your full name, and extract the vowels.
You might be interested: What Does It Mean A Guy Ask His Friend For Your Number Yahoo?
## What is your Soul Urge number?
What is a Soul Urge Number? Your soul number reveals your core desires and destiny. This number allows your personal numerology to uncover vital details from your numerology chart. Each person has different numbers that help them identify who they are and what their purpose in life is.
## What personality is 9?
The best ways to describe Sun Number 9 individuals are that they are idealists. They are very dedicated to the cause of their people and it comes as no surprise when they choose to be politicians, law enforcement officers or serve in public interest jobs.
## What are the core numbers in numerology?
Traditionally there are five Core Numbers: the Life Path Number, Birth Day Number, Expression Number, Personality Number, and Heart’s Desire Number.
## How do you find your heart’s desire number?
To discover your heart’s true desire number, follow these instructions.
1. Add the value of the vowels in your full birth name: A and I = 1, E = 5, U = 6, O = 7. FYI, use your full name—first, middle, and last.
2. If this gives you a double digit, add the digits together until you get to a single digit.
## How do I find my master number?
How do I find my Master Number? Calculate your Life Path Number by adding up your birthday numbers until you get one single-digit number (unless the number is 11, 22, or 33, which are all Master Numbers). For example, take the birthdate May 18, 1970.
## How do I find my name number?
Reduce the sum of your name’s numbers into a single digit. To reduce the sum, add together the 2 digits within it. For example, if the sum of your letters is a 25, split the 25 and add 2+5 to equal 7. The 7 is your true name number.
You might be interested: FAQ: What Does The Top Number In A Time Signature Mean?
## What is my path number?
Calculating Your Life Path Number To find your Life Path Number, simply reduce the digits of your full birth date until you reach a single-digit number —excluding 11 and 22, which are considered Master Numbers (we’ll get back to this). This number is your Life Path Number.
## Why is the number 3 so powerful?
Throughout human history, the number 3 has always had a unique significance, but why? The ancient Greek philosopher, Pythagoras, postulated that the meaning behind numbers was deeply significant. In their eyes the number 3 was considered as the perfect number, the number of harmony, wisdom and understanding.
## What are the angel numbers?
Angel numbers are recurring sequences of numbers that have spiritual significance, according to numerology. They are trying to get your attention, and the numbers they send have meanings. Think of it like a little marker saying that you are on the right direction in life, like a highway sign.” | 864 | 3,951 | {"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-2022-27 | latest | en | 0.906988 |
https://ethereum.stackexchange.com/questions/23430/how-to-calculate-the-amount-to-transfer-to-completely-empty-an-account | 1,708,605,007,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00384.warc.gz | 250,623,016 | 40,389 | # How to calculate the amount to transfer to completely empty an account?
So let's say my wallet is 1 eth, I need to completely empty it and I'm going to do it this way
``````Var value = web3.toWei(1) - gas
``````
Is that right?
• – eth
Aug 1, 2017 at 6:00
Yes. You will need to figure out how much gas (in ETH) the transaction will cost, but luckily simple ETH transfers are exactly 21,000 gas (I think, double check this number). This means to know how much ETH will be spent on gas you will need to multiply that by the chosen gas price (e.g., 10^10).
So your calculation will need to look something like this:
``````var gasPrice = 20*10**9;
var gas = 21000;
var attoethForGas = gasPrice * gas;
var amountToSend = web3.toWei(1) - attoethForGas;
``````
• If I take this gas as 5000 and I can't completely empty my account（unused gas is refunded）
– ali
Aug 1, 2017 at 3:10
• @D.Doy I updated the answer to include a code example, let me know if that helps clear things up. Aug 1, 2017 at 3:13
• Thank you for your answer, but it seems that it didn't prevent the refund from gas too big. If it were to be cleared, do I have to make gas small enough?
– ali
Aug 1, 2017 at 5:43
• Hmm, I'm not sure what your problem might be, I'm reasonably confident that the above is the way to send everything. Can you share the code you have that isn't working? There should be no gas refund for an ETH transfer, even if it zeros out the balance. Aug 1, 2017 at 14:20
In order to calculate an exact value to send, you must use precise calculations which JavaScript doesn't give you out of the box. Luckily, web3.js already returns a special `BigNumber` object from `eth.getBalance()`. In order to execute precise math, you need to use `BigNumber` methods instead of using default arithmetic operators. For example:
``````gasPrice = new BigNumber(web3.toWei('0.6', 'gwei'))
cost = gasPrice.mul(21000)
value = eth.getBalance(eth.accounts[0]).sub(cost)
``````
Constants were pulled from these sources at time of posting:
• ethgasstation.info is actually down at the moment, 0.6 gwei was just a fairly stable price I saw before it went down. Aug 1, 2017 at 17:44 | 602 | 2,154 | {"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-2024-10 | latest | en | 0.918508 |
https://www.math-only-math.com/worksheet-on-comparison-of-integers.html | 1,721,701,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00728.warc.gz | 740,061,355 | 13,359 | # Worksheet on Comparison of Integers
I. Compare the given numbers and put the right sign >, < or =. You may think of a number line when considering the answers:
(i) +7 …….. +17(ii) +53 …….. -53(iii) 0 …….. -25(iv) -108 …….. -208 (v) -12 …….. +14(vi) -26 …….. 0(vii) +5 …….. -60(viii) -10 …….. 0
II. Mark the integers on the number line and arrange the given numbers in increasing order. One has been solved to get the idea.
(i) +20, -2, -5, 0, +16
Ascending order: -5, -2, 0, +16, +20
(ii) +30, -10, +20, -60, -70
(iii) +18, 0, -5, -17, +19
(iv) -9, -12, +55, -15, +60
(v) -3, -6, +9, -8, +10
(vi) -5, +6, +1, +8, -4
III. Mark the integers on the number line and arrange the given numbers in decreasing order.
(i) -2, -5, +8, -9, +11
(ii) -11, +21, -31, +41, -51
(iii) +4, -10, 0, +6, -9
(iv) -6, -2, -8, -15, -22
(v) -16, +19, +34, -47, -88
(vi) -10, +6, -50, +10, 0
IV. Write:
(i) 4 numbers greater than – 5 but less than 0
(ii) 5 numbers greater than – 6 but less than 1
V. Write the integers which come just before and just after the given integers:
(i)_____ -7 _____
(ii) _____ -1000 _____
(iii) _____ 24 _____
(iv) _____ -665 _____
(v) _____ -39 _____
(vi) _____ 0 _____
Answers for the worksheet on comparison of integers are given below to check the exact answers of the above questions.
I. (i) <
(ii) >
(iii) >
(iv) >
(v) <
(vi) <
(vii) >
(viii) <
II. (ii) -70, -60, -10, +20, +30
(iii) -17, -5, 0, +18, +19
(iv) -15, -12, -9, +55, +60
(v) -8, -6, -3, +9, +10
(vi) -5, -4, +1, +6, +8
III. (i) +11, +8, -2, -5, -9
(ii) +41, +21, -11, -31, -51
(iii) +6, +4, 0, -9, -10
(iv) -2, -6, -8, -15, -22
(v) +34, +19, -16, -47, -88
(vi) +10, +6, 0, -10, -50
IV. (i) -4, -3, -2, -1
(ii) -4, -3, -2, -1, 0
V. (i) -8, -6
(ii) -1001, -999
(iii) 23, 25
(iv) -666, -664
(v) -40, -38
(vi) -1, +1
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When we write a decimal number with three places, we are representing the thousandths place. Each part in the given figure represents one-thousandth of the whole. It is written as 1/1000. In the decim… | 1,124 | 3,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-30 | latest | en | 0.553254 |
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2607 | 1,561,604,304,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000610.35/warc/CC-MAIN-20190627015143-20190627041143-00346.warc.gz | 5,426,887 | 2,855 | Welcome to ZOJ
Problem Sets Information Select Problem Runs Ranklist
ZOJ Problem Set - 2607
Laboratory
Time Limit: 10 Seconds Memory Limit: 32768 KB Special Judge
John is working in a secret laboratory. Now he is preparing a special experiment involving four dangerous liquid substances with code names 'A', 'B', 'C', and 'D'.
A special container is used to keep the substances. The container is made of a special strong glass. It consists of four sections: 'A', 'B', 'C', and 'D' as shown on the picture.
Each substance is to be kept in its own part, marked with the letter that coincides with the code name of the substance. Since the container must be transported using standard carts, the height of the container is fixed and must be equal to one inch.
John knows, that he needs a, b, c, and d cubic inches of substances 'A', 'B', 'C', and 'D' respectively. Now he wonders what can be the minimal possible total square of the container so that the required volume of each substance would fit its section. Help him to find it out.
Input
Each test case contains four positive integral numbers a, b, c, and d, not exceeding 106. Proceed until the end of file.
Output
For each case, output the answers in two consecutive lines. On the first line print S - the minimal possible square of the container. On the next line print four real numbers p1, p2, q1, and q2 - the dimensions of the container sections as shown on the picture. Print at least six digits after the decimal point. Do not print any blank lines.
Sample Input
```5 6 8 7
2 1 1 1
```
Sample Output
```27.8571428571
1.6641005886 1.9414506867 4.1206300292 3.6055512754
5.8284271247
1.41421356237309505000 1.00000000000000000000 1.41421356237309505000 1.00000000000000000000
```
Author: Andrew Stankevich
Source: Andrew Stankevich's Contest #7
Submit Status | 488 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-26 | latest | en | 0.844617 |
https://phys.libretexts.org/Core/Quantum_Mechanics/Fowler's_Quantum_Mechanics/1%3A_Reviewing_Elementary_Material/1.4%3A_Wave_Equations%2C_Wavepackets_and_Superposition | 1,508,758,499,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00442.warc.gz | 808,464,552 | 24,850 | $$\require{cancel}$$
# 1.4: Wave Equations, Wavepackets and Superposition
### A Challenge to Schrödinger
De Broglie’s doctoral thesis, defended at the end of 1924, created a lot of excitement in European physics circles. Shortly after it was published in the fall of 1925 Pieter Debye, a theorist in Zurich, suggested to Erwin Schrödinger that he give a seminar on de Broglie’s work. Schrödinger gave a polished presentation, but at the end Debye remarked that he considered the whole theory rather childish: why should a wave confine itself to a circle in space? It wasn’t as if the circle was a waving circular string, real waves in space diffracted and diffused, in fact they obeyed three-dimensional wave equations, and that was what was needed. This was a direct challenge to Schrödinger, who spent some weeks in the Swiss mountains working on the problem, and constructing his equation.
There is no rigorous derivation of Schrödinger’s equation from previously established theory, but it can be made very plausible by thinking about the connection between light waves and photons, and construction an analogous structure for de Broglie’s waves and electrons (and, later, other particles).
### Maxwell’s Wave Equation
Let us examine what Maxwell’s equations tell us about the motion of the simplest type of electromagnetic wave—a monochromatic wave in empty space, with no currents or charges present.
As we discussed in the last lecture, Maxwell found the wave equation
$\nabla ^2 \vec {E} - \dfrac {1}{c^2} \dfrac {\partial ^2 \vec E}{\partial t^2} = 0$
which reduces to
$\dfrac {\partial ^2 \vec {E}}{\partial x^2} - \dfrac {1}{c^2} \dfrac {\partial ^2 \vec {E}}{\partial t^2} = 0$
for a plane wave moving in the x-direction, with solution
$\vec {E} (x,t) = \vec {E} _0 e^{i(kx - wt)}$
Applying the wave equation differential operator to this plane wave solution
$\left ( \dfrac {\partial ^2}{\partial x^2} - \dfrac {1}{c^2} \dfrac {\partial ^2}{\partial t^2} \right ) \vec {E} _0 e^{i(kx - wt)} = \left ( k^2 - \dfrac {\omega ^2}{c^2} \right ) \vec {E}_0 e^{i(kx - wt)} = 0$
so
$\omega = ck$
This is just the familiar statement that the wave must travel at $$c$$.
### What does the Wave Equation tell us about the Photon?
We know from the photoelectric effect and Compton scattering that the photon energy and momentum are related to the frequency and wavelength of the light by
$E = hv = \hbar \omega$
$p = \dfrac {h}{\lambda} = \hbar k$
Notice, then, that the wave equation tells us that $$\omega = ck$$ and hence $$E = cp$$. To put it another way, if we think of $$e^{i(kx -wt)}$$ as describing a particle (photon) it would be more natural to write the plane wave as
$\vec {E}_0 e^{\dfrac {i}{k} (px -Et)}$
that is, in terms of the energy and momentum of the particle. In these terms, applying the (Maxwell) wave equation operator to the plane wave yields
$\left ( \dfrac {\partial ^2}{\partial x^2} - \dfrac {1}{c^2} \dfrac {\partial ^2}{\partial t^2} \right ) \vec {E} _0 e^{\dfrac {i}{k} (px -Et)} = \left ( p^2 - \dfrac {E^2}{c^2} \right ) \vec {E} _0 e^{\dfrac {i}{k} (px -Et)} = 0$
or
$E^2 = c^2 p^2$
The wave equation operator applied to the plane wave describing the particle propagation yields the energy-momentum relationship for the particle.
### Constructing a Wave Equation for a Particle with Mass
The discussion above suggests how we might extend the wave equation operator from the photon case (zero rest mass) to a particle having rest mass $$m_0$$. We need a wave equation operator that, when it operates on a plane wave, yields
$E^2 = c^2 p^2 + m^2_0 c^4$
Writing the plane wave function
$\phi (x,t) = A e^{\dfrac {i}{k} (px -Et)}$
where $$A$$ is a constant, we find we can get $$E^2 = c^2 p^2 + m^2_0 c^4$$ by adding a constant (mass) term to the differentiation terms in the wave operator:
$\left ( \dfrac {\partial ^2}{\partial x^2} - \dfrac {1}{c^2} \dfrac {\partial ^2}{\partial t^2} - \dfrac {m^2_0 c^2}{\hbar ^2} \right ) A e^{\dfrac {i}{k} (px -Et)} = \dfrac {1}{\hbar ^2} \left ( p^2 - \dfrac {E^2}{c^2} + m^2_0 c^2 \right ) A e^{\dfrac {i}{k} (px -Et)} = 0$
This wave equation is called the Klein-Gordon equation and correctly describes the propagation of relativistic particles of mass $$m_0$$. However, it’s a bit inconvenient for nonrelativistic particles, like the electron in the hydrogen atom, just as $$E^2 = c^2 p^2 + m^2_0 c^4$$ is less useful than $$E= p^2/2m$$ for this case.
### A Nonrelativistic Wave Equation
Continuing along the same lines, let us assume that a nonrelativistic electron in free space (no potentials, so no forces) is described by a plane wave:
$\psi (x,t) = Ae^{\dfrac {i}{k}(px - Et)}$
We need to construct a wave equation operator which, applied to this wave function, just gives us the ordinary nonrelativistic energy-momentum relationship, $$E = \dfrac {p^2}{2m}$$. The p2 obviously comes as usual from differentiating twice with respect to x, but the only way we can get E is by having a single differentiation with respect to time, so this looks different from previous wave equations:
$i \hbar \dfrac {\partial \psi (x,t)}{\partial t} = - \dfrac {\hbar ^2}{2m} \dfrac {\partial ^2 \psi (x,t)}{\partial x^2}$
This is Schrödinger’s equation for a free particle. It is easy to check that if $$\psi (x,t)$$ has the plane wave form given above, the condition for it to be a solution of this wave equation is just $$E = \dfrac {p^2}{2m}$$.
Notice one remarkable feature of the above equation—the i on the left means that $$\psi$$ cannot be a real function.
### How Does a Varying Potential Affect a de Broglie Wave?
The effect of a potential on a de Broglie wave was considered by Sommerfeld in an attempt to generalize the rather restrictive conditions in Bohr’s model of the atom. Since the electron was orbiting in an inverse square force, just like the planets around the sun, Sommerfeld couldn’t understand why Bohr’s atom had only circular orbits, no Kepler-like ellipses. (Recall that all the observed spectral lines of hydrogen were accounted for by energy differences between circular orbits.)
De Broglie’s analysis of the allowed circular orbits can be formulated by assuming at some instant in time the spatial variation of the wave function on going around the orbit includes a phase term of the form $$e^{\dfrac {ipq}{k}}$$, where here the parameter q measures distance around the orbit. Now for an acceptable wave function, the total phase change on going around the orbit must be $$2n \pi$$, where n is an integer. For the usual Bohr circular orbit, p is constant on going around, q changes by $$2 \pi r$$, where r is the radius of the orbit, giving
$\dfrac {1}{\hbar} p 2 \pi r = 2 n \pi$
so
$pr = n \hbar$
the usual angular momentum quantization.
What Sommerfeld did was to consider a general Kepler ellipse orbit, and visualize the wave going around such an orbit. Assuming the usual relationship $$p = \dfrac {h}{\lambda}$$, the wavelength will vary as the particle moves around the orbit, being shortest where the particle moves fastest, at its closest approach to the nucleus. Nevertheless, the phase change on moving a short distance $$\Delta q$$ should still be $$\dfrac {p \Delta q}{\hbar}$$, and requiring the wave function to link up smoothly on going once around the orbit gives
$\oint pdq = nh$
Thus only certain elliptical orbits are allowed. The mathematics is nontrivial, but it turns out that every allowed elliptical orbit has the same energy as one of the allowed circular orbits. That is why Bohr’s theory gave all the energy levels. Actually, this whole analysis is old fashioned (it’s called the “old quantum theory”) but we’ve gone over it to introduce the idea ofa wave with variable wavelength, changing with the momentum as the particle moves through a varying potential.
The reader may well be wondering at this point why it is at all useful to visualize a real wave going round an orbit, when we have stated that any solution of Schrödinger’s equation is necessarily a complex function. As we shall see, it is often possible to find solutions, including those corresponding to Bohr’s energy levels, in which the complex nature of the wave function only appears in a time varying phase factor, $$e^{-\dfrac {iEt}{k}}$$. We should also add that if the spatial dependence is a real function, such as sinkx, it represents a standing wave, not a particle circling in one direction, which would be $$e^{ikx}$$, or $$e^{\dfrac {ipq}{k}}$$. Bearing all this in mind, it is still often instructive to sketch real wave functions, especially for one-dimensional problems.
### Schrödinger’s Equation for a Particle in a Potential
Let us consider first the one-dimensional situation of a particle going in the x-direction subject to a “roller coaster” potential. What do we expect the wave function to look like? We would expect the wavelength to be shortest where the potential is lowest, in the valleys, because that’s where the particle is going fastest—maximum momentum.
With a nonzero potential present, the energy-momentum relationship for the particle becomes the energy equation
$E = \dfrac {p^2}{2m} + V (x)$
We need to construct a wave equation which leads naturally to this relationship. In contrast to the free particle cases discussed above, the relevant wave function here will no longer be a plane wave, since the wavelength varies with the potential. However, at a given x, the momentum is determined by the “local wavelength”, that is,
$p = -i \hbar \dfrac {\partial \psi}{\partial x}$
It follows that the appropriate wave equation is:
$i\hbar \dfrac {\partial \psi (x,t)}{\partial t} = - \dfrac {\hbar ^2}{2m} \dfrac {\partial ^2 \psi (x,t)}{\partial x^2} + V(x) \psi (x,t)$
This is the standard one-dimensional Schrödinger equation.
In three dimensions, the argument is precisely analogous. The only difference is that the square of the momentum is now a sum of three squared components, for the x, y and z directions, so $$\dfrac {\partial ^2}{\partial x^2} \text {becomes} \dfrac {\partial ^2}{\partial x^2} + \dfrac {\partial ^2}{\partial y^2} + \dfrac {\partial ^2}{\partial z^2} = \nabla ^2$$,
so now
$i \hbar \dfrac {\partial \psi (x,y,z,t)}{\partial t} = - \dfrac {\hbar ^2}{2m} \nabla ^2 \psi (x,y,z,t) + V (x,y,z)\psi (x,y,z,t)$
This is the complete Schrödinger equation. So far, of course, it is based on plausibility arguments and hand-waving. Why should anyone believe that it really describes an electron wave? Schrödinger’s test of his equation was the hydrogen atom. He looked for Bohr’s “stationary states”: states in which the electron was localized somewhere near the proton, and having a definite energy. The time dependence would be the same as for a plane wave of definite energy, $$e^{-\dfrac {iEt}{k}}$$, the spatial dependence would be a time-independent function decreasing rapidly at large distances from the proton. That is, he took
$\psi (x,y,z,t) = e^{-\dfrac {iEt}{k}} \psi (x,y,z)$
He took advantage of the spherical symmetry by re-expressing the spatial wave function in spherical polar coordinates, and found his equation became a standard differential equation solved in the nineteenth century. The solution gave the shape of possible wave functions, and also allowed values of energy and angular momentum. These values were exactly the same as Bohr’s (except that the lowest allowed state in the new theory had zero angular momentum): impressive evidence that the new theory was correct.
### Current Conservation
When Schrödinger published this result in 1926, he also wrote down the complex conjugate equation, and proved that taking them together it was not difficult to deduce a continuity equation:
$\dfrac {\partial \rho}{\partial t} + div \vec {j} = 0$
where $\rho = \psi ^* \psi = | \psi |^2$
and $\vec {j} = \dfrac {\hbar}{2mi} (\psi ^* \vec {\nabla} \psi - \psi \vec {\nabla} \psi ^*)$
### But What Do These Equations Mean?
Schrödinger believed the above continuity equations represented the conservation of electric charge, and had no further significance. He thought that after all his own equation showed the electron to be just a smooth classical wave at the deepest level. In fact, he succeeded in solving the three-dimensional equation with a Coulomb potential and he found the Bohr energy levels of the hydrogen atom. Obviously, he was on the right track! This classical revival approach, however, couldn’t deal with the unpredictability of quantum mechanics, such as where a single photon—or electron—would land in a two-slit diffraction pattern.
The truth is, Schrödinger didn’t understand his own equation. Another physicist, Max Born, published a paper a few days after Schrödinger’s in which he suggested that $$|\psi (x,y,z,t)|^2 \,dx \,dy\, dz$$ was the relative probability of finding the electron in a small volume dxdydz at (x,y,z) at time t. This interpretation was based directly on the analogy with light waves and photons, and has turned out to be correct.
Notation note: $$\psi$$ is called the “amplitude” or sometimes the “probability amplitude”.
One clue to this probabilistic interpretation is fact that like most waves, solutions to Schrödinger’s equation tend to spread out. Stretch out a string or spring and attach a mass to the middle. If it is one of those "snaky springs" often found in physics classrooms, use duct tape and let the mass be a wooden block with linear dimensions between 1 and 2 inches. Sharply hit the spring near one end with a rod and observe that a pulse travels towards the center. This pulse vaguely resembles a particle...until it hits the block and splits in two. With a block of the right mass, the reflected and transmitted pulses can be made to be roughly equal in amplitude. What property of a particle can be in two places at once? It's hard to think of an answer except "probability", or where the particle might be. Light and sound waves also split when striking an object, which explains how the image of a setting sun can be seen if you look into a window from the outside of a building.
### Photons and Electrons
We have seen that electrons and photons behave in a very similar fashion—both exhibit diffraction effects, as in the double slit experiment, both have particle like or quantum behavior. As we have already discussed, we now have a framework for understanding photons—we first figure out how the electromagnetic wave propagates, using Maxwell’s equations, that is, we find E as a function of x,y,z,t. Having evaluated E(x,y,z,t), the probability of finding a photon in a given small volume of space dxdydz, at time t, is proportional to |E(x,y,z,t)|2dxdydz, the energy density.
Born assumed that Schrödinger’s wave function for the electron corresponded to the electromagnetic wave for the photon in the sense that the square of the modulus of the Schrödinger wave amplitude at a point was the relative probability density for finding the electron at that point. So the routine is the same: for given boundary conditions and a given potential, Schrödinger’s differential equation can be solved and the wave function $$\psi (x,y,z,t)$$ evaluated. Then, $$|\psi (x,y,z,t)|^2 dx dy dz$$ gives the relative probability of finding the electron at (x,y,z) at time t.
Notice, though, that this interpretation of the wave function is not essential in finding the allowed energy levels in a given potential, such as the Bohr orbit energies, which Schrödinger derived before the physical significance of his wave function was understood.
### How Wave Amplitude Varies in a Roller Coaster Potential
We mentioned above that for an electron traveling along a roller coaster potential, the local wavelength is related to the momentum of the electron as it passes that point.
Perhaps slightly less obvious is that the amplitude of the wave varies: it will be largest at the tops of the hills (provided the particle has enough energy to get there) because that’s where the particle is moving slowest, and therefore is most likely to be found.
### Keeping the Wave and the Particle Together?
Suppose following de Broglie we write down the relation between the “particle properties” of the electron and its “wave properties”:
$\dfrac {1}{2} mv^2 = E = h\nu$
and from de Broglie's relationship
$mv = p =\dfrac {h}{\lambda}$
It would seem that we can immediately figure out the speed of the wave, just using
$\lambda \nu = c'.$
We find:
$\lambda \nu = \dfrac {h}{mv} \cdot \dfrac {\dfrac {1}{2} mv^2}{h} = \dfrac {1}{2} v$
So the speed of the wave seems to be only half the speed of the electron! How could they stay together? What’s wrong with this calculation?
### Localizing an Electron
To answer this question, it is necessary to think a little more carefully about the wave function corresponding to an electron traveling through a cathode ray tube, say. The electron leaves the cathode, shoots through the vacuum, and impinges on the screen. At an intermediate point in this process, it is moving through the vacuum and the wave function must be nonzero over some volume, but zero in the places the electron has not possibly reached yet, and zero in the places it has definitely left.
However, if the electron has a precise energy, say exactly a thousand electron volts, it also has a precise momentum. This necessarily implies that the wave has a precise wavelength. But the only wave with a precise wavelength $$\lambda$$ has the form
$\psi (x,t) = A e^{i(kx - \omega t)}$
where $$k = \dfrac {2 \pi}{\lambda} \text {and} \omega = 2 \pi f$$. The problem is that this plane sine wave extends to infinity in both spatial directions, so cannot represent a particle whose wave function is nonzero in a limited region of space.
Therefore, to represent a localized particle, we must superpose waves having different wavelengths. Now, the waves representing electrons, unlike the light waves representing photons, travel at different speeds for different energies. Any intuition gained by thinking about superposing light waves of different wavelengths can be misleading if applied to electron waves!
Fortunately, there are many examples in nature of waves whose speed depends on wavelength. A simple example is water waves on the ocean. We consider waves having a wavelength much shorter than the depth of the ocean. What is the $$\omega , k$$ relationship for these waves? We know it’s not $$\omega = Ck$$, with a constant C, because waves of different wavelengths move at different speeds. In fact, it’s easy to figure out the $$\omega , k$$, relationship, known as the dispersion relation, for these waves from a simple dimensional argument. What physical parameters can the wave frequency depend on? Obviously, the wavelength $$\lambda$$. We will use $$k = \dfrac {2 \pi}{\lambda}$$ as our variable. k has dimensions L-1
These waves are driven by gravity, so g, with dimensions LT-2, is relevant. Actually, that’s all. For ocean waves, surface tension is certainly negligible, as is the air density, and the water’s viscosity. You might think the density of the water matters, but these waves are rather like a pendulum, in that they are driven by gravity, so increasing the density would increase both force and inertial mass by the same amount.
For these deepwater waves, then, dimensional analysis immediately gives:
$\omega ^2 = C g k$
where C is some dimensionless constant we cannot fix by dimensional argument, but actually it turns out to be 1.
### Wavepackets and the Principle of Superposition
To return momentarily to the electron traveling through a vacuum, it is clear physically that it must have a wave function that goes to zero far away in either direction (we’ll still work in one dimension, for simplicity). A localized wave function of this type is called a “wavepacket”. We shall discover that a wavepacket can be constructed by adding plane waves together. Now, the plane waves we add together will individually be solutions of the Schrödinger equation.
But does it follow that the sum of such solutions of the Schrödinger equation is itself a solution to the equation? The answer is yes—in other words, the Schrödinger equation
$i\hbar \dfrac {\partial \psi (x,y,z,t)}{\partial t} = - \dfrac {\hbar ^2}{2m} \nabla ^2 \psi (x,y,z,t) + V(x,y,z) \psi (x,y,z,t)$
is a linear equation, that is to say, if $$\psi _1(x,y,z,t), \psi _2(x,y,z,t)$$ are both solutions of the equation, then so is
$\psi (x,y,z,t) = c_1 \psi _1(x,y,z,t) + c_2 \psi _2(x,y,z,t)$
where c1 and c2 are arbitrary constants, as is easy to check. This is called the Principle of Superposition.
The essential point is that in Schrödinger’s equation every term contains a factor $$\psi$$, but no term contains a factor $$\psi ^2$$ (or a higher power). That’s what is meant by a “linear” equation. If the equation did contain a constant term, or a term including $$\psi ^2$$, superposition wouldn’t work—the sum of two solutions to the equation would not itself be a solution to the equation.
In fact, we have been assuming this linearity all along: when we analyze interference and diffraction of waves, we just add the two wave amplitudes at each spot. For the double slit, we take it that if the wave radiating from one slit satisfies the wave equation, then adding the two waves together will give a new wave which also satisfies the equation.
### The First Step in Building a Wavepacket: Adding Two Sine Waves
If we add together two sine waves with frequencies close together, we get beats. This pattern can be viewed as a string of wavepackets, and is useful for gaining an understanding of why the electron speed calculated from $$\lambda f = c'$$ above is apparently half what it should be.
We use the trigonometric addition formula:
$\sin ((k - \Delta k) x - (\omega - \Delta \omega) t) + \sin ((k + \Delta k) x - ( \omega + \Delta \omega )t) = 2 \sin (kx - \omega t ) \cos ((\Delta k) x - (\Delta \omega)t)$
This formula represents the phenomenon of beats between waves close in frequency. The first term, $$\sin (kx - \omega t)$$, oscillates at the average of the two frequencies. It is modulated by the slowly varying second term, often called the “envelope function”, which oscillates once over a spatial extent of order $$\dfrac {\pi}{\Delta k}$$. This is the distance over which waves initially in phase at the origin become completely out of phase. Of course, going a further distance of order $$\dfrac {\pi}{\Delta k}$$, the waves will become synchronized again.
That is, beating two close frequencies together breaks up the continuous wave into a series of packets, the beats. To describe a single electron moving through space, we need a single packet. This can be achieved by superposing waves having a continuous distribution of wavelengths, or wave numbers within of order $$\Delta k$$, say, of k. In this case, the waves will be out of phase after a distance of order $$\dfrac {\pi}{\Delta k}$$, but since they have many different wavelengths, they will never get back in phase again.
### Phase Velocity and Group Velocity
The best way to understand how these waves add up is to view my applet. It will immediately become apparent that there are two different velocities in the dynamics: first, the velocity with which the individual peaks move to the right, and second the velocity at which the slowly varying envelope function—the beat pattern—moves. The $$\lambda f = c'$$ individual peak velocity is determined by the term $$\sin (kx - \omega t)$$, it is $$\dfrac {\omega}{k}$$ this is called the phase velocity. The speed with which the beat pattern moves, on the other hand, is determined by the term $$\cos ((\Delta k) x - (\Delta \omega)t)$$, this speed is $$\dfrac {\Delta \omega}{\Delta k} = \dfrac {d \omega}{d k}$$, for close frequencies.
Going back one more time to the electron wavepacket, armed with this new insight, we can see immediately that the wave speed we calculated from $$\lambda f = c'$$ was the phase velocity of the waves. The packet itself will of course move at the group velocity—and it is easy to check that this is just v.
We’ve seen how two sine waves of equal amplitude close together in frequency produce beats: if the waves are in phase at the origin, as we go along the x-axis they gradually fall out of phase, and cancel each other at a distance $$x = \dfrac {\pi}{2 \Delta}$$, where $$2 \Delta$$ is the difference in k of the two sinkx waves. (For the moment, we are ignoring the time development of these waves: we’re just looking at t = 0.). If we continue along the x-axis to $$\dfrac {\pi}{\Delta}$$, the two waves will be back in phase again, this is the next beat. Now, if instead of adding two waves, we add many waves, all of different k, but with the k’s taken from some small interval of size of order $$\Delta k$$, and all these waves are in phase at the origin, then, again, they will all stay more or less in phase for a distance of order $$x = \dfrac {\pi}{2 \Delta}$$.
However, as we proceed past that point, the chances of them all getting back in phase again get rapidly smaller as we increase the number of different waves.
This suggests a way to construct a wavepacket: add together a lot of waves from within a narrow frequency range, and they will only be in phase in a region containing the origin.
Adding waves in this way leads to a more general derivation of the formula $$\dfrac {d \omega }{d k}$$ for the group velocity. The standard approach is to replace the sum over plane waves by an integral, with the wavenumber k as the variable of integration, and the convention is to put a factor $$2 \pi$$ in the denominator:
$\psi (x,t) = \int \limits _{-\infty}^{+\infty} \dfrac {dk}{2 \pi} e^{ikx - i\omega (k) t} \phi (k)$
Since we are constructing a wavepacket with a fairly well-defined momentum, we will take the function $$\phi (k)$$ to be strongly peaked at k0, and going rapidly to zero away from that value, so the only significant contribution to the integral is from the neighborhood of k0. Therefore, if $$\omega (k)$$is reasonably smooth (which it is) it is safe to put
$\omega (k) = \omega (k_0) + (k - k_0) \omega ' (k_0)$
in the exponential.
This gives
$\psi (x,t) = \int \limits _{-\infty}^{+\infty} \dfrac {dk}{2 \pi} e^{ikx - i\omega (k_0) t - i(k-k_0) \omega ' (k_0) t} \phi (k)$
$= e^{i(k_0 x - \omega (k_0) t )} \int \limits _{-\infty}^{+\infty} \dfrac {dk}{2 \pi} e^{i (k -k_0)(x-\omega ' (k_0)t)} \phi (k)$
The first term just represents a single wave at k0, and the peaks move at the phase velocity
$v_{phase} = \dfrac {\omega}{k}$
The second term, the integral, is the envelope function: here x only appears in the combination
$x - \omega ' (k_0) t$
so the envelope, and hence the wavepacket, moves to the right at the group velocity: $$v_{phase} = \dfrac {\omega '}{k_0}$$.
Note that if the next term in the Taylor expansion of $$\omega (k)$$ is also included, that amounts to adding wavepackets with slightly different group velocities together, and the initial (total) wavepacket will gradually widen.
### The Gaussian Wavepacket
Fortunately, there is a simple explicit mathematical realization of the addition of plane waves to form a localized function: the Gaussian wavepacket,
$\psi (x,t = 0) = A e^{ik_0x} e^{-x^2}{2\Delta ^2}$
where $$p_0 = \hbar k_0$$. For this wavepacket to represent one electron, with the probability of finding the electron in a small section of length dx at x equal to $$|\psi |^2 dx$$, and the total probability of finding the electron somewhere equal to one, the constant A is uniquely determined (apart from a possible phase multiplier $$e^{i\delta}$$, which would not affect the probability).
Using the standard result
$\int \limits _{-\infty}^{+\infty} e^{-ax^2} dx = \sqrt {\dfrac {\pi}{a}}$
we find $$|A|^2 = (\pi \Delta ^2)^{-\dfrac {1}{2}}$$ so
$\psi (x,t =0) = \dfrac {1}{(\pi \Delta ^2)^{\dfrac {1}{4}}} e^{ik_0x} e^{-\dfrac {x^2}{2\Delta ^2}}$
But how do we construct this particular wavepacket by superposing plane waves? That is to say, we need a representation of the form:
$\psi (x) = \int \limits _{-\infty}^{+\infty} \dfrac {dk}{2 \pi} e^{ikx} \phi (k)$
The function $$\phi (k)$$ represents the weighting of plane waves in the neighborhood of wavenumber k. This is a particular example of a Fourier transform—we will be discussing the general case in detail a little later in the course. Note that if $$\phi (k)$$ is a bounded function, any particular k value gives a vanishingly small contribution, the plane-wave contribution to y(x) from a range dk is $$\dfrac {\phi (k) dk}{2 \pi}$$. In fact, $$\phi (k)$$ is given in terms of $$\psi (x)$$ by
$\phi (k) = \int \limits _{-\infty}^{+\infty} dx e^{-ikx} \psi (x)$
It is perhaps worth mentioning at this point that this can be understood qualitatively by observing that the plane wave prefactor $$e^{ikx}$$ will interfere destructively with all plane wave components of $$\psi (x)$$except that of wavenumber k, where it may at first appear that the contribution is infinite, but recall that as stated above, any particular k component has a vanishingly small weight—and, in fact, this is the right answer, as we shall show in more convincing fashion later.
In the present case, the above handwaving argument is unnecessary, because both the integrals can be carried out exactly, using the standard result:
$\int \limits _{-\infty}^{\infty} e^{-ax^2 + bx} dx = e^{\dfrac {b^2}{4a}} \sqrt {\dfrac {\pi}{a}}$
giving
$\phi (k) = ( 4 \pi \Delta ^2)^{\dfrac {1}{4}} e^{\dfrac {-\Delta ^2 (k - k_0)}{2}}$
Putting this back in the integral for $$\psi (x)$$ shows that the integral equations are consistent. Note the normalization integrals in x-space and k-space are:
$\int \limits _{-\infty}^{\infty} |\psi (x)|^2 dx = 1, \int \limits _{-\infty}^{\infty} | \phi (k) |^2 \dfrac {dx}{2 \pi} = 1$
The physical significance of the second equation above is that if the wavepacket goes through a diffraction grating so that the different k-components are dispersed in different directions, like the colors in white light, and a detector is arranged to register the electron if it has wavenumber between k and k + dk, the probability of finding it in that wavenumber range is $$|\phi (k)|^2 \dfrac {dk}{2 \pi}$$.
### Expectation Values and the Uncertainty Principle
It is clear from the expressions for $$\psi (x)$$ and its Fourier transform $$\phi (k)$$ above that the spreading of the wave function in x-space is inversely related to its spreading in k-space: the x-space wavefunction has spread $$\sim \Delta$$, the k-space wavefunction $$\sim \dfrac {1}{\Delta}$$, This is perhaps the simplest example of Heisenberg’s famous Uncertainty Principle: in quantum mechanics, both the position and momentum of a particle cannot be known precisely at the same moment; the more exactly one is specified the less well the other is known. This is an inevitable consequence of the wave nature of the probability distribution. As we have already seen, a particle with an exact momentum has a wave of specific wavelength, and the only such wave is a plane wave extending from minus infinity to infinity, so the position of the particle is completely undetermined. A particle with precisely defined position is described by a wavepacket having all wavelengths included with equal weight—the momentum is completely undefined. We shall give more examples of the Uncertainly Principle, of efforts to evade it and of its uses in estimates, in the next lecture. | 8,092 | 31,521 | {"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} | 3.203125 | 3 | CC-MAIN-2017-43 | latest | en | 0.908567 |
https://howmanyis.com/length/121-yd-in-cm/98090-2-yards-in-centimeters | 1,610,898,892,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00227.warc.gz | 364,173,183 | 6,080 | How many is
Conversion between units of measurement
You can easily convert 2 yards into centimeters using each unit definition:
Yards
yard = 3 ft = 0.9144 m
Centimeters
centi m = 0.01 m
With this information, you can calculate the quantity of centimeters 2 yards is equal to.
## ¿How many cm are there in 2 yd?
In 2 yd there are 182.88 cm.
Which is the same to say that 2 yards is 182.88 centimeters.
Two yards equals to one hundred eighty-two centimeters. *Approximation
### ¿What is the inverse calculation between 1 centimeter and 2 yards?
Performing the inverse calculation of the relationship between units, we obtain that 1 centimeter is 0.0054680665 times 2 yards.
A centimeter is zero times two yards. *Approximation | 193 | 734 | {"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} | 3.828125 | 4 | CC-MAIN-2021-04 | latest | en | 0.924536 |
https://www.experts-exchange.com/questions/24166197/Problem-adding-sumed-fields-from-separate-nested-select-statements.html | 1,513,348,311,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948572676.65/warc/CC-MAIN-20171215133912-20171215155912-00737.warc.gz | 746,167,125 | 34,907 | Solved
# Problem adding sumed fields from separate nested select statements
Posted on 2009-02-22
Medium Priority
279 Views
The below code works fine. But when I try to perform math funtions + * / - on the last 6 columns it does not work. Maybe because they are sumed fields? The column I want to add is:
.....,T2.RREM*(T3.RDT/(T3.RDT + T4.SDE + T5.ADS + T6.URE + T7.GDE + T8.ALLIED))
or
.....,T2.RREM*(T3.RDT/(T3.RDT + T4.SDE + T5.ADS + T6.URE + T7.GDE + T8.ALLIED+.01)) to avoid dividing by zero (maybe there is another way around that separate issue?)
``````SELECT T2.RCUST, T2.CNME, T2.CTYPE, T2.CCON, T2.CPHON, T2.CXFAX, T2.CXEML, T2.SNAME, T2.CRDOL, T2.CXSTPF, T2.CSTE, T2.CLPDT, T2.CLPAM, T2.RINVC, T2.RAMT, T2.RIDTE, T2.RREM, T2.RDDTE, T2.CTERM, T2.ARCPO, T2.RREF, T2.AXDSPF, T3.RDT, T4.SDE, T5.ADS, T6.URE, T7.GDE, T8.ALLIED
FROM (((((((SELECT * FROM RCX, RCM, SSM, RARL01 LEFT JOIN (SELECT * FROM RAXL01 WHERE AXDSPF<>'') AS T1 ON RINVC = T1.AXIVNM WHERE RCUST=CXCUSN AND RCUST=CCUST AND RREM<>0 AND CTYPE<>'INTC' AND CTYPE<>'INTB' AND SSAL = CSAL) AS T2 LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS RDT FROM SSD, ICX WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC='RDT' GROUP BY SDINV) AS T3 ON T2.RINVC = T3.SDINV) LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS SDE FROM SSD, ICX WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC='SDE' GROUP BY SDINV) AS T4 ON T2.RINVC = T4.SDINV) LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS ADS FROM SSD, ICX WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC='ADS' GROUP BY SDINV) AS T5 ON T2.RINVC = T5.SDINV) LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS URE FROM SSD, ICX WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC IN ('LHD','TME') GROUP BY SDINV) AS T6 ON T2.RINVC = T6.SDINV) LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS GDE FROM SSD, ICX WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC ='GDE' GROUP BY SDINV) AS T7 ON T2.RINVC = T7.SDINV) LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS ALLIED FROM SSD, ICX WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC IN ('CTO','HCN','RCA','LCC','AIP','AII','AIF','CTS','ASR') GROUP BY SDINV) AS T8 ON T2.RINVC = T8.SDINV)
``````
0
Question by:atprato
• 9
• 6
• 4
• +1
LVL 75
Expert Comment
ID: 23705204
>> it does not work<<
This is meaningless. Give us an error message.
0
LVL 75
Expert Comment
ID: 23705210
Or if no error, what are the results and what are the desired results.
0
LVL 1
Expert Comment
ID: 23705215
What error are you getting?
0
Author Comment
ID: 23705552
There is no error, the result column I added is blank. The result should be some fraction of RREM. I'm actually suprised the first version of the result column does not error since it would be dividing by zero for some records.
0
LVL 41
Expert Comment
ID: 23706098
Is this a round of problem? Are you getting blank value or zero for your computed column. try like this.
``````SELECT T2.RCUST, T2.CNME, T2.CTYPE, T2.CCON, T2.CPHON, T2.CXFAX, T2.CXEML, T2.SNAME, T2.CRDOL, T2.CXSTPF, T2.CSTE, T2.CLPDT, T2.CLPAM,
T2.RINVC, T2.RAMT, T2.RIDTE, T2.RREM, T2.RDDTE, T2.CTERM, T2.ARCPO, T2.RREF, T2.AXDSPF, T3.RDT, T4.SDE, T5.ADS, T6.URE,
T7.GDE, T8.ALLIED,
VAL(FORMAT(T2.RREM*(T3.RDT/(T3.RDT + T4.SDE + T5.ADS + T6.URE + T7.GDE + T8.ALLIED+.01)))) AS Computed_Column
FROM (((((((SELECT * FROM RCX, RCM, SSM, RARL01
LEFT JOIN (SELECT * FROM RAXL01 WHERE AXDSPF<>'') AS T1
ON RINVC = T1.AXIVNM
WHERE RCUST=CXCUSN AND RCUST=CCUST AND RREM<>0 AND CTYPE<>'INTC' AND CTYPE<>'INTB' AND SSAL = CSAL) AS T2
LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS RDT FROM SSD, ICX
WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC='RDT'
GROUP BY SDINV) AS T3
ON T2.RINVC = T3.SDINV)
LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS SDE FROM SSD, ICX
WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC='SDE'
GROUP BY SDINV) AS T4
ON T2.RINVC = T4.SDINV)
LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS ADS FROM SSD, ICX
WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC='ADS'
GROUP BY SDINV) AS T5
ON T2.RINVC = T5.SDINV)
LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS URE FROM SSD, ICX
WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC IN ('LHD','TME')
GROUP BY SDINV) AS T6
ON T2.RINVC = T6.SDINV)
LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS GDE FROM SSD, ICX
WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS AND CXPPLC ='GDE'
GROUP BY SDINV) AS T7 ON T2.RINVC = T7.SDINV)
LEFT JOIN (SELECT SDINV, SUM(SDSVAL) AS ALLIED FROM SSD, ICX
WHERE SDRCID='SI' AND SUBSTRING(SDHPRF,1,2)<>'98' AND SDICLS = CXCLAS
AND CXPPLC IN ('CTO','HCN','RCA','LCC','AIP','AII','AIF','CTS','ASR')
GROUP BY SDINV) AS T8 ON T2.RINVC = T8.SDINV)
``````
0
LVL 75
Expert Comment
ID: 23706262
Perhaps you have Null values. But unfortunately without seeing the actual result all we can do is hazard guesses.
0
Author Comment
ID: 23706596
acperkins, T2 has records that are not in T3-T8, so I guess T3-T8 would be null for those cases. Does that give you any ideas? Also, the fields I am calling from T3-T8 are sumed fields, does that affect anything?
Sharath, sorry, when I add ,VAL(FORMAT(T2.RREM*(T3.RDT/(T3.RDT + T4.SDE + T5.ADS + T6.URE + T7.GDE + T8.ALLIED+.01)))) AS Computed_Column I get an OBDC error. I am using VBA to push the sql to AS400. You didn't change anything else did you?
acperkins,
0
LVL 75
Expert Comment
ID: 23706636
>>T2 has records that are not in T3-T8, so I guess T3-T8 would be null for those cases. <<
That is correct. They would be Null and the result of dividing by a NULL value is Null. You can use ISNULL(YourColumn, 0) to ensure that it is 0
0
Author Comment
ID: 23706771
OK, to keep it very simple, I replaced .....,T3.RDT,..... in the original query that works with .....,ISNULL(T3.RDT,0) AS TEST_NULL,..... I got another ODBC error. It seems like whenever I put these sumed fields into a function they fail? When I have just T3.RDT in the working query it results NULL or a value depending on if the record was in T2. There must be some reason I can't put these fields in a function......
0
LVL 41
Expert Comment
ID: 23707163
Can you just add this column to your query and check what values you are getting?
SELECT ......,
(T3.RDT + T4.SDE + T5.ADS + T6.URE + T7.GDE + T8.ALLIED+.01) AS Computed_1
0
Author Comment
ID: 23707278
I did it, the entire excel column comes up blank(not zero). But there is no record where all 6 have a value. So I tried
(T4.SDE + T6.URE) AS Computed_1
This worked for records where SDE and URE had a value. But was blank where neither had a value or one has a value. I think this confirms our Null theory, but I don't know how to get around it because these fields error out when I put them in a function....
0
LVL 41
Expert Comment
ID: 23707297
try this
``````ISNULL(T2.RREM,0)*(ISNULL(T3.RDT,0)/(ISNULL(T3.RDT,0) + ISNULL(T4.SDE,0) + ISNULL(T5.ADS,0) + ISNULL(T6.URE,0) + ISNULL(T7.GDE,0) + ISNULL(T8.ALLIED,0)+.01))
``````
0
Author Comment
ID: 23707393
I get an ODBC error, but as I said above, even if I just do:
ISNULL(T3.RDT,0) AS TEST_NULL
I get an ODBC error. Yet with just T3.RDT, I get a column with some values and some Null. This is really strange....
0
LVL 41
Expert Comment
ID: 23707407
try this
``````IIF(T2.RREM IS NOT NULL,T2.RREM,0)*(IIF(T3.RDT IS NOT NULL,T3.RDT,0)/(IIF(T3.RDT IS NOT NULL,T3.RDT,0) + IIF(T4.SDE IS NOT NULL,T4.SDE,0) + IIF(T5.ADS IS NOT NULL,T5.ADS,0) + IIF(T6.URE IS NOT NULL,T6.URE,0) + IIF(T7.GDE IS NOT NULL,T7.GDE,0) + IIF(T8.ALLIED IS NOT NULL,T8.ALLIED,0)+.01))
``````
0
Author Comment
ID: 23707468
I get a sql syntax error. That is a step in the right direction, any idea what the error is?
0
Author Comment
ID: 23707730
oK IIF did not work but I got CASE to work in the same way to return 0s for null values. So now my issues is, how do I add case statements together in the select statement.
0
LVL 41
Accepted Solution
Sharath earned 2000 total points
ID: 23707769
check this
``````SELECT ...,
(CASE WHEN T2.RREM IS NOT NULL THEN T2.RREM ELSE 0 END)*((CASE WHEN T3.RDT IS NOT NULL THEN T3.RDT ELSE 0 END)/((CASE WHEN T3.RDT IS NOT NULL THEN T3.RDT ELSE 0 END) + (CASE WHEN T4.SDE IS NOT NULL THEN T4.SDE ELSE 0 END) + (CASE WHEN T5.ADS IS NOT NULL THEN T5.ADS ELSE 0 END) + (CASE WHEN T6.URE IS NOT NULL THEN T6.URE ELSE 0 END) + (CASE WHEN T7.GDE IS NOT NULL THEN T7.GDE ELSE 0 END) + (CASE WHEN T8.ALLIED IS NOT NULL THEN T8.ALLIED ELSE 0 END)+.01)) AS Computed_Column
``````
0
Author Comment
ID: 23707810
I had to remove a set of () but that did the trick, thanks!!!!!!!!!
0
Author Closing Comment
ID: 31549780
Just had to remove one set of () to get the desired math, but the sql solution was perfect!
0
LVL 41
Expert Comment
ID: 23707910
you are welcome!
0
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That darn Conway is persistent, isn't he?! That's right, Day 18 is looking like the return of Conway, but in 2 dimensions, and with 3 possible states (Open Ground, Wooded Acre, and Lumberyard) as opposed to the less complicated Alive or Dead we saw a few days ago.
It'll be interesting to see if the repeating semi-stable shapes that we see in a normal 2D Conway's Game of Life are present here or how having 3 possible states affects them.
Good luck! I wood love to see your solutions! 🎄
JavaScript solution
I'm gonna omit reader.js which is the same as the other solutions and jump to the point:
18-common.js
const MAP = {
OPEN_GROUND: '.',
TREES: '|',
LUMBERYARD: '#'
};
const tick = originalMap => {
const n = originalMap.length;
const nextMap = Array.from({length: n}, row => Array.from({length: n}));
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
const originalAcre = originalMap[i][j];
// For an open ground acre
if (originalAcre === MAP.OPEN_GROUND) {
nextMap[i][j] = adjacentTrees >= 3 ? MAP.TREES : MAP.OPEN_GROUND;
}
// For a trees acre
else if (originalAcre === MAP.TREES) {
nextMap[i][j] = adjacentLumberyards >= 3 ? MAP.LUMBERYARD : MAP.TREES;
}
// For a lumberyard acre
else if (originalAcre === MAP.LUMBERYARD) {
nextMap[i][j] = adjacentLumberyards >= 1 && adjacentTrees >= 1 ? MAP.LUMBERYARD : MAP.OPEN_GROUND;
}
}
}
return nextMap;
};
const getAdjacents = (originalMap, i, j, n) => {
const positions = [[i-1, j-1], [i-1, j], [i-1, j+1], [i, j-1], [i, j+1], [i+1, j-1], [i+1, j], [i+1, j+1]];
return positions
.filter(([i, j]) => i >= 0 && j >= 0 && i < n && j < n)
.map(([i, j]) => originalMap[i][j])
.filter(acre => acre !== undefined);
};
const serialize = map => map.map(row => row.join('')).join('');
const printSolution = solution => {
const serializedMap = (Array.isArray(solution) ? serialize(solution) : solution).split('');
const trees = count(serializedMap, MAP.TREES);
const lumberyards = count(serializedMap, MAP.LUMBERYARD);
console.log(The total resource value of the lumber collection area is ${trees * lumberyards}); }; const count = (map, type) => { return map.reduce((total, acre) => total + (acre === type ? 1 : 0), 0); }; module.exports = { MAP, tick, serialize, printSolution }; 18a.js const { readFile } = require('./reader'); const { tick, printSolution } = require('./18-common'); (async () => { let outskirts = (await readFile('18-input.txt')).map(row => row.split('')); for (let i = 0; i < 10; i++) { outskirts = tick(outskirts); } console.log(outskirts.map(row => row.join('')).join('\n')); printSolution(outskirts); })(); 18b.js const { readFile } = require('./reader'); const { MAP, tick, serialize, printSolution } = require('./18-common'); (async () => { let outskirts = (await readFile('18-input.txt')).map(row => row.split('')); const previousStates = new Map(); let elapsedMinutes = 0; let serialized; let hasDetectedLoop = false; do { elapsedMinutes++; outskirts = tick(outskirts); serialized = serialize(outskirts); if (previousStates.has(serialized)) { hasDetectedLoop = true; } else { previousStates.set(serialized, elapsedMinutes); } } while (!hasDetectedLoop); console.log(Loop detected at minute${elapsedMinutes}!);
const firstRepetitionMinutes = previousStates.get(serialized);
const loopDurationMinutes = elapsedMinutes - firstRepetitionMinutes;
const equivalentMinute = ((1000000000 - firstRepetitionMinutes) % loopDurationMinutes) + firstRepetitionMinutes;
console.log(The minute 1000000000 is equivalent to the minute ${equivalentMinute}); const solution = [...previousStates.entries()].find(([state, minute]) => minute === equivalentMinute)[0]; printSolution(solution); })(); For Part 2 I logged out the results and manually counted period length that I used to calculate my solution. I guess I could calculate the period programmatically, but I'm just too lazy :) <?php$input = require_once 'readFile.php';
function calcNext($area) {$next = [];
for ($y=0;$y < count($area);$y++) {
for ($x=0;$x < count($area[0]);$x++) {
$adjacent = [];$tl = $tm =$tr = $cl =$cr = $bl =$bc = $br = false;$curr = $area[$y][$x]; !empty($area[$y-1][$x-1]) && $adjacent[] =$area[$y-1][$x-1];
!empty($area[$y-1][$x]) &&$adjacent[] = $area[$y-1][$x]; !empty($area[$y-1][$x+1]) && $adjacent[] =$area[$y-1][$x+1];
!empty($area[$y][$x-1]) &&$adjacent[] = $area[$y][$x-1]; !empty($area[$y][$x+1]) && $adjacent[] =$area[$y][$x+1];
!empty($area[$y+1][$x-1]) &&$adjacent[] = $area[$y+1][$x-1]; !empty($area[$y+1][$x]) && $adjacent[] =$area[$y+1][$x];
!empty($area[$y+1][$x+1]) &&$adjacent[] = $area[$y+1][$x+1];$count = array_reduce($adjacent, function($acc, $var) {$acc[$var]++; return$acc;
}, ['.' => 0, '|' => 0, '#' => 0]);
if ($curr == '.') {$next[$y][$x] = $count['|'] >= 3 ? '|' : '.'; } else if ($curr == '|') {
$next[$y][$x] =$count['#'] >= 3 ? '#' : '|';
} else if ($curr == '#') {$next[$y][$x] = $count['#'] >= 1 &&$count['|'] >= 1 ? '#' : '.';
}
}
}
return $next; } function countWood($area) {
$wood =$lumber = 0;
for ($y=0;$y < count($area);$y++) {
for ($x=0;$x < count($area[0]);$x++) {
$area[$y][$x] == '|' &&$wood++;
$area[$y][$x] == '#' &&$lumber++;
}
}
return $wood *$lumber;
}
function generateTen($area,$time = 10) {
$next =$area;
for ($i=0;$i < $time;$i++) {
$next = calcNext($next);
}
return countWood($next); } function generateMany($input) {
$next =$input;
$count =$countRep = $repeats = 0; for ($i=0; ; $i++) {$next = calcNext($next);$count = countWood($next); if ((1000000000 - 1 -$i) % 28 == 0 && $count) {$countRep == $count &&$repeats++;
$countRep =$count;
if ($repeats > 1) { return$countRep;
}
}
}
}
echo "Part 1: " . generateTen($input) . "\n"; echo "Part 2: " . generateMany($input) . "\n";
?>
<?php
$file = fopen("input.txt", "r") or exit("Unable to open file"); while(!feof($file)) {
$array[] = fgets($file);
}
fclose($file); return array_map(function($str) {
$str = preg_replace("/\r|\n/", "",$str);
return str_split($str); }, array_filter($array));
?>
Very similar to Conway's Life, but the result is kind of more life-like.
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my @grid;
while (<>) {
chomp;
push @grid, [ split // ];
}
for (1.. 10) {
my @next;
for my $y (0 ..$#grid) {
for my $x (0 ..$#{ $grid[$y] }) {
my @coords = grep $_->[0] >= 0 &&$_->[0] <= $#{$grid[$y] } &&$_->[1] >= 0 && $_->[1] <=$#grid,
[$x - 1,$y - 1], [$x,$y - 1], [$x + 1,$y - 1],
[$x - 1,$y ], [$x + 1,$y ],
[$x - 1,$y + 1], [$x,$y + 1], [$x + 1,$y + 1];
++$adjacent{$grid[ $_->[0] ][$_->[1] ] } for @coords;
$next[$x][$y] = { '.' => sub { ($adjacent{'|'} // 0) >= 3 ? '|' : '.' },
'|' => sub { ($adjacent{'#'} // 0) >= 3 ? '#' : '|' }, '#' => sub { ($adjacent{'#'} && $adjacent{'|'}) ? '#' : '.' }, }->{$grid[$x][$y] }->();
}
}
@grid = @next;
}
my %count;
++$count{$_} for map @$_, @grid; say +($count{'|'} // 0) * (\$count{'#'} // 0);
A python solution very similar to that of day 12.
import functools
class Simulation:
def __init__(self, fname):
self.height = len(self.map)
self.width = len(self.map[0]) if self.height > 0 else 0
self.old_map = [[0 for _ in range(self.width)] for _ in range(self.height)]
def step(self):
self.old_map, self.map = self.map, self.old_map # double buffering
for y, row in enumerate(self.old_map):
for x, current in enumerate(row):
neighborhood = self.neighborhood(x, y)
trees = sum(1 for (xx, yy) in neighborhood if self.old_map[yy][xx] == '|')
lumberyard = sum(1 for (xx, yy) in neighborhood if self.old_map[yy][xx] == '#')
self.map[y][x] = current
if current == '.' and trees >= 3:
self.map[y][x] = '|'
elif current == '|' and lumberyard >= 3:
self.map[y][x] = '#'
elif current == '#' and (lumberyard == 0 or trees == 0):
self.map[y][x] = '.'
def run(self, steps=1):
last_seen = {}
step = 0
while step < steps:
hashable = self.show()
if hashable in last_seen:
remaining = (steps - step) % (step - last_seen[hashable])
step = steps - remaining
last_seen = {}
else:
last_seen[hashable] = step
self.step()
step += 1
@functools.lru_cache(maxsize=None)
def neighborhood(self, x, y):
neighbors = []
for dx in [-1, 0, +1]:
xx = x + dx
if 0 > xx or xx >= self.width:
continue
for dy in [-1, 0, +1]:
yy = y + dy
if 0 > yy or yy >= self.height or (xx == x and yy == y):
continue
neighbors.append((xx, yy))
return neighbors
def show(self):
return '\n'.join(''.join(c for c in row) for row in self.map)
with open(fname, 'r') as file:
return [[c for c in row.strip()] for row in file]
def part(fname, steps):
simulation = Simulation(fname)
simulation.run(steps)
trees = sum(1 for row in simulation.map for c in row if c == '|')
lumber = sum(1 for row in simulation.map for c in row if c == '#')
return trees * lumber
expected = list(map(list, ['.#.#...|#.',
'.....#|##|',
'.|..|...#.',
'..|#.....#',
'#.#|||#|#|',
'...#.||...',
'.|....|...',
'||...#|.#|',
'|.||||..|.',
'...#.|..|.']))
def test_one_step():
expected = """\
.......##.
......|###
.|..|...#.
..|#||...#
..##||.|#|
...#||||..
||...|||..
|||||.||.|
||||||||||
....||..|."""
simulation = Simulation('test_input.txt')
simulation.step()
assert expected == simulation.show()
def test_ten_steps():
expected = """\
.||##.....
||###.....
||##......
|##.....##
|##.....##
|##....##|
||##.####|
||#####|||
||||#|||||
||||||||||"""
simulation = Simulation('test_input.txt')
simulation.run(10)
assert expected == simulation.show()
def test_part1():
assert 1147 == part('test_input.txt', 10)
if __name__ == '__main__':
print('Part1', part('input.txt', 10))
print('Part2', part('input.txt', 1000000000))
Yawn. It was a relief to get something solvable after the problems from Saturday and Monday which were incredibly hard, but another Conway's Game of Life, with another huge iteration count? Just cranking the handle here, nothing new to solve.
I modelled it with a 2D array. Getting more familiar with Kotlin arrays and sequences now.
enum class Acre(val symbol: Char) {
OPEN('.'),
TREES('|'),
LUMBERYARD('#')
}
data class Pos(val x: Int, val y: Int)
val Pos.neighbours: Sequence<Pos> get() = sequence {
(-1..1).flatMap { dy ->
(-1..1).map { dx ->
if (dy != 0 || dx != 0) yield(Pos(x+dx,y+dy))
}
}
}
class Landscape(val width: Int, val height: Int) {
val acres: Array<Array<Acre>> = Array<Array<Acre>>(height) { Array<Acre>(width) { Acre.OPEN }}
val positions: Sequence<Pos> get() = sequence {
(0..height-1).flatMap { y -> (0..width-1).map { x -> yield(Pos(x, y)) }}
}
fun validPos(p: Pos): Boolean = 0 <= p.y && p.y < height && 0 <= p.x && p.x < width
fun neighbourCount(p: Pos, a: Acre): Int =
p.neighbours.count { n -> this[n] == a }
fun totalCount(a: Acre): Int = positions.count { pos -> this[pos] == a }
fun resourceValue(): Int = totalCount(Acre.TREES) * totalCount(Acre.LUMBERYARD)
}
operator fun Landscape.get(p: Pos): Acre = if (validPos(p)) acres[p.y][p.x] else Acre.OPEN
operator fun Landscape.set(p: Pos, a: Acre) { if (validPos(p)) acres[p.y][p.x] = a }
I made an infinite sequence of states again:
fun Landscape.tick(): Landscape {
val next = Landscape(width, height)
positions.forEach { pos ->
val neighbours = pos.neighbours.map(this::get).toList()
next[pos] = when (this[pos]) {
Acre.OPEN ->
if (neighbourCount(pos, Acre.TREES) >= 3)
Acre.TREES
else
Acre.OPEN
Acre.TREES ->
if (neighbourCount(pos, Acre.LUMBERYARD) >= 3)
Acre.LUMBERYARD
else
Acre.TREES
Acre.LUMBERYARD ->
if (neighbourCount(pos, Acre.LUMBERYARD) >= 1 && neighbourCount(pos, Acre.TREES) >= 1)
Acre.LUMBERYARD
else
Acre.OPEN
}
}
return next
}
fun Landscape.timeline(): Sequence<Landscape> = sequence {
var next = this@timeline
while (true) {
yield(next)
next = next.tick()
}
}
So part 1 was trivial:
fun part1(input: Landscape): Int =
input.timeline().drop(10).first().resourceValue()
For part 2 I made a generic loop finder:
inline fun <reified T> findLoop(ts: Sequence<T>): Pair<Int, Int> {
val seen = mutableMapOf<T, Int>()
val firstRepeat = ts.zip(indices).dropWhile { (t, i) ->
if (seen.containsKey(t))
false
else {
seen[t] = i
true
}
}.first()
return Pair(seen[firstRepeat.first]!!, firstRepeat.second)
}
Then used this to do the old find-loop-run-last-few-iterations dance:
fun part2(input: Landscape): Int {
val (loopStart, loopEnd) = findLoop(input.timeline())
val iterations = 1000000000
val loops = (iterations - loopStart) / (loopEnd - loopStart)
val lastLoop = (iterations - loopStart) % loops
val lastLoopTimeline = input.timeline().drop(loopStart)
val lastIteration = lastLoopTimeline.drop(lastLoop).first()
return lastIteration.resourceValue()
}
Also, Conway’s Game of Life? More like Conway’s Game of Logs! 😬 | 3,929 | 12,607 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-40 | longest | en | 0.669241 |
https://www.electrotechnik.net/2017/08/specific-heat-capacity_2.html | 1,686,048,138,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652494.25/warc/CC-MAIN-20230606082037-20230606112037-00131.warc.gz | 798,493,008 | 14,616 | # Specific Heat Capacity
Specific Heat Capacity
Specific Heat Capacity of a substance is defined as the amount of heat required to increase the temperature of unit mass (1 kg) the substance through one degree Celsius.
Its unit is kJ/kgK
Specific Heat Capacity is denoted by C
Specific Heat Capacity at constant volume
Specific heat capacity at constant volume is the amount of heat required to raise the temperature of a unit mass of a substance by one degree celsius at constant volume
Q = m.Cv.dT
where
Q is the amount of heat transferred in Joules
m is the mass in kg
Cv is the specific heat capacity at constant volume in kJ/kgK
dT is the difference in temperature in K
Specific Heat Capacity at constant pressure
Specific heat capacity at constant pressure is the amount of heat required to raise the temperature of a unit mass of a substance by one degree celsius at constant pressure
Q = m.Cp.dT
where
Q is the amount of heat transferred in Joules
m is the mass in kg
Cp is the specific heat capacity at constant pressure in kJ/kgK
dT is the difference in temperature in K | 243 | 1,091 | {"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-2023-23 | longest | en | 0.918662 |
tinder-unlimited.website | 1,571,381,478,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677964.40/warc/CC-MAIN-20191018055014-20191018082514-00179.warc.gz | 194,811,152 | 6,073 | # MODELO LOTKA VOLTERRA PDF
Before moving on to general n−species Lotka-Volterra systems, we will in detail some Lotka-Volterra systems that model the dynamics of two Missing: modelo | Must include: modelo. Seis técnicas de projeto de controladores são aplicados ao modelo Lotka-Volterra, o qual é utilizado como um padrão ou "benchmark'' para avaliar e comparar. #Lotka-Volterra single species discrete population growth. time modelos interesantes de interacción inter-específica.
Author: Vernice Corwin III Country: Armenia Language: English Genre: Education Published: 25 September 2016 Pages: 294 PDF File Size: 36.71 Mb ePub File Size: 49.53 Mb ISBN: 331-4-97436-136-5 Downloads: 10685 Price: Free Uploader: Vernice Corwin III
Think of the two species as rabbits and foxes or moose and wolves or little fish in big fish.
## Matplotlib: lotka volterra tutorial — SciPy Cookbook documentation
Y1 represents the prey, who would live peacefully by themselves if there were no modelo lotka volterra. I've chosen the units of time and population so that the coefficients in front of the leading linear terms are one.
So y1 prime equals y1 represents exponential growth of the prey in the absence of any predators. The predators need the prey, live on the prey. So in the absence of any prey, this minus sign is all important. So y2 prime equals minus y2 represents exponential decay.
And the predators die off exponentially in the absence of any prey. But then here are the non-linear terms. These are like logistics terms, except with the interaction between the two species. The growth of Y1 is limited by the presence of y2. So y1 will grow until this term becomes modelo lotka volterra one y2 read reaches mu2.
On the other hand, the decay of y2 becomes 0 when y1 reaches mu1.
To complete this specification, we need the initial conditions. So we have two values eta 1 and eta 2, which are the initial values of y1 and y2.
So these four parameters, two mus and two etas, are the four parameters we have in our predator prey model. Don't worry about the fact that modelo lotka volterra are continuous variables and that we can have non-integer numbers of individuals.
## Plotting - The Lotka-Volterra predator-prey model - Mathematica Stack Exchange
We can have half of a rabbit or a tenth of a moose. These are, after all, models that are modelo lotka volterra versions of what's happening in nature.
The critical points are when the derivatives become 0. There's a critical point at the origin. But the interesting one is when these terms become 0. So that's the point where y1 and y2 are equal to the mu1 and mu2.
We have to look at the Jacobian. As differential equations are used, the solution is deterministic and continuous.
### Prey predator model
This, in modelo lotka volterra, implies that the generations of both the predator and prey are continually overlapping. Equilibrium occurs when the growth rate is equal to 0. This gives two fixed points: We have to define the Jacobian matrix: | 706 | 3,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-43 | latest | en | 0.753933 |
https://minuteshours.com/14-44-hours-in-hours-and-minutes | 1,642,509,514,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00176.warc.gz | 448,948,082 | 5,061 | # 14.44 hours in hours and minutes
## Result
14.44 hours equals 14 hours and 26.4 minutes
You can also convert 14.44 hours to minutes.
## Converter
Fourteen point four four hours is equal to fourteen hours and twenty-six point four minutes. | 63 | 245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.851371 |
https://arcomalaga.org/and-pdf/1256-numerical-methods-in-finance-and-economics-pdf-654-902.php | 1,653,607,022,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00721.warc.gz | 148,836,208 | 9,775 | # Numerical Methods In Finance And Economics Pdf
File Name: numerical methods in finance and economics .zip
Size: 1918Kb
Published: 03.05.2021
Search this site. A kitten is, in the animal world, what a rosebud is in the garden PDF. About Face PDF.
Financial economics is the branch of economics characterized by a "concentration on monetary activities", in which "money of one type or another is likely to appear on both sides of a trade". It has two main areas of focus: [2] asset pricing and corporate finance ; the first being the perspective of providers of capital, i. It thus provides the theoretical underpinning for much of finance. The subject is concerned with "the allocation and deployment of economic resources, both spatially and across time, in an uncertain environment".
## Numerical Methods in Finance and Economics A MATLAB based Introduction
Financial Mathematics Pdf A third section highlights reasons why workers. Actual investment programs began to be run on our Bayesian-based systems in ISBN: The American Mathematical Monthly, Vol. In the process, we discuss both what we know about the linkage between marketing and financial outcomes as well as what remains to be uncovered. There are 45 credits of Core modules.
Balanced coverage of the methodology and theory of numerical methods in finance Numerical Methods in Finance bridges the gap between financial theory and computational practice while helping students and practitioners exploit MATLAB for financial applications. Paolo Brandimarte covers the basics of finance and numerical analysis and provides background material that suits the needs of students from both financial engineering and economics perspectives. Classical numerical analysis methods; optimization, including less familiar topics such as stochastic and integer programming; simulation, including low discrepancy sequences; and partial differential equations are covered in detail. Extensive illustrative examples of the application of all of these methodologies are also provided. The text is primarily focused on MATLAB-based application, but also includes descriptions of other readily available toolboxes that are relevant to finance.
## Financial Mathematics Pdf
Items in MacSphere are protected by copyright, with all rights reserved, unless otherwise indicated. Show full item record. Applications of Numerical Methods in Economics and Finance. Alan, Sule. Magee, Lonnie.
Search this site. A Plague of Texts? Address unknown book PDF. Agrobiodiversity: Volume 24 PDF. Aidan PDF. Among the Arabs PDF. An Exposition Of St.
Numerical Methods in. Finance and Economics. A MATLAB-Based Introduction. Second Edition. Paolo Brandimarte. A Wiley-Interscience Publication.
## Numerical Methods in Finance
It seems that you're in Germany. We have a dedicated site for Germany. The use of mathematical models and numerical techniques in finance is a growing practice, and an increasing number of applied mathematicians are working on applications in finance and business. This book presents some exciting developments arising from the combination of mathematics, numerical analysis, and finance. It covers a wide range of topics, from portfolio management and asset pricing, to performance, risk, debt and real option evaluation.
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А если и знал, подумала Сьюзан, то зачем ему мешать ее поискам парня по имени Северная Дакота. Вопросы, не имеющие ответов, множились в голове. А теперь все по порядку, - произнесла она вслух. К Хейлу можно вернуться чуть позже.
- Он улыбнулся. - Возвращайся домой. Прямо. - Встретимся в Стоун-Мэнор. Она кивнула, и из ее глаз потекли слезы.
Молчание. Хейл сразу же растерялся, не зная, как истолковать примирительный тон коммандера, и немного ослабил хватку на горле Сьюзан. - Н-ну, - заикаясь начал он, и голос его внезапно задрожал. - Первым делом вы отдаете мне пистолет. И оба идете со .
Давайте скорее.
Совершенный квадрат? - переспросил Джабба. - Ну и что с. Спустя несколько секунд Соши преобразовала на экране, казалось бы, произвольно набранные буквы. Теперь они выстроились в восемь рядов по восемь в каждом.
Люди на соседних койках приподнялись и внимательно наблюдали за происходящим. В дальнем конце палаты появилась медсестра и быстро направилась к. - Хоть что-нибудь, - настаивал Беккер.
Только туда ей и оставалось идти в наглухо запертом помещении. Поднявшись по ступенькам, она обнаружила, что дверь в кабинет шефа открыта, поскольку электронный замок без электропитания бесполезен. Она вошла.
Беккер вздохнул, взвешивая свои возможности. Где ей еще быть в субботний вечер. Проклиная судьбу, он вылез из автобуса. К клубу вела узкая аллея. Как только он оказался там, его сразу же увлек за собой поток молодых людей.
Разумеется, на ее экране замигал значок, извещающий о возвращении Следопыта. Сьюзан положила руку на мышку и открыла сообщение, Это решит судьбу Хейла, - подумала. - Хейл - это Северная Дакота. - На экране появилось новое окошко.
Бринкерхофф застонал, сожалея, что попросил ее проверить отчет шифровалки. Он опустил глаза и посмотрел на ее протянутую руку. - Речь идет о засекреченной информации, хранящейся в личном помещении директора.
Коммандер послал ее жениха, преподавателя, с заданием от АНБ и даже не потрудился сообщить директору о самом серьезном кризисе в истории агентства.
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Numerical Techniques in Finance is an innovative book that shows how to create, and how to solve problems in a wide variety of complex financial models. | 1,521 | 6,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | latest | en | 0.922266 |
https://socratic.org/questions/if-24-l-of-a-gas-at-room-temperature-exerts-a-pressure-of-8-kpa-on-its-container | 1,638,069,360,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00132.warc.gz | 607,652,225 | 6,143 | # If 24 L of a gas at room temperature exerts a pressure of 8 kPa on its container, what pressure will the gas exert if the container's volume changes to 15 L?
Aug 11, 2017
$12.8 \text{kPa}$
#### Explanation:
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$
$1 L = 0.001 {m}^{3}$
$8000 \cdot 0.024 = 0.015 {P}_{2}$
${P}_{2} = \frac{8000 \cdot 0.024}{0.015} = 12800 \text{Pa"=12.8"kPa}$ | 158 | 379 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-49 | latest | en | 0.643957 |
http://mathematicssolution.com/author/mathematicssolution/page/4/ | 1,503,351,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00533.warc.gz | 278,053,619 | 12,420 | BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Tuesday , August 22 2017
Home / Shah Alam (page 4)
## Shah Alam
I am a student of mathematics department of Begum Rokeya University, Rangpur. I want to provide all kinds of solution of mathematics problems by this web site.
## Let X be a topological vector space.If A ⊂ X then A = n (A + V), where V runs through all neighborhoods of O (zero).
Theorem: (x +v) ∩ A ≠ Φ if and only if x ⊂ A – V . Proof: Suppose a ∈ (x + v) ∩ A. Then a ∈ x + V and a ∈ A. But, a ∈ x + V ⟹ a = x + v ⟹ x ...
## Every topological vector space is a Housdorff space
Proof: Let X be a topological vector space and let x, y ∈, such that x ≠ y. Set K = {x} and C = {y} Then K is a compact and C is a closed and that K ∩ C = Φ. Hence there exist a neighborhood v of ...
## If ℬ is a local base for a topological vector space X, then every member of ℬ contains the closure of some member of ℬ.
Proof: Let W ∈ ℬ. Set K = { 0 } and C = WC. Then C is closed set and K is compact set and C ∩ K = Ф. Hence ∃ a neighborhood V of 0 such that (C+ V) ∩ (K+V) = Ф. Since B is a ...
## Suppose K and C are subsets of a topological vector space X, K is compact, C is closed, and K n C = ∅. Then 0 has a neighborhood of V such that (K + V) n (C + V) = ∅.
Proof: If K = ∅, then K + V = ∅, and the conclusion of the theorem is obvious. We therefore assume that K # ∅, and consider a point x ∈ K. Then x ∉ C ⟹ x ∈ CC. Since C is closed, ∃ a neighborhood W of ...
## Useful definitions for functional analysis
Useful definitions for functional analysis Normed spaces: A vector space X is said to be a normed space if to every x∊ X there is associated a nonnegative real number l l x ll, called the norm of x, in such a way that (a) || x + Y || ...
## Short biography of Isaac Newton
PDF FILE OF SHORT BIOGRAPHY OF ISAAC NEWTON Newton Birth and family Isaac Newton was born on January 4, 1643 in the tiny village of Woolsthorpe-by-Colsterworth, Lincolnshire, England. His father, whose name was also Isaac Newton, was a farmer who died before Isaac Junior was born. His mother, Hannah ...
## PDF| Solution of Exercise- 8.1(Circle) of class 9 and 10
PDF| Solution of Exercise: 8.1(Circle) of class 9 and 10 PDF file of Chapter 8.1: CLICK HERE Exercise 8.1 _Class – 9 and 10_ Problem – 1 : Prove that if two chords of a circle bisect each other, their point of intersection is the centre of the circle. ...
## Biography of Pythagoras
Born: Approximately 569 BC Birth place: Samos Greece Founder of a philosophical and religious school Died: Approximately 500 – 475 BC Died At Age – About 70 – 95 Location of death: Metapontum Italy Pythagoras is repeatedly referred to as the first pure mathematician. He was born on ... | 807 | 2,792 | {"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-2017-34 | latest | en | 0.918496 |
http://dopegearinc.com/british-columbia/how-to-find-ions-in-a-compound.php | 1,558,731,948,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257767.49/warc/CC-MAIN-20190524204559-20190524230559-00235.warc.gz | 70,691,573 | 8,498 | ## How To Find Ions In A Compound
#### How do I calculate the coordination number of a compound?
By putting the positive ion and the negative ion in the respective boxes, you can find the scientific name of that polyatomic. In an ionic compound, electrons are transferred to atoms and molecules imparting a charge on both the donor and recipient of the electron.
#### how to calculate number of ions in a compound Brainly.in
3/04/2017 · In an ionic compound XY, the ionic radii of X+ and Y- are 150 pm and 189pm respectively. Calculate the coordination number of cation X+. Our Official Faceboo...
#### Ionic Compounds Names and Formulas Chapter 8 Flashcards
Chloride ion in AlCl3 has a charge of -1. The compound AlCl3 isaluminum chloride, which is an inorganic compound that can reactwith water.
#### How do I calculate the coordination number of a compound?
Formulae of ionic compounds Ionic compounds contain positive and negative ions. The number of positive charges must equal the number of negative charges so that the compound has no charge overall. When the positive ion has the same number of charges as the negative ion, it is easy to work out the formula of the compound formed. Sodium chloride contains sodium ions, Na+, and chloride ions…
How to find ions in a compound
#### Ionic Compounds Names and Formulas Chapter 8 Flashcards
Let's take an example of Trisodium phosphate Write the ionic formula Na3PO4 which is (Na+)3PO43-So in each formula unit there are 3 sodium ions and 1 phosphate ion
#### How do I calculate the coordination number of a compound?
Determining how many ions in a compound. Did you find this useful? Re: Determining how many ions in a compound. Permalink Submitted by Chemistry Tutor on Sat, 2007-09-15 23:06 . ok just this once but we really like for you to show at least what you tried. see my other post about how amu and grams per mole are the same . so first , follow the other post to convert 3.5 mg to moles …
#### how to calculate number of ions in a compound Brainly.in
In an ionic compound there are two different types of ions present, the positively charged cations and the negatively charged anions. An ionic compound is composed of a network of ions that results in a three-dimensional matrix of cations and anions.
#### How to use ionic compound in a sentence WordHippo
3/04/2017 · In an ionic compound XY, the ionic radii of X+ and Y- are 150 pm and 189pm respectively. Calculate the coordination number of cation X+. Our Official Faceboo...
#### how to calculate number of ions in a compound Brainly.in
3/04/2017 · In an ionic compound XY, the ionic radii of X+ and Y- are 150 pm and 189pm respectively. Calculate the coordination number of cation X+. Our Official Faceboo...
#### how to calculate number of ions in a compound Brainly.in
We can calculate the concentration of free ions found in the solution of ionic compounds. The use of mole ratio is the best way to calculate the concentration of ions in the solution. We know that anions are negatively charged ions while cations are positively charged ions. The dissociation equation for a certain compound can be written as given below; xA --> yB + zC. Here A stands for the
#### how to calculate number of ions in a compound Brainly.in
3/04/2017 · In an ionic compound XY, the ionic radii of X+ and Y- are 150 pm and 189pm respectively. Calculate the coordination number of cation X+. Our Official Faceboo...
#### how to calculate number of ions in a compound Brainly.in
We can calculate the concentration of free ions found in the solution of ionic compounds. The use of mole ratio is the best way to calculate the concentration of ions in the solution. We know that anions are negatively charged ions while cations are positively charged ions. The dissociation equation for a certain compound can be written as given below; xA --> yB + zC. Here A stands for the
#### How to use ionic compound in a sentence WordHippo
3/04/2017 · In an ionic compound XY, the ionic radii of X+ and Y- are 150 pm and 189pm respectively. Calculate the coordination number of cation X+. Our Official Faceboo...
### How to find ions in a compound - How do I calculate the coordination number of a compound?
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Australian Capital Territory: Chapman ACT, Bimberi ACT, Richardson ACT, Forrest ACT, Majura ACT, ACT Australia 2664
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Wales: Neath WAL, Neath WAL, Wrexham WAL, Wrexham WAL, Swansea WAL, WAL United Kingdom CF24 7D1 | 2,338 | 8,969 | {"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-2019-22 | latest | en | 0.887841 |
https://stats.stackexchange.com/questions/136842/multivariate-cram%C3%A9r-rao-inequality-intuition-for-positive-semidefiniteness | 1,725,942,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00343.warc.gz | 514,096,778 | 43,362 | # Multivariate Cramér-Rao inequality: intuition for positive semidefiniteness
Here's what Wikipedia says about the Multivariate Cramér-Rao inequality:
If $\boldsymbol{T}(X)$ is an unbiased estimator of $\boldsymbol{\theta}$, then the Cramér–Rao bound reduces to $\mathrm{cov}_{\boldsymbol{\theta}}\left(\boldsymbol{T}(X)\right) \geq I\left(\boldsymbol{\theta}\right)^{-1}$.
The matrix inequality $A \ge B$ is understood to mean that the matrix $A-B$ is positive semidefinite.
I understand everything above. But I started playing around with examples and came up with something that doesn't make sense to me. Suppose we have two unbiased estimators of some two-dimensional $\boldsymbol{\theta}$, one with covariance matrix
$A=\left[\begin{array}{cc} 3 & 1.5\\ 1.5 & 3 \end{array}\right]$
and the other with covariance matrix
$B=\left[\begin{array}{cc} 2 & 0.3\\ 0.3 & 2 \end{array}\right]$.
Now $A-B$ is not positive semidefinite. So even though each parameter estimate has a lower variance in $B$, and there's less covariance between the two of them, $B$ doesn't 'count' as having lower variance than A in the psd sense.
Can someone give an intuitive explanation why? (I guess maybe I'm really asking why is the generalized inequality with respect to the psd cone the 'right' comparison here? Or is it just the only one for which we can prove this result?)
• When the difference of two covariance matrices is not definite, it means that some linear combinations will have larger variance with one of them, other linear combinations will have larger variance with the other one. So there is no domination. Commented Feb 9, 2015 at 8:34
• Thanks. But: why does it matter about linear combinations? And can you give an example of such a linear combination in this case? (I'm having trouble seeing one.)
– mww
Commented Feb 10, 2015 at 1:07
• For an example, take the diagonal case: $A=\begin{smallmatrix} 2&0\\0&3\end{smallmatrix}, B=\begin{smallmatrix}3&0\\0&2\end{smallmatrix}$. Then the linear comb of variables $(X,Y)$ taken with that covariance matrix, with coefficients $(1,0)$ has bigest variance for $B$, while lincomb with coef $(0,1)$ has bigest varaince for $A$. Commented Feb 11, 2015 at 18:15
• But surely my examples of A and B don't present that problem?
– mww
Commented Feb 11, 2015 at 19:47
• $A$ has an eigenvalue, $3/2$, which is smaller than the smallest eigenvalue of $B$ (which is $17/10$). Thus ellipses defined by $x^\prime A x=c$ are actually a little narrower than ellipses $x^\prime B x = c$. Equivalently, there exist linear combinations of the (implied) underlying random variables which have a smaller variance for $A$ than for $B$.
– whuber
Commented Jun 5, 2017 at 19:17
It says that (asymptotically), for any regular estimator $$T$$ we have $$T(X)-\theta= Z+ \Delta$$ where $$Z$$ has covariance matrix $$I^{-1}(\theta)$$ and $$\Delta$$ is independent of $$Z$$. That is, any estimator is the efficient estimator plus pure noise. Which is a nice property for an efficiency bound to have. [Another version says this holds for any estimator $$T(X)$$ at almost all $$\theta$$]
If you look at what this says about $$\mathrm{cov}[T(X)]$$, it is exactly that $$\mathrm{cov}[T(X)]\geq I^{-1}(\theta)$$
So the covariance matrices are $$A=(\begin{smallmatrix} 3 & 1.5 \\ 1.5 & 3\end{smallmatrix})$$ and $$B=(\begin{smallmatrix} 2 & 0.3 \\ 0.3 2\end{smallmatrix})$$. Yes, visually it seems like $$A$$ is "bigger than $$B$$, and really, that holds for the determinants "generalized variances" which are 6.75 and 3.91. But that impression is deceiving. Calculating the eigendecomposition of $$A - B$$, we find a negative eigenvalue -0.2 for the eigenvector $$v=(\begin{smallmatrix} -\sqrt{2}/2 \\ \sqrt{2}/2 \end{smallmatrix})$$. So, in the direction given by that eigenvector, $$A$$ has smaller variance than $$B$$, as the following calculation gives. If $$X$$ is a random variable with covar matrix $$A$$, and $$Y$$ has covar matrix $$B$$, then $$\DeclareMathOperator{\V}{\mathbb{V}} \V v^T X = v^T A v = 1.5 \\ \V v^T Y = v^T B v = 1.7$$ so indeed, in that direction $$A$$ is smaller than $$B$$ despite that the overall impression is the opposite. | 1,212 | 4,177 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.882088 |
https://khg.kname.edu.ua/index.php/khg/article/view/5600 | 1,603,589,417,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00603.warc.gz | 378,974,731 | 9,590 | # MATHEMATICAL MODELING OF INFLUENCE THE OPERATION MODE DIESEL ENGINE ON THE CONTENT OF PARTICULATE MATTER IN EXHAUST GASES
• А. Polivyanchuk O.M. Beketov National University of Urban Economy in Kharkiv
Keywords: diesel, exhaust gases, particulate matter, concentration, test mode, mathematical model, accuracy
### Abstract
The work is devoted to solving an urgent scientific and practical task – establishing mathematical models that describe the effect of operating modes of diesel engines on the content in the exhaust gases of a dangerous pollutant – particulate matter (PM). The purpose of the work was to create and study the accuracy and practical suitability of the calculation method estimating concentrations and emissions with exhaust gases PM by means of mathematical modeling of the influence on them of parameters that determine steady state and unsteady diesel operation modes. The studies were carried out on the basis of the motor stand of a 4ChN12/14 autotractor diesel equipped with a partial-flow system for diluting EG with air – MKT-2 microtunnels. Measurements mass and volume concentrations – cpt (g/kg) and Cpt (g/mn3), mass – PTmass (g/h) and specific – PTp (g/kWh) PM emissions was carried out by the gravimetric method with errors of ± 3 .. 10% in accordance with the requirements of regulatory documents – ISO8178 standard, UNECE Rules R-49, R-96, etc. As parameters which determine the mode of operation of the diesel engine, were considered: during the study steady-state modes – the number of revolutions of the engine crankshaft (n, min-1) and the load (L,%); in the study of unsteady modes – the parameters n and L and the rate of change over time – Δn/Δt and ΔL/Δt. A dependence has been established for indirectly determining the mass concentration of PM in the exhaust gases at steady and unsteady diesel operating modes, which are characterized by a duration of 10 ... 30 s and ranges of vari-ation of the parameters n, L, Δn/Δt and ΔL/Δt, given in dimensionless form: 0.4 ... 0.8, 0.3 ... 1.0, -0.2 ... 0.2 and -0.35 ... 0.35, respectively. The deviation of the calculated and experimental data when using this dependence is ± 0.005-0.006 g/kg, which is comparable with the sensitivity limit of MKT-2.
### Author Biography
А. Polivyanchuk, O.M. Beketov National University of Urban Economy in Kharkiv
Doctor of Technical Sciences, Professor
### References
1. Zvonov, V.A. (2004). Ekologiya avtomobil'nykh dvigateley vnutrennego sgoraniya [Ecology of automobile internal combustion engines]. Lugansk: VNU named after V. Dahl. – 268. (in Russian).
2. Bielaczyc, P., Woodburn Szczotka, J. (2016). Exhaust Emissions of Gaseous and Solid Pollutants Measured over the NEDC, FTP-75 and WLTC Chassis Dynamometer Driving Cycles. SAE Technical Paper, 2016-01-1008, 13.
3. Foote, E., Maricq, M., Sherman, M., Carpenter, D. et al. (2013). Evaluation of Partial Flow Dilution Methodology for Light Duty Particulate Mass Measurement. SAE Technical Pape, 2013-01-1567, 10.
4. Littera, D., Cozzolini, A., Besch, M., Velardi, M. et al. (2013). Comparison of Particulate Matter Emissions from Different Aftertreatment Technologies in a Wind Tunnel. SAE Technical Paper, 2013-24-0175, 17.
5. Alkidas, A. C. (1984). Relationship Between Smoke Measure-ments and Particulate Measurements. SAE Technical Paper. 840412, 316–322.
6. Kuharenok, G.M. (2016). Otsenka soderzhaniya dispersnykh chastits v otrabotavshikh gazakh dizel'nykh dvigateley [Estimation of the content of dispersed particles in the exhaust gases of diesel engines]. Science and technology, 15, 5, 371–379. (in Russian).
7. Kittelson, D., Kraft, M. (2015). Particle Formation and Models. Encyclopedia of Automotive Engineering, 1(23), 107–130.
8. Muntean, G. G. (1999). A Theoretical Model for the Correlation of Smoke Number to Dry Particulate Concentration in Diesel Exhaust. SAE Technical Paper. 1999-01-0515, 316–322.
9. Burtscher, H. (2001). Literature Study on Tailpipe Particulate Emission Measurement for Diesel Engines. Done for the Particle Measurement Programme (PMP) for BUWAL/GRPE. Fachhochschule Aargau, University of Applied Science, Windisch, Switzerland, 45.
10. Anderson, J.D. (2003). UK Particle Measurement Programme. Phase 2. Heavy Duty Methodology Development. Final Report. Ricardo Consulting Engineers Ltd, 222.
11. Seito, K., Shinozaki, O. (1990). The measurement of diesel particulate emissions with tapered element oscillating microbalance and an opacimeter. SAE Technical Paper. 900644, 1-5.
12. Abe, T., Sato, T., Hayashida, M. (1989). Particulate matter emission characteristics under transient pattern driving. SAE Technical Paper. 890468, 151-163.
13. Schraml, S., Will, S., Leipertz, A. (1999). Simultaneous measurement of soot mass concentration and primary particle size in the exhaust of DI diesel engine by time-resolved laser-induced incandescence (TIRE-LII). SAE Technical Paper, 1999-01-0146, 8.
Published
2020-07-01
How to Cite
PolivyanchukА. (2020). MATHEMATICAL MODELING OF INFLUENCE THE OPERATION MODE DIESEL ENGINE ON THE CONTENT OF PARTICULATE MATTER IN EXHAUST GASES. Municipal Economy of Cities, 3(156), 62-68. Retrieved from https://khg.kname.edu.ua/index.php/khg/article/view/5600
Section
статьи | 1,426 | 5,236 | {"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-2020-45 | latest | en | 0.861467 |
http://mathhomeworkanswers.org/21383/what-is-pi-3-14-rounded-to-the-nearest-whole-number?show=223923 | 1,488,398,843,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174276.22/warc/CC-MAIN-20170219104614-00217-ip-10-171-10-108.ec2.internal.warc.gz | 166,375,629 | 14,765 | pi-3.14 often rounded to which number
3.14 rounded to the nearest whole number
= 3
answered Mar 10, 2012 by Level 4 User (8,440 points)
edited Mar 10, 2012 by mshelton
What is the next whole number after three?
answered Feb 14, 2013 by anonymous
what is 26310.06 rounded to the nearest whole number
answered May 13, 2014 by verneshia white
Three because it is nearest whole. Hope it helps
answered Jun 29, 2014 by Samy
75 because if u add up all the numbers u get 75
answered Jul 8, 2014 by BobMarley
answered Jul 8, 2014 by Kieth lemon
Pi to the nearest whole number is 3
As far as I can remember pi equals 3.1415926535897932384
answered Jul 8, 2014 by Level 9 User (42,120 points)
4 because if you add up the numbers and the divide them by three it is 4
answered Jul 11, 2014 by Kieth lemon
2909.466
answered Sep 9, 2014 by anonymous
round 15,698 to the nearest whole number
answered Sep 20, 2014 by anonymous
IF YOU ROUND OF 0.39 AND 0.68 WHAT IS THE NEAREST WHOLE NUMBER?
answered Oct 16, 2014 by anonymous
4499.677419
answered Oct 16, 2014 by anonymous
Pi is often rounded to 3, if you ask me.
answered Feb 18, 2015 by MathDude
3.14 rouned to the nearest whole numer is 3. 5 or more raise the score, 4 or less let it rest.
answered Jun 2, 2015 by anonymous
answered Jun 4, 2016 by anonymous | 419 | 1,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-09 | latest | en | 0.957739 |
http://onlinelibrary.wiley.com/doi/10.1002/ird.1821/abstract | 1,409,612,917,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535920694.0/warc/CC-MAIN-20140909051739-00291-ip-10-180-136-8.ec2.internal.warc.gz | 384,793,613 | 10,417 | ## SEARCH BY CITATION
### Keywords:
• subsurface drainage;
• unsaturated zone;
• horizontal flow;
• groundwater table;
• drain spacing equation
• drainage souterrain;
• zone non saturée;
• écoulement horizontal;
• nappe phréatique;
• équation de l'espacement des drains
### ABSTRACT
Most drain spacing calculations do not take the horizontal flow in the unsaturated zone above the groundwater table into consideration. In this paper, a solution is presented that includes the contribution of unsaturated flow above the groundwater table. Drain spacing calculated with the newly derived equation is compared to that calculated with the Hooghoudt equation and the two-dimensional Hydrus-2D model. Results show that drain spacing calculated with the new equation results in a wider value. The effects are most pronounced for tiles located close to the impervious layer, particularly in coarse, sandy soils. These effects rapidly decrease if the depth of the impervious layer increases. The effect of the unsaturated zone flow contribution is limited to sandy soils, for low infiltration ratios and tiles placed on top of an impervious layer. The maximum increase in drain spacing calculated by the new formula is about 30% higher, suggesting that inclusion of the contribution of the unsaturated zone flow in the computation of drain spacing may result in greater economy in the design of subsurface drainage systems. Moreover, the new equation is more general and is applicable for tiles lying on or far from an impervious layer. Copyright © 2014 John Wiley & Sons, Ltd.
### RÉSUMÉ
La plupart des équations utilisées pour le calcul d'écartement des drains ne tiennent pas compte des écoulements horizontaux vers les drains provenant de la zone non saturée. Dans ce présent travail, nous proposons une nouvelle équation tenant compte de la contribution au débit des drains de la zone non saturée au-dessus du niveau de la nappe. L'écoulement vers un drain est la somme de trois composantes: les deux premières représentent la contribution des flux dans l'écoulement vers le drain provenant de la zone saturée au-dessus et au-dessous du drain, le troisième représente la contribution au flux provenant de la zone non saturée. Les deux premières sont identiques à celles trouvées dans les équations ordinaires pour calculer l'écartement des drains, alors que le troisième représente notre contribution.
Pour examiner l'effet de la zone non saturée sur l'écartement des drains, ces derniers sont calculés par notre nouvelle formule et celle développée par Hooghoudt. Les résultats montrent que l'écartement calculé par notre formule donne des valeurs supérieures à celles de Hooghoudt. La différence est plus grande pour des drains reposant sur le substratum imperméable et pour des sols de texture sableuse. Cette différence décroit rapidement avec l'accroissement de la profondeur du substratum par rapport aux drains. L'effet de la zone non saturée est limité aux sols sableux avec un rapport (|q0|/Ks) faible et des drains reposant ou peu éloignés du substratum imperméable. La différence maximale obtenue entre écartement calculé par notre formule et celle calculée par la formule de Hooghoudt peut atteindre 30%. Ceci suggère l'importance d'utiliser notre formule pour minimiser les coûts d'investissement dans les projets de drainage agricole. En plus, la formule développée s'applique pour des drains reposant ou non sur un substratum imperméable ce qui n'est pas le cas pour les autres formules. Copyright © 2014 John Wiley & Sons, Ltd. | 819 | 3,546 | {"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-2014-35 | latest | en | 0.51895 |
http://lincs.ed.gov/pipermail/numeracy/2010/000197.html | 1,462,299,556,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121737.31/warc/CC-MAIN-20160428161521-00081-ip-10-239-7-51.ec2.internal.warc.gz | 167,236,200 | 7,724 | [Numeracy 187] Re: The double negative language-math link
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steinkedb at q.com steinkedb at q.com
Tue Feb 16 17:52:40 EST 2010
I see a problem with using English grammer as a model for math syntax.
We must remember that not all our students are native English speakers. What
we consider a "double-negative means a positive" in English is a simple
negative in Spanish. A literal translation of a simple negative sentence in
Spanish (No quiero nada.) looks like a "double-negative" in English (I don't
want nothing.) We must be careful that students take in the information in
the way that we intend it.
Dorothea Steinke
-----Original Message-----
From: numeracy-bounces at nifl.gov [mailto:numeracy-bounces at nifl.gov]On
Behalf Of Carol King
Sent: Tuesday, February 16, 2010 2:27 PM
To: The Math and Numeracy Discussion List
Subject: [Numeracy 186] Re: The double negative language-math link
The rule in teaching English is that a double negative statement creates a
positive statement, so, for a few students, it makes sense to hang my hat on
the hook they have. Since they know in English " to not not go" creates a
positive statement that you are going (and must be rewritten as such) it
bridges their mental block about double negatives in math changing to
Carol King
cking at lyon.k12.nv.us
----------------------------------------------------------------------------
--
From: numeracy-bounces at nifl.gov [mailto:numeracy-bounces at nifl.gov] On
Behalf Of Michael Gyori
Sent: Thursday, February 11, 2010 8:58 AM
To: The Math and Numeracy Discussion List
Subject: [Numeracy 148] The double negative language-math link
Greetings everone,
Carol King stated,
If I am taking out taking out 8, as in 10 - (-8), then I must be adding
it.
I read it a few times and find myself perplexed by it, as much as I
believe I understand its intent.
"Taking out" is a positive statement and regardless of how many times you
say it, it remains positive, and what changes - perhaps, depending on how I
choose to understand it - is the number of times you (***yes***) "take
out." If I take out once, I have 2 left, and I cannot take take out again,
because I can't take another 8 out of 2.
Alternatively, I can understand the meaning to be that I am "taking out"
the taking out of 8, which then could leave me to believe that I wanted to
take out, then decided against it, such that I end up doing nothing. I
still have 10.
The problem, as I see it, is that we are getting into integers. Negative
values have no meaning in the world of the concrete, because once you have 0
left, that's it. On the other hand, if we deal with negative balances (such
as when you overdraw your balance in your checking account), you create
meaning because it can and does happen. In other words, negatives carry
meaning in mathematical, but not physical (reality) terms...
Thoughts?
Michael
Michael A. Gyori
Maui International Language School
www.mauilanguage.com
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dissemination, distribution or copying of this e-mail and any attachments is | 848 | 3,476 | {"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-2016-18 | longest | en | 0.877622 |
https://issuu.com/johnpaulmarceaux/docs/songoftime | 1,526,980,023,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00536.warc.gz | 581,866,589 | 21,015 | Song Image of the Song of Time JP Marceaux December 20, 2017
Dedicated to and written for Conway LeBleu One of the most interesting innovations of the 20th century abstract art movement was the development of pictorial representations of music, seen particularly in the correspondence between Wassily Kandinsky and Arnold Schoenberg. Here, I use quantum wavefunctions to achieve the same goal of creating an image that reflects the qualities of a piece of music. I do this by defining wavefunctions for the different notes present in a song and graphing the superposition of all notes. The song chosen in this case was the "Song of Time". The central object of the art is The Ledgend of Zelda Song of Time, depicted in musical notation below.
Figure 1: https://www.youtube.com/watch?v=SEBybsl_k1U). A construction of a numeric encodage of the song is necessary in order to arrange the song’s information in a form that can be interpreted by the computer. I choose to do this with a structural definition of each note, wherein each note is associated with a block that contains numeric description of its essence. Memory blocks containing a note name, a note index, a chromatic index, a temporal location, a temporal length, and a frequency represent a note. The algorithm uses this information to construct the image that is displayed on the monitor. Here, I choose to omit the frequency of a note and give only its mathematical definition. Note A D F A D F A C’ B G F G A D C E D Note index 5 1 3 5 1 3 5 7 6 4 3 4 5 1 0 2 1 Chromatic index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 1 1 1 1 Length 1 2 1 1 2 1 12 1 1 1 1 1 2 2 2 2 2 The prime over the C denotes that the note occurs an octave above. The note number is an integer enumeration defined by setting the first C equal to 0 and continuing in integer steps such that the second C is equal to 7. The note length is in units of quarter-notes. The frequency is determined according to the equation: n
fn ≡ f0 2 7 , where n is the note number and f0 is an arbitrary definition of the center frequency of the scale. The form is such that we are working over a scale of 7 notes. The chromatic index of a note of index n is defined by: cn ≡ mod7 (n), (1) 1
such that the chromatic index takes on the 7 integer values between 0 and 6 inclusive. We map each integer cn ∈ [0, 6] to a color. In the final image, I use the concept of overtones to add detail and interest. Overtones, or harmonics, are physical phenomenon that occur in all real instruments. When a musician plucks a string, blows a note, or strikes a membrane, the resulting waveform contains a superposition harmonics or basis states, normally taken as sines and cosines. This superposition can be defined by its fundamental note, which is of lowest frequency and often of highest amplitude. We assume that all harmonics of the notes occur in integer multiples of the fundamental frequency. Thus, an overtone of the note f0 is mf0 , where m = 1, 2, 3.... To a good approximation, all the integer multiples of a given note will be equal in frequency to notes of higher frequency. However, the note name will change. For example, in the chromatic scale, the 3rd harmonic of the note C0 is G2. The note name, not the frequency, is used to define the color index. We now consider the chromatic index of an overtone. Let us label the j th overtone of a note of index n by its frequency with the notation: fnj , where
(2) n
fnj = jf0 2 7 .
(3)
The chromatic index of a note is defined in terms of its note index. The frequency of an overtone can be used to define the overtone’s note index. We may associate the frequency of an overtone with a new note index m and impose the equivalence: fm = fnj , which is not valid for all j because
n
m
2 7 = j2 7
is not true for all {n, j, m} ∈ Z3 . We shall impose the equivalence by rounding fnj to the nearest fm . This frequency fnj will be near the frequency of some other note of index m. Setting the two frequency equal: m n f0 2 7 = jf0 2 7 m n = + log(j), 7 7 where the passage between the two lines was accomplished with a logarithm of base 2. The note index of the jth overtone of a note of index m is thus: n = 7logm + j,
(4)
where the RHS is rounded to the nearest integer. We employ the notation: j Im
(5)
to represent the note index of the j th overtone of the note of index m. We use Equation (1) to find the chromatic index, remembering that we must round the index to the nearest whole integer: c(fnj ) = mod7 (round(7log(j) + m)).
(6)
In words, the above equation says that the chromatic index of the j th overtone of the note of index n is give by the modulus base 2 of the sum of the note index plus 8 times the logarithm base 2 of the harmonic number. We will make use of the notation cjn ≡ c(fnj ) 2
(7)
We now consider the image of the song, and we invoke techniques used in the definition of a quantum mechanical wavefunction. We will consider the screen as a space-time graph, where the space values x are graphed on the ordinate and the time values t are graphed on the abscissa. We seek to define an overall wavefunction Ψ that represents the space-time development of the song. A convenient analogy is that the image produced is a graph of the temporal development of a string that plays the song of time. We associate each note index with a quantum number n, where n = 0, 1, 2.... We associate each chromatic index with a quantum number l, such that l takes on integer values ∈ [0, 6]. This choice of definition of l is similar to a numbering of the states of a particle of spin 3. The n quantum number will define the shape of the wavefunction and the l quantum number will define the color. The background of the image begins as black. The computer samples the overall wavefunction Ψ, which produces a color that the computer creates on the monitor. Ψ is composed of superposition of individual wavefunctions. I write the 3 wavelets plus a gabor wavelet that I will use in my definition of Ψ. I will pass from wavefunctions defined in only space to a space-time wavefunction. The first wavelet forms the fundamental basis upon which all superposition are drawn. It is: ϕln (x) = sin(nπx)χl , (8) where x has a domain ∈ [0, 1] and χl is the quantum color vector associated with the color value l. The second wavelet defines the harmonic superposition. It is: ψn (x, t) =
X 1 ck n k=1
k
ϕI k (x, t),
(9)
n
where we use Equations (5) and (7) to define the note index and color index of the overtone respectively. Let us define the gabor wavelet of the block b by: gb = e
(t−τb )2 αγb
,
where α is the base width of each note and τb and γb are the temporal displacement and the augmented width of the block of index b respectively. This wavelet will be used to define the temporal description of the wave. The third wavelet defines the wavefunction of a single block Φb (x, t) = ψb.n (x) ∗ gb (t) =
X 1 ck b.n k=1
k
!
ϕI k (x) e
(t−τb )2 αγb
,
(10)
b.n
where the symbol b.n denotes the note index contained in the block of index b. This is C++ code for the content n of the structure b. The wavefunction for the entire piece of music is finally the sum of each individual block’s wavefunction: Ψ(x, t) =
16 X b=0
3
Φb (x, t)
(11)
Songoftime
Songoftime | 1,938 | 7,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-22 | latest | en | 0.8893 |
http://math.stackexchange.com/questions/83791/convergence-of-v-n-sum-k-1n-n2k-1-2 | 1,469,434,373,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824217.36/warc/CC-MAIN-20160723071024-00306-ip-10-185-27-174.ec2.internal.warc.gz | 151,556,941 | 18,039 | # Convergence of $v_n = \sum_{k=1}^n (n^2+k)^{-1/2}$
Good day, I would like to solve this problem using the theorem of the comparison, but I do not understand how to proceed, can anyone help me?
How do I prove that $(v_n)$, defined by $$v_n:= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}},$$ is convergent?
-
I don't really feel like it right now. – Tyler Nov 20 '11 at 1:09
The point Tyler is trying to make is that you shouldn't state your questions as commands. It's better to ask "How can I prove this" instead of saying "Prove this." – Aaron Nov 20 '11 at 1:10
Hint: There are $n$ terms, so $1>n/\sqrt{n^2+1}>v_n>n/\sqrt{n^2+n}$. What are the limits of the two sides of the inequality? – Aaron Nov 20 '11 at 1:13
Which grammar you use is really not relevant. What is relevant is that you're just copying an exercise without adding any thoughts of your own. That will make many members much less likely to want to help you, even if you rephrase the question with different grammar but no new content. – Henning Makholm Nov 20 '11 at 1:13
The story does not end at the command Aaron mentioned: some people even consider good manners to explain what you know, what you tried and why it failed. Amazing, no? – Did Nov 20 '11 at 1:14
$$\frac{n}{\sqrt{n^2+n}} = \frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+\cdots+\frac{1}{\sqrt{n^2+n}} \leq$$
$$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}}$$
$$\leq \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+\cdots+\frac{1}{\sqrt{n^2}} = \frac{n}{\sqrt{n^2}} = 1$$
The inequalities hold since bigger denominators make smaller fractions and obviously: $n^2+n \geq n^2+i \geq n^2$ for $i=1,\dots,n$.
Next, $\lim\limits_{n\to\infty} \frac{n}{\sqrt{n^2+n}}= 1$. So your summation is squeezed between 1 and 1. Thus its limit is 1. | 631 | 1,826 | {"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} | 3.84375 | 4 | CC-MAIN-2016-30 | latest | en | 0.908606 |
thompson955.madelineinteriors.com | 1,627,406,047,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00107.warc.gz | 572,916,445 | 6,606 | # Types of Bets on the Roulette Table
The table where a roulette game is playing is known as the “rolette table.” The numbers on this roulette table range from game to game. The chance of winning about the same spin of a wheel is known as the “roll.” How big is each bet in a casino game is also variable. If you place a bet and win, that bet will undoubtedly be doubled if you choose the same number from another players in a game that you play.
A Roulette table layout is used to help place bets. You can find two forms of Roulette table layouts, even and odd. When playing Roulette the game is a lot easier to take care of when there are even numbers up for grabs. Odd numbers include even numbers and multiples of even numbers. When placing bets, odd numbers like an odd number four always have the highest possibility of being won by you.
The most frequent type of roulette table used in most casinos in the us and Europe may be the push-tabard roulette table. This is a smaller version of a standard roulette table that means it is simpler to place and win bets. This table layout is used more often in European and American casinos than it is in home casinos in America and Europe. Many home casinos in America and Europe have adopted the push toward layout for roulette tables due to the smaller size.
Another type of Roulette table is the non-push layout. In this type of table all of the symbols are shown on the left hand side. The numbers are printed on the right and all of the outcomes of spins of the wheels are read from top to bottom. For example, in a 4 wheel spin the dealer will first show the ace symbol then the numbers, and the 3rd wheel symbol to show that the 3rd spin occurred.
In the end of the information is shown the odds of every result is then given on the left hand side. The chances on a Roulette table payouts table can be extremely complicated if you don’t have some basic knowledge about probability. The best way to find out what the chances are on the Roulette table payouts would be to place your bets on a live game and take time to study the odds. THE WEB also has an abundance of information about the various odds for different game variations.
Before placing your bets on a Roulette table payout you need to learn about the different types of spins that are used in the game. Each of the spins aside from the straight flush require that the winning number is well known beforehand. Once the wheels are spinning at full speed you have two options. You can 조커 바카라 사이트 either win a number that doesn’t represent the odds on the Roulette table, or you can end up with chips which are worth less than the number of chips you have in your bank roll.
The “ringing bet” is the first bet you make when you enter the game. This bet is placed on the proper or the outer corner of your seat. It represents the first non-zero value seen on the Roulette wheel while it spins. If the result of this bet is really a straight flush or an Ace, you have lost all your previous chip burn.
One of the most popular types of bets may be the outside bet or “one number bet”. That is a risky bet, as it can put you at risk for losing more chips than you have in your bankroll. The outside bet, also called the blind or spread bet, is made when the dealer reveals only 1 number that’s on the Roulette wheel. In almost all Roulette games, the dealer will then place this number within an exact location on the Roulette table (externally or the middle of the circle). You need to place your bet prior to the dealer reveals that one number.
Website: | 754 | 3,574 | {"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} | 2.5625 | 3 | CC-MAIN-2021-31 | longest | en | 0.935054 |
https://maimelatct.com/2019/04/13/internal-resistance-and-series-parallel-networks/ | 1,685,791,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00638.warc.gz | 415,451,931 | 42,577 | Knowledge Area: Electricity and Magnetism
Topic: Internal resistance and Series-Parallel networks
Part 1
Aim
To determine the internal resistance of a battery
Precautions
Do not leave the switch closed at all times. The battery will run flat. Close the switch only for taking the readings on the voltmeter and Ammeter. Take your readings accurately. Do not conduct the experiment with wet hands.
Instructions
Place 4 x 1,5V cells in a cell holder in a series connection. Connect a Switch, Ammeter and a Rheostat in series to the battery. Connect the Voltmeter across the battery.
If a rheostat is not available, resistors or bulbs could be used in the place of a rheostat. The resistors or bulbs must be connected in series. One resistor/bulb must be connected first and then the resistors/bulbs be increased to two in series, three in series, etc.
Draw the electrical circuit diagram. Set the Rheostat at 10W. Close the switch and reduce the resistance in the Rheostat step by step, increasing the current at the same time. Take the ammeter and voltmeter readings in each case/step. Open the switch while recording the readings in order to spare the batteries’ life span.
Take and record a minimum of 5 readings. Interpret and analyse the data in order to determine the internal resistance. Write a conclusion and prepare a report (write-up).
Part 2
Aim:
1. To determine the equivalent resistance in a Series-parallel network electrical circuit.
2. To compare the experimental values of the equivalent resistance to the theoretical values.
Instructions:
Place 4 x 1.5V cells in a cell in a cell holder in a series connection. Connect a voltmeter (V1) across the battery. Connect three resistors of different resistance. One resistor must be connected in series with a voltmeter (V2) across and the other two resistors must be connected in parallel with a voltmeter across (V3). Connect a switch and an ammeter.
Draw the electrical circuit. Determine the equivalent resistance experimentally and compare the readings to the calculated values of equivalent resistance. Submit a report (write up) | 464 | 2,116 | {"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-2023-23 | latest | en | 0.848274 |
http://www.merlot.org/merlot/viewMaterial.htm?id=365142 | 1,638,339,536,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00541.warc.gz | 110,737,467 | 15,803 | # Material Detail
## Three dimensional elasticity solutions for plates
A Navier-type method for finding the exact three-dimensional solution for isotropic thick and thin rectangular plates is presented. The extension of this procedure to the case of multilayered plates (composite structures) is straightforward and can be found in the listed references . The method presented in these lectures uses the Mixed Form of Hooke's Law (MFHL) which leads one to write the boundary conditions on the top and bottom surfaces of the plate directly in terms of transverse stresses. Mixed Form of Hooke's Law is obtained from the Classical Form of Hooke's Law (CFHL). The displacements, stresses and external loads are expressed by using trigonometric functions. It is possible to demonstrate that this particular choice satisfies the simply supported boundary conditions. The elasticity solution must satisfy the equilibrium equations, geometric relations and Hooke's law. After some mathematical derivations it is possible to demonstrate that a set of differential equations in the unknown amplitudes of the displacements and stresses has to be solved. An eigenvalue problem is obtained. Only 2 (over a total of 6) eigenvalues are distinct. Therefore, a basis of eigenvectors is not available and 2 generalized eigenvectors have to be found. The solution is a combination of the eigenvectors and generalized eigenvectors multiplied by functions of the out-of-plane coordinate z. The unknown constants of this combination are determined by imposing that the transverse stresses at the top and bottom surfaces of the plate must match the applied pressures. In a general multilayered structure additional conditions on the interlaminar continuity of the displacements and transverse stresses are required. Once the unknown coefficients are calculated, the elasticity solution is complete. These lectures present an algebraic solution for the particular case of a pressure applied at the top surface. The extension to the case of generic load can be obtained by using Fourier analysis. The lectures also present the FREE software that can be downloaded. The program calculates the amplitudes of the displacements and stresses for a sinusoidal load pressure applied at the top surface of the plate. The amplitude of the pressure distribution, material properties, geometric dimensions of the plate and wave numbers are inputs of the software.
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• Accessibility Info | 491 | 2,586 | {"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} | 2.859375 | 3 | CC-MAIN-2021-49 | latest | en | 0.884028 |
http://mathoverflow.net/feeds/question/95557 | 1,371,589,483,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707188217/warc/CC-MAIN-20130516122628-00065-ip-10-60-113-184.ec2.internal.warc.gz | 163,462,417 | 2,582 | Reference wanted for application of Parametric Transversality - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:04:43Z http://mathoverflow.net/feeds/question/95557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95557/reference-wanted-for-application-of-parametric-transversality Reference wanted for application of Parametric Transversality Dick Palais 2012-04-30T06:54:21Z 2012-04-30T18:26:03Z <p>Let $\hbox{Aff(}k,n)$ be the space of $k$-dimensional affine subspaces of $R^n$. The group of Euclidean isometries of $R^n$ (the semi-direct product of rotations and translation) acts transitively on $\hbox{Aff(}k,n)$, so by general Lie theory $\hbox{Aff(}k,n)$ is a smooth manifold and there is a smooth invariant measure on it (called kinematic measure in Integral Geometry) that is unique to within positive scalar multiple. If $M$ is a smooth $(n-m)$-dimensional submanifold of $R^n$, then it is an easy corollary of the Parametric Transversality Theorem that all $F \in \hbox{Aff(}k,n)$ outside a set $E$ of measure zero are transversal to $M$, so that if $F \notin E$ then $F\cap M$ is a smooth $(n-(m+k))$-dimensional submanifold of $R^n$. (Special case: Almost all lines in $R^3$ intersect a given surface in a discrete set.) Of course this has to be a well-known theorem, and I would like to know the appropriate reference for it. </p> http://mathoverflow.net/questions/95557/reference-wanted-for-application-of-parametric-transversality/95560#95560 Answer by alvarezpaiva for Reference wanted for application of Parametric Transversality alvarezpaiva 2012-04-30T07:30:03Z 2012-04-30T07:30:03Z <p>Dear Richard,</p> <p>I think this is just thought of as "folklore" in integral geometry (sorry, never seen it written down). This is probably because the statement you give follows from something that needs almost no technology to prove:</p> <p>The set of $k$-flats that are <em>parallel</em> to a given $k$-dimensional subspace $L$ can be identified $R^n / L$ and provided with the Lebesgue measure. Among these flats, those that intersect the manifold $M$ transversely have full measure in $R^n / L$. </p> <p>One then uses the usual decomposition of the kinematical measure (the set of affine $k$-flats is the total space of a tautological bundle over a Grassmannian) to conclude the statement you seek.</p> http://mathoverflow.net/questions/95557/reference-wanted-for-application-of-parametric-transversality/95600#95600 Answer by Misha for Reference wanted for application of Parametric Transversality Misha 2012-04-30T18:26:03Z 2012-04-30T18:26:03Z <p>This is a special case of a more general theorem on generic transversality of submanifolds in homogeneous spaces: </p> <p>Let $Y$ be a smooth homogeneous space for a Lie group $G$ and $Z_1, Z_2\subset Y$ be smooth submanifolds. Then for any measure in the Lebesgue measure class on $G$, for almost all $g\in G$ the intersection $Z_1\cap g Z_2$ is transversal. </p> <p>Algebraic geometers know this result as <em>Kleiman Transversality Theorem</em> (in the context of smooth algebraic varieties and algebraic group actions), see </p> <p>S.L. Kleiman, The transversality of a general translate, Compositio Math. 28 (1974), 287–297. </p> <p>When we needed this (actually, a bit more general) result in the category of smooth manifolds and actions we could not find this (folklore) theorem in the literature, so we proved it in Proposition 2.3 of M. Kapovich, B. Leeb and J. Millson, "Convex functions on symmetric spaces, side lengths of polygons and the stability inequalities for weighted configurations at infinity," Journal of Differential Geometry, Vol. 81, 2009, p. 297-354. </p> <p>The result that we proved is slightly more general and deals with several smooth submanifolds of $Y$ and their $G$-translates. </p> | 1,075 | 3,842 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2013-20 | latest | en | 0.780242 |
http://www.jiskha.com/members/profile/posts.cgi?name=Marie&page=8 | 1,416,713,891,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400379063.15/warc/CC-MAIN-20141119123259-00219-ip-10-235-23-156.ec2.internal.warc.gz | 695,947,482 | 11,857 | Saturday
November 22, 2014
# Posts by Marie
Total # Posts: 1,271
math
Determine the regular payment amount, rounded to the nearest dollar. The price of a home is \$160,000. the bank requires a 15% down payment. The buyer is offered two mortgage options: 1 year fixed at 8% or 30-year fixed at 8%. Calculate the amount of interest paid for each ...
August 15, 2011
math
Determine the periodic deposit. round to the nearest dollar. how much of the financial goal comes from deposits and how much comes from interest. Periodic deposit %? at the end of each months Rate 8.25% compounded monthly Time 40 years Financial goal \$1,500,000
August 15, 2011
math
find the interest rounded to the nearest dollar periodic deposit \$60 at the end of each month Rate 5% compounded monthly Time 30 years
August 15, 2011
math
Using the present value formula you deposit \$12,000 in an account that pays 6.5% interest compounded quarterly. A. find the future value after one year? B. Use the future value formula for simple interest to determine the effective annual yield?
August 14, 2011
math
a=16,00, r=11.5%, t=5 years determine the present value, p, you must invest to have the future value, a, at simple interest rate r after time t. round uo to nearest cent
August 14, 2011
math
its really easy... just find the area of all the wall except for the floor and ceiling. and subtract that by 140ft
August 12, 2011
math
886 sp. ft.
August 12, 2011
English
I really need help finding all the grammatical errors in the passage below. Look for commas, question marks, capitailzation and colons, semi-colons, commonly misused words. I did most of them I just need help in finishing. Students also realize that a traditional college ...
August 6, 2011
Statics
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 41 male ...
July 19, 2011
Help!!
A catcher stops 92 mi/h pitch in his glove bring it to rest in 0.15 m. If the force exerted by the catcher is 803N, what is the mass of the ball?
July 17, 2011
Probability
if two fair dice are rolled, what is the probablity that the total showing is even or less than seven?
July 11, 2011
human services
what is integrity; sexual harassment; human differences; and the legal definition of insanity in the courtrooms?
July 5, 2011
Algebra
A chess board has 64 squares. If pieces occupy 19 squares, how many squares are unoccupied?
June 12, 2011
need help for medical coding 1
can anybody tell me if they received an examination booklet of medical coding 1, exam 1, lessons 1-3 and medical coding 1, exam 2, lessons 4-6 from penn foster college. because i didn't received any one of them i only received the final examination booklet. i was wondering...
June 5, 2011
math
The base of a triangle is 6 cm greater than twice the height.The area is 28 cm2.Find the height and the base.
June 2, 2011
2. Individual Assignment: Functions of Management Paper • Prepare a 700- to 1,050-word paper in which you define the four functions of management (planning, organizing, leading, and controlling). In your paper, include an explanation of how each function relates to your ...
May 17, 2011
Algebra 2
Identifying Conics. Put equation in standard form and graph. 9x^2-4y^2-90x+189=0
May 12, 2011
Psychology
nevermind, I got it
May 10, 2011
UOP
Nevermind, I got it
May 10, 2011
UOP
I do not want answers! The question is: Evaluate the effectiveness of self-determination theory in explaining various behaviors. My question is: Exactly what is this asking for and how would I start the response? Is it asking for the behaviors themselves or how the behaviors ...
May 10, 2011
Literature
What are some literary devices used in the novel Journey to the Center of the Earth?
May 10, 2011
Algebra
plug in the equation y=-16^2+47X+3 into a graphing calculator under the Y= button, hit graph, and where ever the graph touches the X axis, is your answer.
May 2, 2011
AP World History
what are some differences in racial ideologies in N. America and Latin America between 1500 and 1830
May 2, 2011
Chemistry
1How would each of the following affect the accuracy of the calculated neutralizing power of antaid (too high, too low, no effect) explain a)The actacid tablet was not cut exactly in half before weighing b)The solution mixture of reacted antacid and excess HCL spattered out of...
April 26, 2011
math
5/12 ____________ x/3+x/2
April 20, 2011
chem
i don't know how to solve this. looked through all my notes but i'm not sure what kind of formula i am to use or if i even need one! please help me to understand the theory behind this: "calculate the work need to make room for the products in the combustion of S8...
April 17, 2011
chemistry
i don't know how to solve this. looked through all my notes but i'm not sure what kind of formula i am to use or if i even need one! please help me to understand the theory behind this: "calculate the work need to make room for the products in the combustion of S8...
April 17, 2011
science
thank you, i tried to google the answer myself, but i couldn't find anything. I hope someone can actually help me to understand it, i don't just want to know the answer, i'd like to know the theory behind it too =)
April 16, 2011
science
i don't know how to solve this. looked through all my note but i'm not sure what kind of formula i am to use or if i even need one! please help me to understand this: "calculate the work need to make room for the products in the combustion of S8(s) to SO2(g) at 1 ...
April 16, 2011
chem
i don't know how to solve this. looked through all my note but i'm not sure what kind of formula i am to use or if i even need one! please help me to understand this: "calculate the work need to make room for the products in the combustion of S8(s) to SO2(g) at 1 ...
April 16, 2011
chem
please help me? :) topic: standard enthalpies of formation "Calculate ΔHf° of octane, C8H18(l), given the entalpy of combustion of octane to CO2(g) and H2O(l) is -5471kJ/mol. The standard enthalpies of formation ofCO2 and H2O are given: CO2(g)ΔHf°=-393....
April 16, 2011
chem
oh! i get it now. q mug + q coffee = 0 after typing many numbers on my calculator i got the final temperature to equal 84.65 degrees C. I'm sure this is right, thank you very much :)
April 16, 2011
chem
topic: thermochemistry i'm not really sure how to go about his question. i'm not sure with formula to use. (perhaps Cspht= q/(m x deltaT)?)can you please help me step by step or give me a hint and i'll post my answer, pretty please? "250g of hot coffee at 95 ...
April 16, 2011
chem- acids and bases
you're the best, thank you so much!
April 16, 2011
chem- acids and bases
correction for typo: "- how do i find the concentration of HNO2, is it 0.20M? " i meant NO2^- instead of HNO2
April 16, 2011
chem- acids and bases
sorry to take up your time again, acids and bases just really confuse me, i'd really appreciate your help again. "What is the pH at the end point for the titration of 0.20M HNO2 by 0.20M NaOH? Ka nitrous acid= 4.5 x 10^-4." this is what i do know: - end point, so...
April 16, 2011
chem
thanks again!
April 16, 2011
chem
what will the pH at the end point of 0.0812M Ba(OH)2 be when titrated with HCl? a) 9.0 b) 8.0 c) 12.2 d) 7.0 e) 6.0 for this question, can i just assume the pH will be 7 since it's s titration with a strong base and a strong acid? (or do i have to do calculations?) I'm...
April 16, 2011
chem
oh! i get it now, thank you very much =)
April 16, 2011
chem
what is the pH change after the addition of 10mL of 1.0M sodium hydroxide to 90mL of a 1.0M NH3/1.0M NH4^+ buffer? Kb ammonia= 1.8 x 10^-5 i found the pH for the original buffer to be 9.255, but i'm not sure how to go about the rest of the question, please, someone help me.
April 16, 2011
math
Mr. Forrester has a swimming pool that measures 3 1/3 yards by 8 yards. If the deck around the pool is 2 2/3 yards wide, what is the outside perimeter of the deck?
April 14, 2011
College Physics
nvm I got this one. but thank you! :]
April 14, 2011
College Physics
Two identical conducting spheres each having a radius of 0.500 cm are connected by a light 1.90 m long conducting wire. A charge of 65.0 µC is placed on one of the conductors. Assume the surface distribution of charge on each sphere is uniform. Determine the tension in ...
April 14, 2011
English
In the story "Oedipus" by Sophocles, do you think there was any way that Oedipus could have avoided his fate? How? In what way?
April 10, 2011
Chemistry
HELP!!! An unknown metal having a mass of 14.00g is dropped into 11.00mL of water at 20.0 degrees Celcius. The starting temperature of the chunk of metal is 100.0C. The final temperature of the water and metal is 32.5C. What is the specific heat of the metal?
April 6, 2011
english
Please help me identify the adjectives and adverbs in this: My experiences with credit and credit card started at age 18. I had several credit cards by the time I was in my mid-twenties. I did not realize that I over extended myself until I lost my job. I had to figure out how...
April 5, 2011
math
At kenneht's school, tickets for a concert were sold at the rate of 40 per half-hour. how many tickets will be sold in 7 hours?
April 3, 2011
Physics
A student on a piano stool rotates freely with an angular speed of 2.95 rad/s . The student holds a 1.35 kg mass in each outstretched arm, 0.739 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.13 kg m^2, a...
March 18, 2011
algebra
a roof rises 1.3 feet vertically and 13.1 feet horizontally. wgat is the grade of the roof?
March 17, 2011
Health Care
Why is it when ever someone ask for help someone thinks they are cheating? If I wanted to cheat I would let soeone else do the work. All I want is a formula or example to follow so I can figure this out for myself!!!!!
March 17, 2011
Access
A database has two tables named Products and Orders. The tables are joined by a column called ProductID. You have created a query named uery1, which combines the data from the tables. If the Products table is deleted, what will happen when users run Query1? A. Only the data of...
March 13, 2011
In the video game Corporate Cowboy, your task is to investigate complaints of wrongdoing on the part of corporate directors and officers, decide whether there is a violation of the law, and deal with the wrongdoers accordingly. Jane, a shareholder of Goodly Corporation, ...
March 8, 2011
english
rosaq was nervous around the large dog whose moods were easily changable
March 8, 2011
english
recommendation
March 8, 2011
statistics
2.Children’s heights are normally distributed with a mean of 100 cm and a standard deviation of 6 cm. What proportion of children’s heights is smaller than 94?
March 7, 2011
statistics
2. Children’s heights are normally distributed with a mean of 100 cm and a standard deviation of 6 cm. What proportion of children’s heights is smaller than 94?
March 7, 2011
science
i need help with chapter 5.1 for a chemstry review.
March 3, 2011
clara baton hs
answer for 182.25+ y squared + 8y
March 3, 2011
Algebra 2
I am completely lost trying to solve this problem. I could really use some help and explanation. x/x+1 + 5/x = 1/x^2+x also how do you write (x-5)(x-5) would it be x^2-5
February 28, 2011
english homework
maybe she has seen the message I'm definitely DEAD What can I do? What can you do? I will be eternally grateful!
February 26, 2011
deleting
can you definitely suppress the post (message)or at least suppress the "my incompetent teacher can't really help"? I forgot that she looks on the net. What can you do for me please, she will tell to my headteacher; So she will take an official warning and my ...
February 26, 2011
Personal Finance
From the throne room in your base station on Alpha Centauri-in the video game Galactic Empire-you dispatch your loyal, obedient minions to use their diligence and skill to loot the universe on your behalf and return with the treasure for its accounting and their compensation. ...
February 17, 2011
calculus
Please help me solve this equation. log[2](3)+log[2](x)=log[2](5)+log[2](x-2) 1. Solve equation by setting up algebraically. 2. Use appropriate properties of Logarithm to reduce the number of logarithmic terms to only one. 3. Solve logarithmic equation.
February 16, 2011
Finance
An investor purchases a 10-year U.S. government bond for \$800. The bond's coupon rate is 10 percent and,? at time of purchase, it still had five years remaining until maturity. If the investor holds the bond until it matures and collects the \$1,000 par value from the ...
February 13, 2011
math
how do I write the slope of a line containing points (-2,0) (-2,8)
February 12, 2011
math
a ceiling in a new house rises 8 feet over a distance of 29 feet. Find the slope of this ceiling. rouns=d to two decimal places
February 12, 2011
algebra
A ceiling in a new home rises 8 feet over a distance of 29 feet. Find the slope of this ceiling. Round to two decimal places.
February 12, 2011
fin 200
As the time period until receipt increases, the present value of an amount at a fixed interest rate A. decreases. B. increases. C. remains the same. D. Not enough information to tell.
February 6, 2011
Cal
The demand for Sportsman 5 X 7 tents is given by the following function where p is measured in dollars and x is measured in units of a thousand. (Round your answers to three decimal places.) p = f(x) = −0.1x^2 − x + 40 (a) Find the average rate of change in the ...
February 4, 2011
Finance
Jud Wheeler signed a contract to purchase 10 acres of land in Idaho from the Krauses. If Jud offered the Krauses considerably less for their property than its possible market value and the Krauses accepted the offer, could they avoid the contract on later learning that they ...
February 3, 2011
com/170
• Write several sentences describing a recent interaction with a friend or family member about personal finance or credit cards. This interaction may be imagined or real. • Use at least five different pronouns, adverbs, and adjectives in your sentences. • ...
February 3, 2011
Finance
Is it 12.2? Thanks alot!
February 2, 2011
Finance
Could someone show me how to calculate average tax rate. What would be the average tax rate for a person who paid taxes of \$4,864.14 on a taxable income of \$39,870? Can someone help please?
February 2, 2011
Finance
Compute taxable income. gross salary 46,660 interest earnings 225.00 dividend income 80.00 one personal exemption 3,400 itemized deductions 7,820 adjustments to income 1,150 What amount would be reported as taxable income?
January 31, 2011
Finance
Need help with computing taxable income. Can someone explain to me how this is done. Gross salary, 46,660 Interest earnings 225 Dividend Income 80
January 31, 2011
com/170
• Write several sentences describing a recent interaction with a friend or family member about personal finance or credit cards. This interaction may be imagined or real. • Use at least five different pronouns, adverbs, and adjectives in your sentences. • ...
January 31, 2011
composition
I need help, do i have at least five adverbs and 5 adjectives, 3 comparisons and three superlatives? Also do i have all adverbs and adjectives labeled? Curtis, my children’s father and I are in the process of finding us a larger (adj.), updated (adj.) apartment better (...
January 31, 2011
composition
Do I have five different pronouns, adjectives, and adverbs? Curtis, my children’s father and I are in the process of moving to a larger (adj.) apartment. Our personal (adj.) finances have been the topic of conversation as of late. There is concern; whether or not; his ...
January 31, 2011
Personal Finance
NEED HELP IMMEDIATELY, HAVE UNTIL 10P.M. CENTRAL TO HAVE ANSWERS?? 1.Determining the Future Value of Education. Jenny Franklin estimates that as a result of completing her master’s degree, she will earn \$6,000 a year more for the next 40 years. a.What would be the total ...
January 29, 2011
Question 1: Which side of this debate is more ethical or supportive? Copyright infringement, the tortuous offense, is never OK on my behalf. But the copyright law recognizes exceptions to the generic standard that a copyright holder controls. The Fair Use provisions;with ...
January 29, 2011
At issue in the twenty-first century is the trade-off between the necessity of writers, musicians, artists, and movie studios to profit from their work and the free flow of ideas for the public benefit. Movie (and music) industry participants claim that encryption programs are...
January 29, 2011
chem
Under what conditions does water boil at exactly 100 degrees Celsius?
January 26, 2011
algebra
January 26, 2011
English
Which word is the correct choice? I was in a real/really hurry and forgot some of the ingredients?
January 25, 2011
dr bob
how much heat would be required to warm Earths ocean by 1.0 degrees C assuming that the volume is 137*10^7 km ^3 and the density of sea water is 1.03 g/cm 3. also assume that the heat capacity of seawater is the same as that of water
January 24, 2011
math
I just need a little help I got this far and got stuck again. 0.10 mm=0.010cm V= (4/3) (0.010cm) ^3= 1.3333x10^-06 M= (1.3333*10^-06cm^3)*(4.0*10^14g/cm ^3) =
January 24, 2011
math
how much heat would be required to warm Earths ocean by 1.0 degrees C assuming that the volume is 137*10^7 km ^3 and the density of sea water is 1.03 g/cm 3. also assume that the heat capacity of seawater is the same as that of water
January 24, 2011
math
calculate the mass in kilograms of a small piece of a neutron star the size of a spherical pebble with a radius of 0.10 mm. 0.10 mm=0.010cm V=(4/3)(0.010cm)^3= M= ( i got this far now I am lost please help
January 24, 2011
chem/math
hang on I dont know why it keeps doing that. how much heat would be required to warm Earths ocean by 1.0 degrees C assuming that the volume is 137*10^7 km ^3 and the density of sea water is 1.03 g/cm 3. also assume that the heat capacity of seawater is the same as that of water
January 24, 2011
chem/math
how much heat would be required to warm Earth¡¯s oceans by 1.0 ¡ãC? Assume that the volume of Earth¡¯s oceans is 137 x ¡¼10¡½^7 ¡¼km¡½^3 and that the density of sea water is 1.03 g/cm3. Also...
January 24, 2011
chem/math
How much heat would be required to warm Earth¡¯s oceans by 1.0 ¡ãC? Assume that the volume of Earth¡¯s oceans is 137 x ¡¼10¡½^7 ¡¼km¡½^3 and that the density of sea water is 1.03 g/cm3. Also...
January 24, 2011
chem
A NASA satellite showed that in 2003 the ozone hole was much larger than it was in 2002. The 2003 hole measured 1.04 x ¡¼10¡½^7 ¡¼mi¡½^2 in diameter, while the diameter of the 2002 hole was 6.9 x ¡¼10¡&...
January 24, 2011
chem
Global warming refers to the rise in average global temperature due to the increased concentration of certain gases, called greenhouse gases, in our atmosphere. Earth¡¯s oceans, because of their high heat capacity, can absorb heat and therefore act to slow down ...
January 24, 2011
Astronomy
According to Newton's second law, when the same force acts on two bodies, the body with the larger mass will have the ________ acceleration.
January 23, 2011
Math
sin2x-cosx (0<=x<=180)
January 23, 2011
Biology (immune system)
How does the immune system help maintain homeostasis? Help me please :(
January 13, 2011
biology /cell analogy
Can you help me find an everyday object that has a similar function as : 1-Golgi Bodies an example 2-Lysosome an example kitchen garberator 3- peroximes example a gym 4-Vesicules/vacuoles an example water bottle 5-Cytoskeleton an example 6-Microvilli an example 7- Cilia an ...
January 12, 2011
english
thanks
January 10, 2011
english
Hello! My name's Marie,I'm 16 I live in Paris and I'm bad in english,(so sorry for my neglish)I must do a homework and I would like that you correct my mistakes, please... It's your language (mercy) :) I used pictures to tell the story of a slave One June day ...
January 10, 2011
Chemistry (acids and bases)
Calculate pH, pOH, [H+], [OH-], and Ka for a 0.026 M solution of HI (a strong acid). Please help me :(
January 7, 2011
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Members | 5,656 | 20,497 | {"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-2014-49 | longest | en | 0.924613 |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-13-probability-mid-chapter-quiz-page-843/18 | 1,716,768,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00815.warc.gz | 684,318,294 | 13,079 | ## Geometry: Common Core (15th Edition)
Order does matter, so we must use a permutation. We know that: $_nP_r = \frac{n!}{(n-r)!}$ $_9P_5 = \frac{9!}{(9-5)!}$ $\frac{9!}{4!}$ We also know that: $x! = x(x-1)(x-2)...(1)$ Thus, we have: 15120 | 96 | 240 | {"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} | 3.8125 | 4 | CC-MAIN-2024-22 | latest | en | 0.809364 |
https://techwhiff.com/learn/1-complete-the-table-below-for-fx3-use-exact/309336 | 1,685,623,605,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00154.warc.gz | 637,395,151 | 12,861 | # 1. Complete the table below for f(x)=3". Use exact values. No work needed. [1.25 points) -...
###### Question:
1. Complete the table below for f(x)=3". Use exact values. No work needed. [1.25 points) - 101 y = f(x) 2. Consider the functions gtx)=(13)" and h(x) =--() +4 12.5,1.25, 1.5 points) a) The points given below (in the first column) belong to g(x)= - Perform two b) Use the point found in part a) to sketch a graph of y=h(x). Include the horizontal asymptote as a dashed line. Approximate point placement where necessary. transformations (and show how the points change) to get points that belong to h(x)=- (-2,4) ►( , )+(,) (-1,2)-( , ) ► , ) (0,1) -( , )+(,) ()-( , )(,) (2,7 ) c) Complete the table. Function Domain (interval notation) Range (interval notation) Equation of horizontal asymptote Parent_and_Piece zip
3. Find the exact value of each expression. Show your process and use proper mathematical notation. It using a variable, make sure your final answer does not contain it. [2 points b. In(e) c. log, (1) d. log, ( 4. Find the equation of the exponential function that passes through the points (-1,10) form y = c.a'. Show all work, not just the final equation. [1.5 points) Use the
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https://robotics.stackexchange.com/questions/12838/how-to-move-a-robot-to-a-point-while-avoiding-obstacles | 1,718,475,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00216.warc.gz | 432,148,570 | 41,884 | # How to move a robot to a point while avoiding obstacles?
I am trying to move a robot in a straight line from point A, to point B. The robot's primary sensor is a Hokuyo URG-04LX-UG01 LIDAR that gives me the magnitude and direction of each point it detects in the form of two arrays.
I also have the wheel encoders on each motor of the robot, so I can obtain some odometry of the robot even though its accuracy will diminish over time.
The problem that I have is that there are obstacles in the path of point A and point B that the robot must go around. I am not sure how to take the readings from the LIDAR, and convert them into movements that go around the obstacle, while also heading to B. Sorry if that doesn't make sense, but I am really confused on how to solve this problem.
Is this an easy problem to solve and I am just over complicating it? Does anyone know of a way to do this? Any help or guidance would be greatly appreciated.
• I would recommend looking up RRT (rapidly exploring random trees). There is likely an open source implementation for your software stack (ROS has many). Commented Jul 19, 2017 at 19:23
First off, I would highly recommend checking ROS (Robot Operating System) out, the ROS community includes researchers and enthusiasts as well as professionals from all around the world and thus most of what you need to solve the problem is already available there. You would have to just implement what others have done, on your system.
I, rather me and my team, have done the exact same thing as you want and the problem becomes much easier if you are using ROS.
There are pre-built SLAM packages available open source which you can use and for making maps and localizing your robot. Once you have the robot's position in a map that also contains all of the obstacles, you just have to do the path planning so that you do not hit any of the obstacles and reach your goal as fast as possible. Then break your path into neared goal points which are simpler to reach so that just by getting to these points your robot can get to the final goal.
As the problem you have put forth is a rather large problem, I feel it is hard to explain it completely here, but I will try to do as much as I can (When I get time I will add more links to help).
Let us break this problem into several parts,
(Don't worry about some occasional ROS jargon, you'll catch them during the tutorials).
1. If you are unaware of ROS, I suggest taking a look at it and here is a place where you can learn it hands-on.
2. Next you will have to get your sensors working with ROS, this should be easy as for most of them you will have packages already made and open source.
3. Once all your sensors are working with ROS, you will have topics which contain have current sensor data, now you can do anything with them, i would use filters on the data and fuse to find your robots estimated pose. (You should also look into Kalman and Extended Kalman Filters at this point.)
4. Now is the time when you can use some SLAM (Simultaneous Localization and Mapping) algorithms to have a map with obstacles and your robot at all times.
5. Do the Motion Planning, break the path into smaller pieces and feed them to your robot in the format it wants (You will need some knowledge about Control Systems at this point).
From my experience you can use these packages : gmapping, robot-poe-ekf, moveit. (Google these by adding ros)
I hope this helps, and all the best man.
What you are talking about is Robotic Path Planning. | 781 | 3,526 | {"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} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.972547 |
http://neilsculthorpe.com/icfp09/BoolOrd1.html | 1,696,189,899,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00527.warc.gz | 28,931,413 | 7,864 | ```open import NeilPrelude1
open import BoolProps1
open import Ord1
module BoolOrd1 where
infix 3 _<₂_ -- _>₂_
≤false : {b : Bool} → false ≤ b
≤false {false} = refl
≤false {true} = f≤t
≤true : {b : Bool} → b ≤ true
≤true {false} = f≤t
≤true {true} = refl
≤antisym : {a b : Bool} → a ≤ b → b ≤ a → a ≡ b
≤antisym refl _ = refl
≤antisym f≤t ()
≤trans : {a b c : Bool} → a ≤ b → b ≤ c → a ≤ c
≤trans refl lt2 = lt2
≤trans f≤t lt2 = ≤false
≤total : {a b : Bool} → Either (a ≤ b) (b ≤ a)
≤total {false} = left ≤false
≤total {true} = right ≤true
{-
import TotalOrder
open module ℕTO = TotalOrder _≤_ ≤antisym ≤trans ≤total public
-}
----------------------------------------------------------------
data _<₂_ : Bool → Bool → Set where
f<t : false <₂ true
<trans : {a b c : Bool} → a <₂ b → b <₂ c → a <₂ c
<trans f<t ()
<trich : {a b : Bool} → Either (a ≡ b × Not (a <₂ b) × Not (b <₂ a))
(Either (Not (a ≡ b) × a <₂ b × Not (b <₂ a))
(Not (a ≡ b) × Not (a <₂ b) × b <₂ a))
<trich {false} {false} = left (refl & fork (\ ()))
<trich {false} {true} = right (left ((\ ()) & f<t & \ ()))
<trich {true} {false} = right (right ( (\ ()) & (\ ()) & f<t))
<trich {true} {true} = left (refl & fork (\ ()))
{-
import StrictTotalOrder
open module ℕSPO = StrictTotalOrder _<₂_ <trans <trich public
_>₂_ : Bool → Bool → Set
_>₂_ = _>'_
-}
-----------------------------------------------------------------
∨max : {a b : Bool} → a ≤ a ∨ b
∨max {false} = ≤false
∨max {true} = refl
∨maxR : {b : Bool} → (a : Bool) → b ≤ a ∨ b
∨maxR false = refl
∨maxR true = ≤true
∧min : {a b : Bool} → a ∧ b ≤ a
∧min {false} = refl
∧min {true} = ≤true
∧minR : {b : Bool} → (a : Bool) → a ∧ b ≤ b
∧minR true = refl
∧minR false = ≤false
-----------------------------------
≤resp∧ : {a b c d : Bool} → a ≤ c → b ≤ d → (a ∧ b) ≤ (c ∧ d)
≤resp∧ refl refl = refl
≤resp∧ {a} refl f≤t = ≤trans (∧minR a) ≤false
≤resp∧ f≤t refl = ≤false
≤resp∧ f≤t f≤t = f≤t
≤resp∧L : {a b c : Bool} → a ≤ c → (a ∧ b) ≤ (c ∧ b)
≤resp∧L = flip ≤resp∧ refl
≤resp∧R : {b c : Bool} → (a : Bool) → b ≤ c → (a ∧ b) ≤ (a ∧ c)
≤resp∧R a = ≤resp∧ {a} refl
{-
≤resp∨ : {a b c d : Bool} → a ≤ c → b ≤ d → (a ∨ b) ≤ (c ∨ d)
≤resp∨ refl refl = refl
≤resp∨ .{_} .{_} {c} refl f≤t = {!!} -- ≤substL ? (∨max {c})
≤resp∨ f≤t refl = ≤true
≤resp∨ f≤t f≤t = f≤t
≤resp∨L : {a b c : Bool} → a ≤ c → (a ∨ b) ≤ (c ∨ b)
≤resp∨L = flip ≤resp∨ refl
≤resp∨R : {b c : Bool} → (a : Bool) → b ≤ c → (a ∨ b) ≤ (a ∨ c)
≤resp∨R a = ≤resp∨ {a} refl
-}
≤false≡min : {b : Bool} → b ≤ false → b ≡ false
≤false≡min refl = refl
≥max≡max : {b : Bool} → b ≥ true → b ≡ true
≥max≡max refl = refl
≤∧L : {a b c : Bool} → a ≤ c → a ∧ b ≤ c
≤∧L = ≤trans ∧min
≤∧R : {b c : Bool} → (a : Bool) → b ≤ c → a ∧ b ≤ c
≤∧R a = ≤trans (∧minR a)
≤∨L : {a b c : Bool} → a ≤ b → a ≤ b ∨ c
≤∨L = flip ≤trans ∨max
≤∨R : {a c : Bool} → (b : Bool) → a ≤ c → a ≤ b ∨ c
≤∨R b = flip ≤trans (∨maxR b)
``` | 1,216 | 2,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-40 | latest | en | 0.262322 |
https://edurev.in/t/183804/Physics-CBSE-Sample-Question-Paper--2020-21--1 | 1,718,877,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00226.warc.gz | 196,300,936 | 81,154 | Physics: CBSE Sample Question Paper (2020-21)- 1
# Physics: CBSE Sample Question Paper (2020-21)- 1 | Physics Class 12 - NEET PDF Download
Class-XII
Physics Theory
TIME: 3 Hrs.
M.M: 70
### General Instructions:
(1) All questions are compulsory. There are 33 questions in all.
(2) This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
(3) Section A contains ten very short answer questions and four assertion reasoning MCQs of 1 mark each, Section B has two case-based questions of 4 marks each, Section C contains nine short answer questions of 2 marks each, Section D contains five short answer questions of 3 marks each and Section E contains three long answer questions of 5 marks each.
(4) There is no overall choice. However, an internal choice is provided. You have to attempt only one of the choices in such questions.
### Section A
All questions are compulsory. In the case of internal choices, attempt any one of them.
Q.1. The figure shows electric field lines in which an electric dipole P is placed as shown. In which direction, the dipole will experience a force? (1 Mark)
Ans: We know the electric field emerges radially outward from the positive point charge. In the figure given above, space between field lines increases (or the density of the electric field line is decreasing). In other words, the electric force is decreasing while moving from left to right. Thus, the force on charge – q is greater than the force on charge + q in turn, a dipole will experience a force towards the left direction.
Q.2. A capacitor of 4 μF is connected, as shown in the figure. The internal resistance of the battery is 0.5 Ω.
Find the amount of charge on the capacitor plates. (1 Mark)
Ans: As the capacitor offers infinite resistance for the DC circuit. The cells current will not flow across the branch of 4 mF, and 10 W. So current will flow across 2-ohm branches. So current flows through across 2 W resistance from left to right is,
So Potential Difference (PD) across 2 W resistance V = RI = 2 × 1 = 2 Volt As battery,
capacitor and 2 branches are in parallel. So PD will remain the same across all three
branches. As current does not flow through the capacitor branch, so no potential drop will be across 10 Ω. So PD across 4 μF capacitor = 2 Volt
Q = CV = 4 μF× 2 V = 8 μC.
OR
What is the value of the phase difference between two points on the same wavefront? (1 Mark)
Ans: Zero
Detailed answer: The phase difference between two points on a wavefront is zero. The phase difference is defined as the difference in the phase angle between the two waves.
Q.3. The output of a step-down transformer is measured to be 24 V when connected to a 12 W Light bulb. Find the value of the peak current. (1 Mark)
Ans: Given, Power associated with secondary, Ps = 12 W Secondary voltage, Vs = 24 V Current in the secondary, I= = 0.5 A Peak value of the current in the secondary,
A.
Q.4. Two charged particles traverse identical helical paths in an opposite sense in a uniform magnetic field . What will be their charge to the mass ratio? (1 Mark)
Ans: When the charge/mass ratio of these two particles are the same and charges on them are of opposite nature, then the charged particles will traverse identical helical paths in a opposite sense.
OR
Name the phenomenon which shows the quantum nature of electromagnetic radiation. (1 Mark)
Ans: Photoelectric effect and the Compton effect.
Q.5. If the number of turns per unit length of a coil of a solenoid is doubled, how will its self-induction change? (1 Mark)
Ans: The self-inductance of coil or solenoid is given by where μ0 is the permeability of free space, n is the number of turns per unit length of a coil, A is the cross-sectional area of the coil, and l is the length of the coil.
Q.6. What do you mean by the sensitivity of the meter bridge? (1 Mark)
Ans: The meter bridge is more sensitive when all the resistances are in the same order or their ratio is unity.
Q.7. A ray of light incident at an angle q on a refracting face of a prism emerges
from the other face normally. If the angle of the prism is 5° and the prism is made of
a material of refractive index 1.5, what is the angle of incidence? (1 Mark)
Ans: Given that,
A = 5°
μ = 1.5
i= 0°
r2 = 0°
As we know,
Since,
From Snell’s law :
∴ i1 angle of incidence = 7.5°.
OR
What is the maximum number of spectral lines emitted by a hydrogen atom in the third excited state? (1 Mark)
Ans: If n is the quantum number of the highest energy level, then the total number of possible spectral lines emitted is
Here, third excited state means fourth energy level, i.e., n = 4 ∴ N = 4
Q.8. A cylindrical bar magnet is rotated about its axis in the figure. A wire is connected from the axis and is made to touch the cylindrical surface through contact. Then, find the amount of current flowing through the ammeter. (1 Mark)
Ans: The phenomenon of electromagnetic induction is used in this problem. Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes (or a moving conductor cuts the magnetic flux), an emf is produced in the circuit (or emf induces across the ends of the conductor), is called induced emf. The induced emf persists only as long as there is a change or cutting of flux. When a cylindrical bar magnet is rotated about its axis, no change in flux is linked with the circuit. Consequently, no emf induces, and hence, no current flows through the ammeter A. Hence the ammeter shows no deflection.
OR
Find the ratio of de-Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V. (1 Mark)
Ans: V1 = 25 V, V2 = 36 V
the de-Broglie wavelength of an electron :
Q.9. Name the phenomenon which shows the quantum nature of electromagnetic
Ans: Photoelectric Effect (Raman Effect/Compton Effect)
Q.10. Name the minority charge carriers in n-type silicon. (1 Mark)
Ans: Holes are the minority carrier in n-type Silicon.
For question numbers 11, 12, 13 and 14, two statements are given-one labelled
Assertion (A) and the other labelled Reason (R). Select the correct answer to these
questions from the codes (a), (b), (c), and (d) as given below:
Q.11. Assertion (A): It is impossible to use 35Cl for fusion.
Reason (R): Binding energy of 35Cl is minimal. (1 mark)
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is NOT the correct explanation of A
(c) A is true, but R is false
(d) A is false, and R is also false
Ans: Correct option is (c)
Explanation: Only lighter atoms are used for fusion. So, 35Cl cannot be used for Fusion. So the assertion is true.
The binding energy of Chlorine binding energy is large. So the reason is false.
Q.12. Assertion(A): If an electron is not deflected when moving through a certain region of space, the only possibility is that no magnetic field is present.
Reason (R): Force on the electron is directly proportional to the strength of the magnetic field (1 mark)
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is NOT the correct explanation of A
(c) A is true, but R is false
(d) A is false, and R is also false
Ans: Correct option is (a)
Explanation: In the absence of a magnetic field, moving electron will not be deflected. This possibility is true.
So, the assertion is true.
So, the electron's force is directly proportional to the magnetic field's strength. So, the reason is true. Reason properly explain the assertion.
Q.13. Assertion (A): When the magnetic flux changes around a metallic conductor, the eddy current is produced.
Reason (R): Electric potential determines the flow of charge. (1 mark)
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is NOT the correct explanation of A
(c) A is true, but R is false
(d) A is false, and R is also false
Ans: Correct option is (b)
Explanation: Change in flux induces emf in the conductor, which generates eddy current. So the assertion is true.
Electric potential determines the flow of charge. So, the reason is also true.
But the reason is not the proper explanation of the generation of eddy current.
Q.14. Assertion (A): Magnetic poles cannot be separated by breaking a bar magnet into two pieces.
Reason (R): When a magnet is broken into two pieces, the magnetic moment will be reduced to half. (1 mark)
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is NOT the correct explanation of A
(c) A is true, but R is false
(d) A is false, and R is also false
Ans: Correct option is (b)
Explanation: Magnetic poles always exist in pairs, even at the atomic level. So the assertion is true.
When a magnet is broken into two pieces, the pole strength remains the same; only the length becomes half. So, the magnetic moment becomes half. So, the reason is also true.
But R is not the proper explanation of A.
### Section B
Questions 15 and 16 are Case Study based questions and are compulsory. Attempt any 4 subparts from each question. Each question carries 1 mark.
Q.15. Faster and smaller: The future of computer technology The Integrated Chip (IC) is at the heart of all computer systems. In fact, ICs are found in almost all electrical devices like cars, televisions, CD players, cell phones etc. The miniaturisation that made the modern personal computer possible could never have happened without the IC. IC's are electronic devices that contain many transistors, resistors, capacitors, connecting wires – all in one package. You must have heard of the microprocessor. The microprocessor is an IC that processes all information in a computer, like keeping track of what keys are pressed, running programmes, games etc. The IC was first invented by Jack Kilby at Texas Instruments in 1958, and he was awarded Nobel Prize for this in 2000. IC's are produced on a piece of semiconductor crystal (or chip) by a photolithography process. Thus, the entire Information Technology (IT) industry hinges on semiconductors. Over the years, the complexity of ICs has increased while the size of its features continued to shrink. In the past five decades, dramatic miniaturisation in computer technology has made modern-day computers faster and smaller. In the 1970s, Gordon Moore, co-founder of INTEL, pointed out that the memory capacity of a chip (IC) approximately doubled every one and a half years. This is popularly known as Moore’s law. The number of transistors per chip has risen exponentially, and each year, computers are becoming more powerful yet cheaper than the year before. It is intimated from current trends that the computers available in 2020 will operate at 40 GHz (40,000 MHz) and would be much smaller, more efficient and less expensive than present-day computers. A famous quote from Gordon Moore best expresses the explosive growth in the semiconductor industry and computer technology: “If the auto industry advanced as rapidly as the semiconductor industry, a Rolls Royce would get half a million miles per gallon, and it would be cheaper to throw it away than to park it”.
1. Full form of IC is: (1 mark)
(a) Indigenous circuit
(b) Improved chip
(c) Isolated circuit
(d) Integrated chip
Ans: Correct option is (d)
Explanation: The full form of IC is an integrated chip or Integrated Circuit.
2. IC was first invented by (1 mark)
(a) Isaac Newton
(b) W. H. Schottky
(c) Charles Babbage
(d) Jack Kilky
Ans: Correct option is (d)
Explanation: The IC was first invented by Jack Kilby at Texas Instruments in 1958, and he was awarded Nobel Prize for this in 2000.
3. Moor’s Law states that: (1 mark)
(a)
The memory capacity of a chip (IC) approximately doubles every one and a half years
(b)
The packing density doubles every year.
(c)
The memory capacity of a chip (IC) approximately doubles every two and a half years
(d)
Operational frequency of computer doubles every one and a half years.
Ans: Correct option is (a)
Explanation: In the past five decades, dramatic miniaturisation in computer technology has made modern-day computers faster and smaller. In the 1970s, Gordon Moore, co-founder of INTEL, pointed out that a chip (IC) memory capacity approximately doubled every one and a half years. This is popularly known as Moore’s law.
4. Which statement is correct? (1 mark)
(a) IC's contain many transistors, resistors, capacitors, connecting wires.
(b) IC's contain many transistors, resistors, capacitors, inductors, connecting wires
(c) IC's contain many transistors, capacitors, inductors, connecting wires.
(d) IC's contain many transistors, resistors, capacitors, crystal oscillator, connecting wires.
Ans: Correct option is (a)
Explanation: IC's are electronic devices that contain many transistors, resistors, capacitors, connecting wires – all in one package. It does not contain any inductor or crystal oscillator.
5. IC's are produced on a piece of semiconductor crystal by a process called (1 mark)
(a)
X-ray imaging
(b)
Ultrasonography
(c)
Magnetic resonance imaging
(d)
photolithography
Ans: Correct option is (d)
Explanation: IC's are produced on a piece of semiconductor crystal (or chip) by a photolithography process.
Q.16. Photocell: A photocell is a technological application of the photoelectric effect. It is a device whose electrical properties are affected by light. It is also sometimes called an electric eye. A photocell consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb. It is connected to the external circuit having a high-tension battery B and microammeter (μA) as shown in the Figure.
Sometimes, instead of plate C, a thin layer of photosensitive material is pasted on the inside of the bulb. A part of the bulb is left clean for the light to enter. When the light of suitable wavelength falls on the emitter C, photoelectrons are emitted. These photoelectrons are drawn to the collector A. Photocurrent of a few microampere order can be normally obtained from a photocell. A photocell converts a change in intensity of illumination into a photocurrent change. This current can be used to operate control systems and light measuring devices.
1. Photocell is also known as (1 mark)
(a) Electric sense
(b) Electric eye
(c) Photo emitter
(d) Photo transducer
Ans: Correct option is (b)
Explanation: A photocell is a technological application of the photoelectric effect. It is a device whose electrical properties are affected by light. It is also sometimes called an electric eye.
2. A photocell consists of (1 mark)
(a) a semi-cylindrical photo-sensitive metal plate called emitter and a wire loop called collector
(b) A metal cylinder called emitter and a filament called the collector.
(c) Two semi-cylindrical photo-sensitive metal plates – one is called emitter, and the other is called collector
(d) A wire mesh called emitter and a photosensitive wire loop called collector
Ans: Correct option is (a)
Explanation: A photocell consists of a semi-cylindrical photosensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb.
3. Which of the following statements is true? (1 mark)
(a) The photocell is totally painted black
(b) A part of the photocell is left clean
(c) The photocell is completely transparent.
(d) A part of the photocell is made black
Ans: Correct option is (b)
Explanation: A part of the bulb is left clean for the light to enter.
4. The photocurrent generated is in the order of (1 mark)
(a) Ampere
(b) Milliampere
(c) Microampere
(d) None of the above
Ans: Correct option is (c)
Explanation: Photocurrent of the order of a few microamperes can be normally obtained from a photocell.
5. A photocell converts a change in ________ of incident light into a change in ___________. (1 mark)
(a) Intensity, photo-voltage
(b) Wavelength, photo-voltage
(c) Frequency, photo-current
(d) Intensity, photo-current
Ans: Correct option is (d)
Explanation: A photocell converts a change in intensity of illumination into a photocurrent change.
### Section C
All questions are compulsory. In the case of internal choices, attempt any one.
Q.17. State the underlying principle of a transformer. How is the large scale
transmission of electric energy over long distances done using
transformers? (2 Mark)
Ans: A transformer is based on the principle of mutual induction, which states that due to
continuous change in the current in the primary coil, an emf gets induced across the
secondary coil.
Electric power generated at the power station is stepped-up to very high voltages using a step-up transformer and transmitted to a distant place. At receiving end, it is
stepped down by a step-down transformer.
Q.18. Given a uniform electric field N/C. Find the flux of this field through a square of a side 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis? (2 Mark)
Ans: Given : N/C along (+) positive direction of x-axis.
Surface area, A = 10 cm × 10 cm
= 0.10 m × 0.10 m
= 10–2 m2
(i) In case of plane parallel to y-z plane, normal to plane will be along x-axis, so ϕ = 0°
Electric flux will be calculated using ϕ =cos θ
5 × 103 × 10–2 × cos 0°
= 50 Nm2 /C
(ii) Since the plane is making an angle of 30° with the x-axis, so normal to its plane will make 60° with the x-axis, so θ = 60°
Now finding Electric flux again with ϕ = cos θ
= 5 × 103 × 10–2 × cos 60°
= 25 Nm2 /C
OR
Why do we prefer carbon brushes to copper in an ac generator? (2 Mark)
Ans: The carbon brushes used in the generator are corrosion-free. On small
expansion on heating, it maintains the proper contact as well.
Q.19. If magnetic monopoles existed, how would the Gauss law of magnetism be modified? (2 Mark)
Ans: Gauss law of magnetism describes that divergence of the magnetic field will be zero while divergence of the electric field is not zero, which shows the non-existence of magnetic monopole. As per Gauss law of magnetism,
If monopole exists, then the right side will be equal to the monopole multiplied by μ0.
Detailed Answer: According to the Gauss law of magnetism,
(Integral form)
(Differential form)
If magnetic monopoles exist, then Gauss's law for magnetism would be modified as :
(Integral form)
(Differential form)
Where, ρm = magnetic charge density μ0 = permeability of free space.
OR
Calculate the amount of work done to dissociate a system of three charges 1 μC, 1 μC and – 4 μC placed (2 Mark)
Ans: Given :
Q.20. The battery remains connected to a parallel plate capacitor, and a dielectric slab is inserted between the plates. What will be the effect on its
(i) potential difference
(ii) capacity
(iii) electric field and
(iv) energy stored (2 Mark)
Ans: When a battery remains connected,
(i) the potential difference V remains constant
(ii) capacity C increases
(iii)the electric field will remain the same
(iv) energy stored 1/2 CV2 increases as C increases
Detailed Answer: When a battery remains connected to a parallel plate capacitor and if a dielectric slab is inserted between the capacitor plates, then
(i) there will be no change in the potential difference as the capacitor remained connected with the battery.
(ii) capacity or capacitance will increase since with the introduction of the dielectric slab, the capacitor's capacitance will result in C = where K > 1, increasing C.
(iii)Electric field will remain the same as there will be no change in potential difference and distance between the plates.
(iv) Energy stored will be increased since from the expression U = 1/2 CV2, potential
difference V remains the same while C increases, which finally increases the capacitor's energy.
Q.21. The diagram below shows a potentiometer set up. The galvanometer pointer deflects to the left on touching the jockey near the end X of the potentiometer wire. On touching the jockey near to end Y of the potentiometer, the galvanometer pointer again deflects to the left but now by a larger amount. Identify the fault in the circuit and explain how it leads to such a one-sided deflection using appropriate equations or otherwise. (2 Mark)
Ans: The positive of E1 is not connected to terminal X.
So, VG (or deflection) will be maximum when l is maximum, i.e., when a jockey is touched
near and Y. Also, VG (or deflection) will be minimum when l is minimum, i.e., when a jockey is touched near end X.
Q.22. Define the distance of the closest approach. An α-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is ‘r’. What will be the distance of the closest approach for an α-particle of double the kinetic energy? (2 Mark)
Ans: It is the distance of the charged particle from the centre of the nucleus, at which the whole of the initial kinetic energy of the (far off) charged particle gets converted into the electric potential energy of the system. 1 Distance of closest approach (rc) is given by rc
'K' is doubled, ∴ rbecomes r/2.
[Alternatively: If a candidate writes directly r/2 without mentioning formula, award the 1 mark for this part.]
Detailed Answer: When an α-particle is bombarded towards the nucleus, it is repelled by electrostatic repulsion. As a result, its kinetic energy is converted into electrostatic potential energy. At a certain distance (rC) between the ∝particle and nucleus, the moving particle loses all its kinetic energy and becomes stationary momentarily. This distance is known as the distance of the closest approach. In this process, the particle's total kinetic energy is converted into potential energy.
Kinetic energy = K =
Let rC’ be the new distance of closest approach when kinetic energy becomes 2K
Q.23. Write two important limitations of Rutherford's nuclear model of the atom. (2 Mark)
Ans: (i) According to Rutherford's model, an electron orbiting around the nucleus continuously radiates energy due to the acceleration; hence the atom will not remain stable.
(ii) As electron spirals inwards, its angular velocity and frequency change continuously; therefore, it will emit a continuous spectrum.
Q.24. Calculate the curvature radius of an equi-concave lens of refractive index 1.5, when kept in a medium of refractive index 1.4, to have a power of –5D. (2 Mark)
Ans: Calculation of focal length
Lens maker's formula
Detailed Answer: Given the refractive index of the biconcave lens = 1.5
Power of biconcave lens = – 5D
Refractive index of medium= 1.4.
OR
A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with an angular speed of 50 rad s–1 in a uniform magnetic field of magnitude 3.0 × 10–2 T. Calculate the maximum value of the current in the coil. (2 Mark)
Ans: Maximum value of emf
= 600 mV
Maximum induced current,
Q.25. A rectangular coil of sides ‘l’ and ‘b’ carrying a current I is subjected to a uniform magnetic field, acting perpendicular to its plane. Obtain the expression for the torque acting on it. (2 Mark)
Ans: Equivalent magnetic moment of the coil,
### Section D
All questions are compulsory. In the case of internal choices, attempt anyone.
Q.26. The figure shows a metallic rod PQ of length l, resting on the smooth horizontal rails AB positioned between a permanent magnet's poles. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Assume the magnetic field to be uniform. Given the resistance of the closed-loop containing the rod is R.
(i) Suppose K is open and the rod is moved with a speed v in the direction shown. Find the polarity and magnitude of induced emf.
(ii) With K open and the rod moving uniformly, there is no net force on the rod PQ electrons even though they do experience a magnetic force due to the rod's motion. Explain.
(iii) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of perpendicular? (3 Mark)
Ans: (i) |ε| = Bvl
P is positive end Q is the negative end
(ii) Magnetic force gets cancelled by electric force that generates an extra charge of opposite sign at rod ends.
(iii)Induced emf is zero as the motion of rod not cutting field lines
Q.27. Define electric flux and write its SI unit. The electric field components in the figure shown are : Ex = αx, Ey = 0, Ez = 0 where α = 100 N/C m
Calculate the charge within the cube, assuming a = 0.1 m. (3 Mark)
Ans: Definition of Electric flux
SI unit
Formula (Gauss's Law)
Calculation of Charge within the cube
Electric Flux is the dot product of the electric field and area vector.
also, accept
SI Unit: Nm2 /C or volt-meter
For a given case
OR
(i) Monochromatic light of wavelength 589 nm is incident from the air on a water surface. If μ for water is 1·33, find the refracted light's wavelength, frequency, and speed.
(ii) A double convex lens is made of a glass of refractive index 1·55, with both faces of the same radius of curvature. Find the radius of curvature required if the focal
length is 20 cm. (3 Mark)
Ans:
Therfore, R = (20 × 1.10) cm = – 22 cm
Q.28. A toroidal solenoid of mean radius 20 cm has 4000 turns of wire wound on a ferromagnetic core of relative permeability 800. Calculate the magnetic field in the core for a current of 3 A passing through the coil. How does the field change when a core of Bismuth replaces this core? (3 Mark)
Ans: Formula for the magnetic field of a toroid
Calculation of magnetic field
Effect of change of core
= 9.6 T
Since Bismuth is diamagnetic, its μr <
Therefore, the magnetic field in the core will get very much reduced.
Given :
Mean radius of toroidal solenoid = 20 cm
Number of turns of wire wound = 4000
Relative permeability of ferromagnetic core = 800
Current passing through the coil = 3 A
The magnetic field in a toroid coil :
As Bismuth is a diamagnetic substance with a relative permeability of less than 1, it will tend to move away from the stronger to the weak part of the external magnetic field, making the core field less than the empty core field.
OR
(i) How are electromagnetic waves produced? Explain.
(ii) A plane electromagnetic wave travels through a medium along the +ve z-direction. Depict the electromagnetic wave showing the oscillating electric and magnetic fields' directions. (3 Mark)
Ans: (i) Production of EM wave: Electromagnetic waves consist of both electric and magnetic fields travelling through space with the speed of light c. These waves oscillate in perpendicular planes with respect to each other and are in phase. An electromagnetic wave can be created by accelerating charges, moving charges back and forth, producing oscillating electric and magnetic fields. When the accelerating charged particle moves with acceleration, magnetic and electric fields change continuously, leading to the production of electromagnetic waves.
(ii)
Q.29. What is relaxation time? Derive an expression for resistivity of a wire in terms
of the number density of free electrons and relaxation time. (3 Mark)
Ans: Definition and Derivation.
Detailed answer: (i) Relaxation time shows the effect of collisions among the electrons
and ions or impurities on electrical conduction in a metal. The time is taken for the drift velocity to decay 1/e of its initial value. As drift velocity increases, relaxation time
decreases since the electrons move the distance they frequently collide faster.
(ii) When a potential difference V is applied across a conductor of length l, then drift speed of electron will result as :
The electric current through the conductor and drift speed are linked as I = neAvd
where,
n = number density of electrons
e = electronic charge
A = area of cross-section
vd = electron drift speed
Q.30. (i) Three-point charges q, – 4q and 2q are placed at the vertices of a 3 equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.
(ii) Find out the amount of the work done to separate the charges at infinite distance. (3 Mark)
Ans: (i) Finding the magnitude of the resultant force on charge q
Force on charge q due to the charge – 4q
The forces F1 and F2 are inclined to each other at an angle of 120°
Hence, resultant electric force on charge q
(ii) Finding the work done
Net P.E. of the system
### Section E
All questions are compulsory. In the case of internal choices, attempt any one.
Q.31. (a) Why are photodiodes preferably operated under reverse bias when the current in the forward bias is known to be more than that in reverse bias?
(b) The two optoelectronic devices: Photodiode and solar cell, have the same working principle but differ in terms of their operation process. Explain the difference between the two devices in terms of :
(i) biasing
(ii) junction area
(iii) I-V characteristics. (5 Mark)
Ans: (a) The fractional change in majority charge carriers is very less compared to the
the fractional change in minority charge carriers on illumination.
(b) The difference in the working of two devices :
Photodiode Solar cell (i) Biasing Used in reverse biasing No external biasing is given (ii) Junction Area Small Large for solar radiation to be incident on it.
(iii)I-V characteristics
OR
(i) Using Bohr’s postulates, derive the expression for the electron's total energy in the stationary states of the hydrogen atom.
(ii) Using Rydberg's formula, calculate the wavelengths of the spectral lines of the first member of the Lyman series and Balmer series. (5 Mark)
Ans:
(ii) Rydberg's formula: For the first member of the Lyman series,
Q.32. (i) State Faraday’s laws of electromagnetic induction.
(ii) The magnetic field through a circular loop of wire 12 cm in radius and 8.5 W resistance changes with time, as shown in the figure. The magnetic field is perpendicular to the plane of the loop. Calculate the induced current in the loop and plot it as a function of time.
(iii) Show that Lenz’s law is a consequence of conservation of energy (5 Mark)
Ans: (i) Faraday’s Laws of Electromagnetic Induction:
Faraday’s First Law of Electromagnetic Induction states that whenever a conductor is placed in a varying magnetic field, emf is induced, which is known as induced emf. If the conductor circuit is closed, the current is also induced, called induced current. Faraday’s Second Law of Electromagnetic Induction states that the induced emf is equal to the rate of change of flux linkage where flux linkage is the product of the number of turns in flux associated with the coil.
eB is the magnetic flux through the circuit as
With N loops of similar area in a circuit and ϕB being the flux through a loop, then emf is induced in every loop, making Faraday law as
where, e = Induced emf [V],
N = number of turns in the coil
∆ϕ= change in the magnetic flux [Wb],
= change in time [s]
The negative sign means that e opposes its cause.
(ii)
(iii) Similarly :
Lenz's Law: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
Explanation :
When the north pole of a bar magnet is pushed towards the close coil, the magnetic flux through the coil increases and the current is induced in the coil in such a direction that it opposes the increase in flux. This is possible when the coil's induced current is in the anti-clockwise direction. The opposite will happen when the north pole is moved away from the coil. In either case, it is the work done against the force of magnetic repulsion/attraction that gets ‘converted‘ into the induced emf. So Lenz's law is a consequence of the conservation of energy.
OR
(a) Write the expression for the equivalent magnetic moment of a planar current loop of area A, having N turns and carrying a current i. Use the expression to find the magnetic dipole moment of a revolving electron.
(b) A circular loop of radius r, having N turns and carrying current I, is kept in the XY plane. It is then subjected to a uniform magnetic field to Obtain an expression for the magnetic potential energy of the coil-magnetic field system. (5 Mark)
Ans: (a) The equivalent magnetic moment is given by μ = NiA
The direction of m is perpendicular to the plane of the current-carrying loop. It is directed along the direction of advance of a right-handed screw rotated along the current flow direction.
Derivation of expression for μ of an electron revolving around a nucleus
For 1 revolution of electron, N = 1
Current =
Putting in the expression,
μ= NiA
(b) for the loop,
Magnetic potential energy =
Q.33. (i) Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm’s law.
(ii) A wire whose cross-sectional area is increasing linearly from one end to another is connected across a V volts battery. Which of the following quantities remain constant in the wire?
(a)
drift speed
(b) current density
(c) electric current
(d) electric field
Ans: (i) Derivation of the expression for drift velocity
Deduction of Ohm's law
Let an electric field E be applied to the conductor. The acceleration of each electron is
Velocity gained by the electron
Let the conductor contain n electrons per unit volume. The average value of time 't', between their successive collisions, is the relaxation time, 't'.
Hence average drift velocity, vd =
The amount of charge, crossing area A, in time Δt, is = neAvdΔt = IΔt
Substituting the value of vd, we get
But I = JA, where J is the current density,
This is Ohm's law
[Note: Credit should be given if the student derives the alternative form of Ohm's law by substituting
(ii) Name of quantity and justification
(b) Current density will remain constant in the wire.
All other quantities depend on the cross-sectional area of the wire.
Detailed Answer: Out of these, current density remains constant in a wire whose cross-sectional area increases linearly from its one end to other as current density is :
It is a current per unit area that depends on the area of cross-section.
Drift speed is given as :
OR
A capacitor of capacitance C1 is charged to a potential V1 while another capacitance C2 is charged to a potential difference V2. The capacitors are now disconnected from their respective charging batteries and connected in parallel to each other
(i) Find the total energy stored in the two capacitors before they are connected.
(ii) Find the total energy stored in the parallel combination of the two capacitors.
(iii) Explain the reason for the difference of energy in parallel combination compared to the total energy before they are connected. (5 Mark)
Ans: (i) Finding the total energy before the capacitors are connected
(ii) Finding the total energy in the parallel combination
(iii)Reason for difference
(i) We have Energy stored in a capacitor = 1/2 CV2
Therefore, Energy stored in the charged capacitors
Therefore, the total energy stored =
(ii) Let V be the potential difference across the parallel combination. Equivalent capacitance = (C+ C2)
Since charge is a conserved quantity, we have (C1 + C2)V = C1V1 + C2V2
Therefore, the total energy stored in the parallel combination
(iii) The total energy of the parallel combination is different (less) from the total energy before the capacitors are connected. This is because some energy gets used up due to the movement of charges.
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- finding points determined parametrically
- Eliminating the Parameter
- Using a graphing calculator in parametric mode
- Horizontal line test
- One-to-One functions
- Finding the Inverse of a Function
- Finding the domain of the inverse of a function
- Graphing the Inverse of a Function
- find the inverse of a non-invertible function by restricting its domain
- Inverse Composition Rule
- Writing a linear equation from 2 ordered pairs
- Use composite functions to determine whether two functions are one-to-one and therefore, inverses of each other.
- Given an equation, Name the Parent Function and Describe the translation
- Sketch a graph of the transformation of functions
- Translations
- Stretches and shrinks
- Reflections
- Combining transformations
- Finding equations for Transformations
- Order of combining transformations
- Graphing absolute value Compositions
- Transforming a given graph of a function
- Writing mathematical expressions from verbal descriptions
- Write the specified quantity as a function of the specified variable
- Comparing Functions
- Maximizing a quantity
- Mixture Problems
- Percent increase/decrease problems
- Unit Conversions
- Scatter plots
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- Systems of equations and break even points
- Constructing a Function from Data
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\$2.00 | 366 | 1,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-47 | latest | en | 0.72757 |
http://www.convertaz.com/convert-1-pt-to-cc/ | 1,527,094,545,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865691.44/warc/CC-MAIN-20180523161206-20180523181206-00632.warc.gz | 361,945,357 | 16,644 | # Convert 1 pt to cc
## Conversion details
To convert pt to cc use the following formula:
1 pt equals 568.26125 cc
For example, to convert 10 pt to cc, multiply 568.26125 by 10 i.e.,
10 pt = 568.26125 * 10 cc = 5682.6125 cc
For conversion tables, definitions and more information on the pt and cc units scroll down or use the related pt and cc quick access menus located at the top left side of the page.
Pt is a common alias of the unit 'pint (Imperial)'
cc is the symbol for cubic centimeter
### From 1.00 to 40.00 pt, 40 entries
1 pt = 568.26125 cc
2 pts = 1136.5225 cc
3 pts = 1704.78375 cc
4 pts = 2273.045 cc
5 pts = 2841.30625 cc
6 pts = 3409.5675 cc
7 pts = 3977.82875 cc
8 pts = 4546.09 cc
9 pts = 5114.35125 cc
10 pts = 5682.6125 cc
11 pts = 6250.87375 cc
12 pts = 6819.135 cc
13 pts = 7387.39625 cc
14 pts = 7955.6575 cc
15 pts = 8523.91875 cc
16 pts = 9092.18 cc
17 pts = 9660.44125 cc
18 pts = 10228.7025 cc
19 pts = 10796.96375 cc
20 pts = 11365.225 cc
21 pts = 11933.48625 cc
22 pts = 12501.7475 cc
23 pts = 13070.00875 cc
24 pts = 13638.27 cc
25 pts = 14206.53125 cc
26 pts = 14774.7925 cc
27 pts = 15343.05375 cc
28 pts = 15911.315 cc
29 pts = 16479.57625 cc
30 pts = 17047.8375 cc
31 pts = 17616.09875 cc
32 pts = 18184.36 cc
33 pts = 18752.62125 cc
34 pts = 19320.8825 cc
35 pts = 19889.14375 cc
36 pts = 20457.405 cc
37 pts = 21025.66625 cc
38 pts = 21593.9275 cc
39 pts = 22162.18875 cc
40 pts = 22730.45 cc
Click here for a list of all conversion tables of pt to other compatible units.
## pint (Imperial)
Pint (imperial) is a unit of measurement of volume. The definition for pint (imperial) is the following:
An Imperial pint is equal to 1/8 Imperial gallon.
The symbol for pint (Imperial) is pt (Imp)
## cubic centimeter
Cubic centimeter is a unit of measurement of volume. The definition for cubic centimeter is the following:
A cubic centimeter is equal to 1/1000th of a liter.
The symbol for cubic centimeter is cc
## Other people are also searching for information on pt conversions.
Following are the most recent questions containing pt. Click on a link to see the corresponding answer.
10 pt = qt
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28c= pt
150pt=c
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Home | Base units | Units | Conversion tables | Unit conversion calculator | 806 | 2,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-22 | latest | en | 0.726661 |
https://www.coursehero.com/file/6814313/Study-Guide-Chapter-1/ | 1,513,398,871,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581053.56/warc/CC-MAIN-20171216030243-20171216052243-00037.warc.gz | 709,096,662 | 42,759 | Study Guide, Chapter 1
# Study Guide, Chapter 1 - In science what is measurement...
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In science, what is measurement information used to describe objects, conditions, and changes? ______ A relationship between two or more numbers is a ________________________. As the value of one property increases or decreases, so does the other in the same direction is known as a(n) ________________________________. As the value of one property increases or decreases, so does the other in the opposite direction is known as a(n) ________________________________. Mass and volume are _______________ related in the property of density. Mass and density are________________ related in the density equation. Volume and density are____________________ related in the density equation. _______________ represents quantities, or measured properties. A(n) _________________ is a statement that describes a relationship in which quantities on one side of the equal sign are identical to the other side. An equation is used to ___________________, ____________________, or _______________________. x=y is a _______________ relationship
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# 1.2 graph, model, expo, ratio1
Course Number: MATH 1014, Spring 2009
College/University: York University
Word Count: 4979
Rating:
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S E C T I O N 1 . 2 M AT H E M AT I C A L M O D E L S : A C ATA L O G O F E S S E N T I A L F U N C T I O N S |||| 25 A characteristic feature of linear functions is that they grow at a constant rate. For instance, Figure 2 shows a graph of the linear function f x 3x 2 and a table of sample values. Notice that whenever x increases by 0.1, the value of f x increases by 0.3. So f x increases three times as fast...
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York University - MATH - 1014
S E C T I O N 1 . 2 M AT H E M AT I C A L M O D E L S : A C ATA L O G O F E S S E N T I A L F U N C T I O N S|25A characteristic feature of linear functions is that they grow at a constant rate. For instance, Figure 2 shows a graph of the linear functi
York University - MATH - 1014
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Taped Interviews by Lowell Fewster Second Gift of Prof. Dean Harper, 7/23/08[Each real: .65 mil tensilized polyester; 3 in. reel: 30 min. at 3/34 ips recording both directions] Ashford, Laplois Negro Leaders Interview No. 017 Cooper, Dr. Walter Negro L
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Stewart, Susan. Columbarium. Chicago: University of Chicago Press, 2004. Cloth. \$25. Hoeing \$. Abe Books from The Legacy Company, Williamsburg, IA. Notified JL Stewart, Susan. Euripedes Andromache. Oxford University Press, 2001. Paperback. \$4.99. Hoeing \$
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Stewart,Susan.Columbarium.Chicago:UniversityofChicagoPress,2004.Cloth.\$25.Hoeing \$.AbeBooksfromTheLegacyCompany,Williamsburg,IA.NotifiedJL Stewart,Susan.EuripedesAndromache.OxfordUniversityPress,2001.Paperback.\$4.99. Hoeing\$.AbeBooksfromMagersandQuinnBook
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Saturday July 2 This day has been the counterpart of yesterday and worse as we are weaker. Had dreadful headache all day. Took some ale and ice and _ a small piece of ham for breakfast. Rain is falling again. Stayed in the upper saloon. Mrs. Morris finall
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Sabbath July 3rd Bright and beautiful. Everything is changed. The sun has brought healing in his wings. We have all been out walling or sitting in the sun and the change has given _ to everything. At 11 o clock, the bell called us down to church. Rev _, a
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D.73 Perkins (Dexter) Papers Box 3 Date 1937 1937 1939 1941 1944 1946 1947 1947-53 1949 1950 1951 1951 1953 1953 1953 1956 1957 1957 1960 1960 1963 1964 1964 1969 1973-74 Title The Constitution and the American Spirit The Love of Knowledge, the Guide of L
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Rochester - LIB - 125
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Old Don Lyon Files 1980s (in two boxes) Box I Admission TV Tips Admissions Radio Spots 1967-1983 CAMEROS 83 CAMEROS Lect 84 CASE Entry Meliora Community-University Event Calendar Current Res. Beat 1979 GSM Annual Economic Round-up for Corporate Executives
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And from thence to ascend the _ to a [larger] which is some 70 miles from Moville as t he mouth and the bay at the head and which London day is outdated. The Irish wash, which we [started] first, is rocky and the hills treetop bleak and sterile. We saw no
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Ay754.B86 DC801.N68.L32 1888 DG70.P78 A21b DK215.8.S57 E672.A1p E727.P34c E727.U59n F593.M32t F1863.5 A12w HD2731.B71r HE1051.B71r HG9711.S53f HQ754.B43m LD586.A1y LD2872.5.c67r PR2944.097s PS1358.C68s PS2242.H25t TG145.H37g U17.T69s U103H18e 1862 U130.C8
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#Standard Jet DB#n#b` Ugr@?#h~ 1y0c F Nbm7#"( #cfw_6#Sb#C+93y[v#|*|# # f_ u\$ g#'De#Fx#`bT#4.0# # # # # # # # # #
Rochester - LIB - 154
Rochester - LIB - 154
CFB account book 1920-1941 [A.B81 47:1] 1920 Wilkinsons (Jo & Katherine) for boys 1921 " (Jo, Marion & Katherine) for boys \$ 500 \$10001922 " (Jo, Marion & Katherine) for boys \$ 800 June, 1922 from A.S. for H.B.? \$ 100 [Spent 6497.81; income from work, ro
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Clemente, Vince April 28, 1932 Clifton, Lucille June 27, 1936 Conners, Peter Freudenberger, Nell 1975 Anthony Giardina 1950 Ignatow, David February 17, 1914 Krapf, Norbert 1943 Leavitt, David June 23, 1961 Mason, David December 11, 1954 Sleigh, Tom 1953 Z
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BRAGDON RESEARCHER, THURSDAY 4/5 AROUND NOON The owners of a Bragdon home in Oswego are supposed to be coming in to look at the drawings of the house this Thursday around noon. It is the Sloan house, tube 17 in D. 87, the Bragdon architectural files. In a
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Alexander Hooker Papers, 1804-1865 Box I Folder 1 Horace A. Hooker Correspondence Folder 2 Alexander Hooker Personal Correspondence (indexed), 1804-1809 Folder 3 Alexander Hooker- Personal Correspondence (indexed), 1810-1815 Folder 4 Alexander Hooker- Per
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Box 37 Folder 1 History Fifty years of NYS Division NYS Division The Seventies NYS Division- The Eighties Folder 2- Calendars of AAUW, 1996-2000 Folder 3- College/Union Representation Materials, 1994-2004 Folder 4- Educational Equity, 1998-2002 Folder 5-
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Box 26 Folder 1- Financial Guidelines and Memos to Branch Treasurers- 1974-1996 Folder 2 Board of Directors Meetings- 1984-1989 Folder 3 Treasurers Report 1961-1991, not complete Folder 4 Financial Reports- 1985-1987 Folder 5- Money Market Folder 6- Budge
E. Kentucky - IDK - 23432
_/ 120 + _/ 30= _ (-_ for lateness)= _Literary Analysis Drafting PacketThesis/ Argument Suggested revisions:Name_On TimeRevision Credit 0 Reason One Suggested revisions: 1 2 3 4 5On TimeRevision Credit 0 Reason Two Suggested revisions: 1 2 3 4 5On
E. Kentucky - IDK - 23432 | 4,991 | 16,331 | {"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-2014-15 | longest | en | 0.838642 |
https://convertoctopus.com/471-4-hours-to-minutes | 1,721,028,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00627.warc.gz | 167,002,134 | 8,280 | ## Conversion formula
The conversion factor from hours to minutes is 60, which means that 1 hour is equal to 60 minutes:
1 hr = 60 min
To convert 471.4 hours into minutes we have to multiply 471.4 by the conversion factor in order to get the time amount from hours to minutes. We can also form a simple proportion to calculate the result:
1 hr → 60 min
471.4 hr → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 471.4 hr × 60 min
T(min) = 28284 min
The final result is:
471.4 hr → 28284 min
We conclude that 471.4 hours is equivalent to 28284 minutes:
471.4 hours = 28284 minutes
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 minute is equal to 3.5355678121906E-5 × 471.4 hours.
Another way is saying that 471.4 hours is equal to 1 ÷ 3.5355678121906E-5 minutes.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred seventy-one point four hours is approximately twenty-eight thousand two hundred eighty-four minutes:
471.4 hr ≅ 28284 min
An alternative is also that one minute is approximately zero times four hundred seventy-one point four hours.
## Conversion table
### hours to minutes chart
For quick reference purposes, below is the conversion table you can use to convert from hours to minutes
hours (hr) minutes (min)
472.4 hours 28344 minutes
473.4 hours 28404 minutes
474.4 hours 28464 minutes
475.4 hours 28524 minutes
476.4 hours 28584 minutes
477.4 hours 28644 minutes
478.4 hours 28704 minutes
479.4 hours 28764 minutes
480.4 hours 28824 minutes
481.4 hours 28884 minutes | 447 | 1,692 | {"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-2024-30 | latest | en | 0.823386 |
http://freemat.sourceforge.net/help/inspection_nnz.html | 1,371,683,642,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709805610/warc/CC-MAIN-20130516131005-00047-ip-10-60-113-184.ec2.internal.warc.gz | 106,001,759 | 1,462 | ## NNZ Number of Nonzeros
Section: Inspection Functions
### Usage
Returns the number of nonzero elements in a matrix. The general format for its use is
``` y = nnz(x)
```
This function returns the number of nonzero elements in a matrix or array. This function works for both sparse and non-sparse arrays. For
### Example
```--> a = [1,0,0,5;0,3,2,0]
a =
1 0 0 5
0 3 2 0
--> nnz(a)
ans =
4
--> A = sparse(a)
A =
1 1 1
2 2 3
2 3 2
1 4 5
--> nnz(A)
ans =
4
``` | 166 | 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} | 2.9375 | 3 | CC-MAIN-2013-20 | longest | en | 0.636617 |
https://forum.babylonjs.com/t/isometric-2d-map-tile/40434 | 1,695,910,809,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00538.warc.gz | 293,371,517 | 6,922 | # Isometric 2d map tile
Hello, I’m learning to use Babylonjs. I wanted to start a project of a small 2d game, I made a tiled map and exported it as a js file. I would like to create an isometric view for my game but I can’t in any way, I’m very new to this xD
I am including an image of the map that I export in js :
I include the tile:
this is the code i use :
export function map(scene, tileMap) {
const grid = {
‘h’: tileMap.height,
‘w’: tileMap.width
};
``````const tileWidth = tileMap.tilewidth;
const tileHeight = tileMap.tileheight;
// Create the multi material
const multiMaterial = new BABYLON.MultiMaterial("multi", scene);
// Create the different materials
const grassMaterial = new BABYLON.StandardMaterial("grass", scene);
grassMaterial.diffuseTexture = new BABYLON.Texture("assets/grass_n.png", scene);
grassMaterial.diffuseTexture.hasAlpha = true;
// Add the materials to the multi material
multiMaterial.subMaterials.push(grassMaterial);
// Iterate over the tile map and create isometric blocks
const data = tileMap.layers[0].data;
for (let i = 0; i < data.length; i++) {
const x = i % grid.w;
const y = Math.floor(i / grid.w);
// Convert matrix coordinates to isometric coordinates
const isoX = (x - y) * tileWidth / 2;
const isoY = (x + y) * tileHeight / 2;
// Create the isometric block
const block = BABYLON.MeshBuilder.CreatePlane(`block\${i}`, { size: tileWidth }, scene);
block.position = new BABYLON.Vector3(isoX, 0, isoY);
// Set the material of the isometric block
block.material = multiMaterial.subMaterials[data[i] - 1];
}
``````
}
my camera :
const isometricCamera = function(scene) {
const camera = new BABYLON.ArcRotateCamera(
“Camera”,
-Math.PI / 2,
Math.PI / 4,
15,
BABYLON.Vector3.Zero(),
scene
);
`````` camera.ortho = true;
// Set Zoom Level
camera.radius = 1000; //this is because if not, the map is not noticeable
camera.setTarget(BABYLON.Vector3.Zero());
return camera
``````
}
and finally the result xD :
Thank you ahead of time
Welcome aboard!
It will be easier for us to help if you can move your code in the Playground.
One wrong thing I can see is that to set an orthographic camera you should set `camera.mode = camera.mode = BABYLON.Camera.ORTHOGRAPHIC_CAMERA`. When it’s done, you need to set the size of the orthographic camera, something like:
2 Likes
Hi, thanks for your response! I wrote my code in the playground and managed to make some changes in the camera to achieve the view I expected
the tiles would not be displaying as expected, instead of taking the shape of a block it is still 2d. I think I don’t understand the tool well and I’m missing something.
I think it will be much easier if you create cubes and position them appropriately + using an orthographic camera.
Is there any reason why these should be planes with a false cube perspective? I feel like it might be harder for you to implement your game that way…
But create a cube is not a 3d figure? my textures are png … I don’t understand well, thanks for the help anyway. I am using excaliburjs that has a function to create isometric maps and I will continue with that. Thanks for the help! | 788 | 3,144 | {"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-2023-40 | latest | en | 0.56436 |
http://mh-audio.nl/Calculators/ABPF.html | 1,611,189,873,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00748.warc.gz | 66,696,178 | 8,512 | ## Active filter design precautions
When building the active filter circuit, high tolerance components must be used to obtain the best performance. Typically they should be 1% or better.
A further item to ensure the optimum operation of the circuit is to ensure that the source impedance is less than about 100 ohms. Additionally the load impedance should be greater than about 2 M Ohms.
Q factor The Q Factor is a measure of how "Selective" or "Un-selective" the band pass filter is towards a given spread of frequencies. The lower the value of the Q factor the wider is the bandwidth of the filter and consequently the higher the Q factor the narrower and more "selective" is the filter.
<<< Back to Lowpass, Bandpass and Highpass Filter Calculators
## Calculated Values
Filter Type : Butterworth Chebyshev 0.1 dB Bessel Capacitors : uF Center Freq : Hz 3dB Bandwidth : Hz Voltage Gain : Vu
C1,C2,C3,C4 = uF
R1 = Ω
R2 = Ω
R3 = Ω
R4 = Ω
R5 = Ω
R6 = Ω
Amplification = dB
Section 1 2 Q Freq
Q values up to about 20 are reasonable.
Above that may result in an unstable circuit.
Filter Response for Butterworth, Chebyshev & Bessel
<<< Back to Lowpass, Bandpass and Highpass Filter Calculators
## Calculated Values
Filter Type : Butterworth Chebyshev 0.1 dB Bessel Capacitors : uF Center Freq : Hz 3dB Bandwidth : Hz Voltage Gain : Vu
C1,C2,C3,C4,C5,C6 = uF
R1 = Ω
R2 = Ω
R3 = Ω
R4 =
R5 =
R6 =
R7 =
R8 =
R9 =
Amplification = dB
Section 1 2 3 Q Freq
Q values up to about 20 are reasonable.
Above that may result in an unstable circuit.
Filter Response for Butterworth, Chebyshev & Bessel
<<< Back to Lowpass, Bandpass and Highpass Filter Calculators
## Computed Values
Capacitors : uF Center Freq : Hz 3dB Bandwidth : Hz Voltage Gain : Vu
C1,C2 = uF nF R1 = kΩ Ω R2 = kΩ Ω R3 = kΩ Ω Amplification = dB Filter Q =
Use a potentiomter for R1 to fine-adjust the gain. Use a potentiomter for R2 to fine-adjust the center frequency.
<<< Back to Lowpass, Bandpass and Highpass Filter Calculators | 570 | 2,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-04 | latest | en | 0.832276 |
http://www.clubdetirologrono.com/2-digit-subtraction-with-regrouping-worksheet/two-digit-subtraction-worksheets-2-with-regrouping-pdf-subtractingonedigitfromtwodigitregrou/ | 1,566,632,171,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00527.warc.gz | 226,715,569 | 12,043 | # Two Digit Subtraction Worksheets 2 With Regrouping Pdf Subtractingonedigitfromtwodigitregrou
Category : Math Worksheet.
Topic : 2 digit subtraction with regrouping worksheets 2nd grade pdf.
Author : Ferreira Craig.
Published : Thu, Jul 18th 2019 03:13 AM.
Format : jpg/jpeg.
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Some sites do offer math online quiz that is sure to bring about an inclination towards math among children. It is a common practice for parents around the world to send their children to special math training centers. Invariably, every parent is unaware of the actual quality of training provided by these centers. To help parents combat this problem, there are a lot of online resources available that offer math assignment help exclusively for children. Parents can go onto the various online portals to find suitable option to help their children learn math solving problem skills. Thorough research over the internet will give a wide range of websites that offers right set of math worksheets for children.
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With a packed curriculum and the increased emphasis on testing, our children are taught tons of procedures - but procedures disconnected from when to use them, and why. Sustained thinking - the key ingredient to math success - is painfully absent in too many math classes. Let us discuss some tangible advantages of Mathematics in todays world. One should also be aware of the wide importance of Mathematics, and the way in which it is advancing at a spectacular rate. Mathematics is about pattern and structure; it is about logical analysis, deduction, calculation within these patterns and structures. When patterns are found, often in widely different areas of science and technology, the mathematics of these patterns can be used to explain and control natural happenings and situations. Mathematics has a pervasive influence on our everyday lives, and contributes to the wealth of the individual. | 767 | 3,689 | {"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} | 2.96875 | 3 | CC-MAIN-2019-35 | latest | en | 0.919999 |
https://brainmass.com/engineering/aerospace-engineering/event-diagrams-poisson-exponential-failure-processes-630074 | 1,575,885,598,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518627.72/warc/CC-MAIN-20191209093227-20191209121227-00081.warc.gz | 303,389,963 | 12,823 | Explore BrainMass
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Event diagrams, Poisson & exponential failure processes
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Note: Please see attachment for more clarification
TQE #5
1. For a given initiating event, I, that occurs once every 5 years, there are three
Hazard barriers to prevent loss of revenue: A, B, and C, triggered in that order. If
Hazard barrier A performs, then there is no loss of revenue. If hazard barrier A fails,
but either B or C performs, then the loss of revenue is \$250. If all three hazard
barriers fail, the loss of revenue is \$500.
a) Draw an event tree showing the possible outcomes of this initiating event. Label
Each scenario with its full sequence of events (in terms of hazard barriers) and the consequence of the scenario.
b) Assume the probability of hazard barrier B failing is 0.25. The probabilities of barriers A and C failing can be calculated with the fault trees and probabilities given below. Calculate the frequency of each scenario in the event tree in part a.
c) Calculate the total risk to the system posed by the initiating event, I. Give appropriate units
2. Suppose you have a system with a repairable component that experiences in-service failure. When the component fails, it is repaired and run to failure. The interarrival times of failure of the component during the first 100 days of operation are given by: 8 days, 6 days, 10 days, 9 days, 13 days, 16 days, 8 days, and 20 days.
a) Assume the failure times follow a homogeneous Poisson process. Estimate the ROCOF
b) Now assume that the failure times follow a nonhomogeneous Poisson process according to
Find the MLE of α and β for the given data.
c) Does the failure intensity have a time trend (α = 0.05)? Based on this result, which ROCOF from above is more likely correct?
3. Suppose you have 10 nominally identical fielded non-repairable systems. Every time a system fails, it is replaced with a new system. The following failure times have been observed over the first 2000 hours of observation: 750, 900, 1250, 1400, 1525, 1620, 1625, 1700, 1710, 1825, 1905. Assuming the population system failure follows an exponential distribution, estimate the failure rate. Give a two sided 95% confidence interval of the failure rate and the associated confidence interval for the reliability at time t = 3000 hours.
© BrainMass Inc. brainmass.com October 10, 2019, 8:30 am ad1c9bdddf
https://brainmass.com/engineering/aerospace-engineering/event-diagrams-poisson-exponential-failure-processes-630074
Solution Summary
3 questions and solutions in reliability engineering concerning Homogenous and Non-Homogeneous Poisson processes, exponential failure, event diagrams and determining such quantities as failure rates & event probabilities
\$2.19 | 665 | 2,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-51 | latest | en | 0.91548 |
https://quantumcomputing.stackexchange.com/questions/38704/optimizing-sympy-implementation-of-prime-factorization-in-form-of-qubo/38716 | 1,721,021,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00090.warc.gz | 452,031,093 | 41,315 | # Optimizing SymPy Implementation of prime factorization in form of QUBO
I'm trying to reproduce a paper on Prime Factorization. This paper converts the problem into a QUBO form, which then we can map it to an Ising Minimizer. I've basically done everything, and I've obtained the final QUBO expression of page 3. My only problem is that my implementation gets slow for larger numbers really fast. The bottleneck of my code is in the reduced_f function. I also suspect that all of these operations are possible without symbolic calculations and only with matrix multiplication, but I don't know how to implement it like that.
from sympy import IndexedBase, expand, Indexed, Mul
import numpy as np
from tqdm import tqdm
x = IndexedBase('x')
def high_degree_f(N, l1 = 2, l2 = 3):
# p and q are binary numbers of length l1 and l2
p = 1
q = 1
for i in range(1, l1):
p += x[i] * 2**i
for idx, l in enumerate(range(l1, l1+l2 - 1)):
q += x[l] * 2**(idx+1)
f = (N - p * q ) **2
f = expand(f)
r = len(f.free_symbols) + 1
# replace x[i]**k by x[i] as x[i] is binary
for i in range(1, r):
for k in range(2, 4):
f = f.subs(x[i]**k, x[i])
return (f, p, q)
def out_degrees_more_than_2(f):
# Iterate over the terms in the expanded polynomial
for term in f.as_ordered_terms():
# Check if the term is a product
if isinstance(term, Mul):
# Count the number of Indexed instances in the term
variable_count = sum(isinstance(factor, Indexed) for factor in term.args)
# Check if there are more than two variables in the product
if variable_count > 2:
# Print the term
yield (variable_count, term)
def max_degree(f):
max_degree = 0
for (variable_count, term) in out_degrees_more_than_2(f):
max_degree = max(max_degree, variable_count)
return max_degree
def reduced_f(N, l1 = 2, l2 = 3):
(f, p, q) = high_degree_f(N, l1 = l1, l2 = l2)
number_of_variables = len(f.free_symbols) - 1
while max_degree(f) > 2:
for (variable_count, term) in tqdm(out_degrees_more_than_2(f)):
if variable_count == 3:
# extract the numerical coefficient of the term:
coefficient, v = term.as_coeff_Mul()
variables = v.args
new_term = coefficient* (variables[2] * x[number_of_variables+1] + 2 * ( variables[0] * variables[1] - 2 * variables[0] * x[number_of_variables+1] - 2 * variables[1] * x[number_of_variables+1] + 3 * x[number_of_variables+1]))
number_of_variables += 1
# substitute the term with the new term
f = f.subs(term, new_term)
elif variable_count == 4:
pass
else:
raise ValueError("Unexpected number of variables in term")
f = expand(f)
return (f, p, q)
N = 15
l1 = 2
l2 = 3
(f, p, q) = reduced_f(N, l1 = l1, l2 = l2)
print("N:\t\t", N)
print("p:\t\t", p)
print("q:\t\t", q)
print("Reduced f:\t", f)
N: 15
p: 2*x[1] + 1
q: 2*x[2] + 4*x[3] + 1
Reduced f: 200*x[1]*x[2] - 48*x[1]*x[3] - 512*x[1]*x[4] - 52*x[1] + 16*x[2]*x[3] - 512*x[2]*x[4] - 52*x[2] + 128*x[3]*x[4] - 96*x[3] + 768*x[4] + 196
expression from paper:
$$\begin{array}{rcl}f^{\prime} (x) & = & 200{x}_{1}{x}_{2}-48{x}_{1}{x}_{3}-512{x}_{1}{x}_{4}+16{x}_{2}{x}_{3}-512{x}_{2}{x}_{4}+128{x}_{3}{x}_{4}\\ & & -52{x}_{1}-52{x}_{2}-96{x}_{3}+768{x}_{4}+\mathrm{196,}\end{array}$$
But for N = 10403, l1 = 7, l2 = 7, it takes minutes to get the final expression. Can anyone suggest any improvement to my implementation?
## 1 Answer
To make it fast, I would guess a good option is to ditch sympy and simply use a dictionary and strings to create the QUBO, probably something like dic[(x_1,x_2,x_3)] = alpha_(x_1,x_2,x_3) which takes the (sorted) tuple of variables to corresponding coeffecient in the QUBO and then worry about making all tuples of length 2.
From a pragmatic point of view, I would define all the 'product variables' from $$p\times q$$ first and sub them into $$(N-pq)^2$$. This way, $$N-p\times q$$ is linear and and no further reductions will be needed. How these variables are chosen may change the final QUBO though which might not be what you are after.
• The whole point of using SymPy is to do the reduction. otherwise, the coefficients are pretty easy to obtain even in a matrix multiplication form (which is even faster than dictionary). I don't understand how your suggestion can skip the reduction. Please read the paper and if you think you can somehow improve the embedding, don't hesitate to write it as a detailed answer. Commented Jun 11 at 11:21
• My point is you are creating a 4th degree polynomial and then reducing it to a quadratic. The alternative I'm suggesting is to make the quadratic expression linear first and then square it, this should result in less terms to deal with. I'm not sure how sympy stores equations but if the long times are being caused by list manipulations. Moving to something like dictionaries and doing the reductions yourself could be a lot better.
– john
Commented Jun 11 at 11:32
• please write your approach as an answer. Commented Jun 11 at 16:16 | 1,489 | 4,914 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.817595 |
https://www.tutorela.com/math/equivalent-ratios | 1,723,249,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00493.warc.gz | 808,397,208 | 22,742 | # Equivalent Ratios
🏆Practice ratio
To easily solve ratio problems and to gain a better understanding of the topic in general, it is convenient to know about equivalent ratios.
Equivalent ratios are, in fact, ratios that seem different, are not expressed in the same way but, by simplifying or expanding them, you arrive at exactly the same relationship.
Think of it this way,
## Test yourself on ratio!
There are 18 balls in a box, $$\frac{2}{3}$$ of which are white.
How many white balls are there in the box?
If you have two fractions in front of you:
$\frac{2000}{4000}$
and
$\frac{2}{4}$
we can simplify the larger fraction and arrive at
and even more, we can simplify the smaller fraction and arrive at:
In fact, we can say that:
$\frac{4000}{2000}=\frac{2}{4}=\frac{1}{2}$
All the expressions are equivalent ratios.
Do you remember we said that a ratio can be shown in the form of a fraction?
Therefore, the same rule also applies to the ratios we have learned.
We can reduce both terms of the ratio or amplify them and arrive at equivalent ratios.
To solve this type of problem easily we will always try to arrive at the smallest ratio.
We will ask ourselves by what number we can divide both terms of the ratio, in this way we will arrive at the most reduced equivalent ratio possible.
## How can you tell if they are equivalent ratios?
We will ask ourselves: Will we arrive at the same ratio by reduction or by amplification?
Let's see some examples:
$2:5$
$6:16$
We have managed to demonstrate that by multiplying both terms by $3$ we arrive at the same ratio. Therefore, they are equivalent ratios!
Are these ratios equivalent?
$1:3$
$2:6$
$6:18$
Yes! The first ratio is equivalent to the second: we multiply both terms by $2$.
The first ratio is also equivalent to the third, we multiply by $6$.
The second ratio is equivalent to the third: multiplication of both terms by $3$.
Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
### Example
José and Dani have notebooks and pencils. José has $4$ notebooks and $8$ pencils.
The ratio between the notebooks and pencils that Dani has is the same as José's. Dani has $6$ notebooks. We are asked to calculate how many pencils Dani has.
We see that the number of pencils José has is double the number of notebooks he has. Since we already know that the ratio between notebooks and pencils that José and Dani have is identical, we can deduce that Dani has $12$ pencils ($6$ times $2$, so that the number of pencils is double the number of notebooks).
## Examples and exercises with solutions of Equivalent Ratios
### Exercise #1
There are 18 balls in a box, $\frac{2}{3}$ of which are white.
How many white balls are there in the box?
12
### Exercise #2
In a box there are 28 balls, $\frac{1}{4}$ of which are orange.
How many orange balls are there in the box?
7
### Exercise #3
There are two circles.
One circle has a radius of 4 cm, while the other circle has a radius of 10 cm.
How many times greater is the area of the second circle than the area of the first circle?
### Video Solution
$6\frac{1}{4}$
### Exercise #4
There are two circles.
The length of the radius of circle 1 is 6 cm.
The length of the diameter of circle 2 is 12 cm.
How many times greater is the area of circle 2 than the area of circle 1?
They are equal.
### Exercise #5
There are two circles.
The length of the diameter of circle 1 is 4 cm.
The length of the diameter of circle 2 is 10 cm.
How many times larger is the area of circle 2 than the area of circle 1?
### Video Solution
$6\frac{1}{4}$ | 894 | 3,656 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-33 | latest | en | 0.939152 |
https://math.libretexts.org/Bookshelves/PreAlgebra/Book%3A_Prealgebra_(Arnold)/08%3A_Graphing/8.03%3A_Graphing_Linear_Equations | 1,685,227,612,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643388.45/warc/CC-MAIN-20230527223515-20230528013515-00324.warc.gz | 441,795,699 | 32,884 | # 8.3: Graphing Linear Equations
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Consider $$y = x + 1$$ an equation in two variables. If we substitute the ordered pair $$(x, y) = (1, 2)$$ into the equation $$y = x + 1$$, that is, if we replace x with 1 and y with 2, we get a true statement.
\begin{aligned} y = x + 1 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 2=1+1 ~ & \textcolor{red}{ \text{ Substitute: 1 for } x \text{ and 2 for } y.} \\ 2=2 ~ & \textcolor{red}{ \text{ Simplify.}} \end{aligned}\nonumber
We say that the ordered pair (1, 2) is a solution of the equation $$y = x + 1$$.
Solution of an Equation in Two Variables
If substituting the ordered pair (x, y)=(a, b) into an equation (replace x with a and y with b) produces a true statement, then the ordered pair (a, b) is called a solution of the equation and is said to “satisfy the equation.”
Example 1
Which of the ordered pairs are solutions of the equation $$y = 2x + 5$$: (a) (−3, −2), or (b) (5, 15)?
Solution
Substitute the points into the equation to determine which are solutions.
a) To determine if (−3, −2) is a solution of $$y = 2x + 5$$, substitute −3 for x and −2 for y in the equation $$y = 2x + 5$$.
\begin{aligned} y = 2x + 5 ~ & \textcolor{red}{ \text{ Original equation.}} \\ −2 = 2(−3) + 5 ~ & \textcolor{red}{ \text{ Substitute: −3 for } x \text{ and } −2 \text{ for } y.} \\ −2 = −6+5 ~ & \textcolor{red}{ \text{ Multiply first: } 2(−3) = −6} \\ −2 = −1 ~ & \textcolor{red}{ \text{ Add: } −6+5= −1.} \end{aligned}\nonumber
Because the resulting statement is false, the ordered pair (−3, −2) does not satisfy the equation. The ordered pair (−3, −2) is not a solution of $$y = 2x + 5$$.
a) To determine if (5, 15) is a solution of $$y = 2x + 5$$, substitute 5 for x and 15 for y in the equation $$y = 2x + 5$$.
\begin{aligned} y = 2x + 5 ~ & \textcolor{red}{ \text{ Original equation.}} \\ 15 = 2(5) + 5 ~ & \textcolor{red}{ \text{ Substitute: 5 for } x \text{ and } 15 \text{ for } y.} \\ 15 = 10 + 5 ~ & \textcolor{red}{ \text{ Multiply first: } 2(5) = 10} \\ 15 = 15 ~ & \textcolor{red}{ \text{ Add: } 10 + 5 = 15.} \end{aligned}\nonumber
The resulting statement is true. The ordered pair (5, 15) does satisfy the equation. Hence, (5, 15) is a solution of $$y = 2x + 5$$.
Exercise
Which of the ordered pairs (1, 7) and (2, 9) are solution of the equation $$y = 3x + 4$$?
(1, 7)
## The Graph of an Equation
We turn our attention to the graph of an equation.
The Graph of an Equation
The graph of an equation is the set of all ordered pairs that are solutions of the equation.
In the equation $$y = 2x + 5$$, the variable y depends on the value of the variable x. For this reason, we call y the dependent variable and x the independent variable. We’re free to make choices for x, but the value of y will depend upon our choice for x.
We will also assign the horizontal axis to the independent variable x and the vertical axis to the dependent variable y (see Figure 8.7).
The graph of $$y = 2x+5$$ consists of all ordered pairs that are solutions of the equation $$y = 2x+ 5$$. So, our first task is to find ordered pairs that are solutions of $$y = 2x + 5$$. This is easily accomplished by selecting an arbitrary number of values, substituting them for x in the equation $$y = 2x + 5$$, then calculating the resulting values of y. With this thought in mind, we pick arbitrary integers −7, −6, . . . , 2, substitute them for x in the equation $$y = 2x + 5$$, calculate the resulting value of y, and store the results in a table.
$\begin{array}{|r|r|r|r|} \hline y = 2x+5 & x & y & (x,~y) \\ \hline y = 2(−7) + 5 = −9 & −7 & −9 & (−7, −9) \\ y = 2(−6) + 5 = −7 & −6 & −7 & (−6, −7) \\ y = 2(−5) + 5 = −5 & −5 & −5 & (−5, −5) \\ y = 2(−4) + 5 = −3 & −4 & −3 & (−4, −3) \\ y = 2(−3) + 5 = −1 & −3 & −1 & (−3, −1) \\ y = 2(−2) + 5 = 1 & −2 & 1 & (−2, 1) \\ y = 2(−1) + 5 = 3 & −1 & 3 & (−1, 3) \\ y = 2(0) + 5 = 5 & 0 & 5 & (0, 5) \\ y = 2(1) + 5 = 7 & 1 & 7 & (1, 7) \\ y = 2(2) + 5 = 9 & 2 & 9 & (2, 9) \\ \hline \end{array}\nonumber$
The result is 10 ordered pairs that satisfy the equation $$y = 2x + 5$$. Therefore, we have 10 ordered pairs that belong to the graph of $$y = 2x + 5$$. They are plotted in Figure 8.7(a).
However, we’re not finished, because the graph of the equation $$y = 2x + 5$$ is the set of all points that satisfy the equation and we’ve only plotted 10 such points. Let’s plot some more points. Select some more x-values, compute the corresponding y-value, and record the results in a table.
$\begin{array}{|r|r|r|r|} \hline y = 2x + 5 & x & y & (x, y) \\ \hline y = 2(−7.5) + 5 = −10 & −7.5 & −10 & (−7.5, −10) \\ y = 2(−6.5) + 5 = −8 & −6.5 & −8 & (−6.5, −8) \\ y = 2(−5.5) + 5 = −6 & −5.5 & −6 & (−5.5, −6) \\ y = 2(−4.5) + 5 = −4 & −4.5 & −4 & (−4.5, −4) \\ y = 2(−3.5) + 5 = −2 & −3.5 & −2 & (−3.5, −2) \\ y = 2(−2.5) + 5 = 0 & −2.5 & 0 & (−2.5, 0) \\ y = 2(−1.5) + 5 = 2 & −1.5 & 2 & (−1.5, 2) \\ y = 2(−0.5) + 5 = 4 & −0.5 & 4 & (−0.5, 4) \\ y = 2(0.5) + 5 = 6 & 0.5 & 6 & (0.5, 6) \\ y = 2(1.5) + 5 = 8 & 1.5 & 8 & (1.5, 8) \\ y = 2(2.5) + 5 = 10 & 2.5 & 10 & (2.5, 10) \\ \hline \end{array}\nonumber$
That’s 11 additional points that we add to the graph in Figure 8.7(b).
Note that we can continue indefinitely in this manner, adding points to the table and plotting them. However, sooner or later, we have to make a leap of faith, and imagine what the final graph will look like when all of the points that satisfy the equation y = 2x+ 5 are plotted. We do so in Figure 8.8, where the final graph takes on the appearance of a line.
Ruler Use
All lines must be drawn with a ruler. This includes the x- and y-axes.
Important Observation. When we use a ruler to draw a line through the plotted points in Figure 8.7(b), arriving at the final result in Figure 8.8, we must understand that this is a shortcut technique for plotting all of the remaining ordered pairs that satisfy the equation. We’re not really drawing a line through the plotted points. Rather, we’re shading all of the ordered pairs that satisfy the equation $$y = 2x + 5$$.
The Result. The graph of the equation $$y = 2x + 5$$, pictured in Figure 8.8, is a line. Actually, the graph is an infinite collection of points satisfying the equation $$y = 2x + 5$$ that takes the shape of a line, but it’s all right to say the graph of $$y = 2x + 5$$ is a line.
Ordered Pairs and the Graph
Because the graph of an equation is the collection of all ordered pairs that satisfy the equation, we have two important results:
1. If an ordered pair satisfies an equation, then the point in the Cartesian plane represented by the ordered pair is on the graph of the equation.
2. If a point is on the graph of an equation, then the ordered pair representation of that point satisfies the equation.
Example 2
Find the value of k so that the point (2, k) is on the graph of the equation $$y = 3x − 2$$.
Solution
If the point (2, k) is on the graph of $$y = 3x−2$$, then it must satisfy the equation $$y = 3x − 2$$.
\begin{aligned} y = 3x − 2 ~ & \textcolor{red}{ \text{ Original equation.}} \\ k = 3(2) − 2 ~ & \begin{array}{l} \textcolor{red}{ \text{ The point (2, } k) \text{ is on the graph.}} \\ \textcolor{red}{ \text{ Substitute 2 for } x \text{ and } k \text{ for } y \text{ in } y = 3x − 2.} \end{array} \\ k = 6 − 2 ~ & \textcolor{red}{ \text{ Multiply: } 3(2) = 6.} \\ k = 4 ~ & \textcolor{red}{ \text{ Subtract: } 6 − 2 = 4.} \end{aligned}\nonumber
Thus, k = 4.
Exercise
Find the value of k so that the point (k, −3) is on the graph of the equation $$y = 4x + 2$$.
k = −5/4
## Linear Equations
Let’s plot the graph of another equation.
Example 3
Sketch the graph of $$y = −2x + 1$$.
Solution
Select arbitrary values of x: −4, −3, . . . , 5. Substitute these values into the equation $$y = −2x + 1$$, calculate the resulting value of y, then arrange your results in a table.
$\begin{array}{|r|r|r|r|} \hline y = −2x + 1 & x & y & (x, y) \\ \hline y = −2(−4) + 1 = 9 & −4 & 9 & (−4, 9) \\ y = −2(−3) + 1 = 7 & −3 & 7 & (−3, 7) \\ y = −2(−2) + 1 = 5 & −2 & 5 & (−2, 5) \\ y = −2(−1) + 1 = 3 & −1 & 3 & (−1, 3) \\ y = −2(0) + 1 = 1 & 0 & 1 & (0, 1) \\ y = −2(1) + 1 = −1 & 1 & −1 & (1, −1) \\ y = −2(2) + 1 = −3 & 2 & −3 & (2, −3) \\ y = −2(3) + 1 = −5 & 3 & −5 & (3, −5) \\ y = −2(4) + 1 = −7 & 4 & −7 & (4, −7) \\ y = −2(5) + 1 = −9 & 5 & −9 & (5, −9) \\ \hline \end{array}\nonumber$
We’ve plotted the points in the table in Figure 8.9(a). There is enough evidence in Figure 8.9(a) to imagine that if we plotted all of the points that satisfied the equation $$y = −2x + 1$$, the result would be the line shown in Figure 8.9(b).
Exercise
Sketch the graph of $$y = 2x − 2$$.
The graph of $$y = 2x + 5$$ in Figure 8.8 is a line. The graph of $$y = −2x + 1$$ in Figure 8.9(b) is also a line. This would lead one to suspect that the graph of the equation $$y = mx+b$$, where m and b are constants, will always be a line. Indeed, this is always the case.
Linear Equations
The graph of $$y = mx + b$$, where m and b are constants, will always be a line. For this reason, the equation $$y = mx+b$$ is called a linear equation.
Example 4
Which of the following equations is a linear equation? 1. $$y = −3x + 4$$, 2. $$y = \frac{2}{3}x + 3$$, and 3. $$y = 2x^2 + 4$$.
Solution
Compare each equation with the general form of a linear equation, $$y = mx + b$$.
1. Note that $$y = −3x + 4$$ has the form $$y = mx + b$$, where m = −3 and b = 4. Hence, $$y = −3x + 4$$ is a linear equation. Its graph is a line.
2. Note that $$y = \frac{2}{3} x+ 3$$ has the form $$y = mx+b$$, where m = 2/3 and b = 3. Hence, $$y = \frac{2}{3} x + 3$$ is a linear equation. Its graph is a line.
3. The equation $$y = 2x^2 + 4$$ does not have the form $$y = mx + b$$. The exponent of 2 on the x prevents this equation from being linear. This is a nonlinear equation. Its graph is not a line.
Exercise
Which of the following equations is a linear equation? a) $$y = 2x^3 + 5$$ b) $$y= -3x-5$$
$$y = −3x − 5$$
The fact that $$y = mx + b$$ is a linear equation enables us to quickly sketch its graph.
Example 5
Sketch the graph of $$y = − \frac{3}{2} x + 4$$.
Solution
The equation $$y = −3 2x+ 4$$ has the form $$y = mx+b$$. Therefore, the equation is linear and the graph will be a line. Because two points determine a line, we need only find two points that satisfy the equation $$y = − \frac{3}{2} x + 4$$, plot them, then draw a line through them with a ruler. We choose x = −2 and x = 2, calculate y, and record the results in a table.
$\begin{array}{|r|r|r|r|} \hline y = − \frac{3}{2} x + 4 & x & y & (x, y) \\ \hline y = − \frac{3}{2} (−2) + 4 = 3 + 4 = 7 & −2 & 7 & (−2, 7) \\ y = − \frac{3}{2} (2) + 4 = −3+4=1 & 2 & 1 & (2, 1) \\ \hline \end{array}\nonumber$
Plot the points (−2, 7) and (2, 1) and draw a line through them. The result is shown in Figure 8.10.
Exercise
Sketch the graph of $$y = − \frac{1}{2} x + 2$$.
You may have noted in Example 5 that are choices of −2 and 2 for x eased the calculation of the corresponding y-values because of the resulting cancellation.
Choosing Strategic Values
When plotting a linear equation, it is a good strategy to choose values of x that simplify the calculation of the corresponding y-values.
Example 6
Sketch the graph of y = \frac{1}{3} x + 3.
Solution
The equation $$y = \frac{1}{3} x + 3$$ has the form $$y = mx + b$$. Therefore, the equation is linear and the graph will be a line. Because two points determine a line, we need only find two points that satisfy the equation y = 1 3x + 3, plot them, then draw a line through them with a ruler. We choose x = −6 and x = 6, calculate y, and record the results in a table.
$\begin{array}{|r|r|r|r|} \hline y = \frac{1}{3} x + 3 & x & y & (x, y) \\ \hline y = \frac{1}{3} (−6) + 3 = −2+3=1 & −6 & 1 & (−6, 1) \\ y = \frac{1}{3} (6) + 3 = 2 + 3 = 5 & 6 & 5 & (6, 5) \\ \hline \end{array}\nonumber$
Plot the points (−6, 1) and (6, 5) and draw a line through them. The result is shown in Figure 8.11.
Exercise
Sketch the graph of $$y = \frac{2}{3} x + 1$$.
## Exercises
1. Which of the points (2, −14), (−1, −6), (−8, 11), and (3, −13) is a solution of the equation $$y = −2x − 8$$?
2. Which of the points (1, −2), (8, 23), (−3, −23), and (8, 24) is a solution of the equation $$y = 4x − 9$$?
3. Which of the points (1, −1), (−2, 20), (−4, 31), and (−9, 64) is a solution of the equation $$y = −6x + 7$$?
4. Which of the points (−8, −61), (4, 42), (−3, −18), and (−6, −46) is a solution of the equation $$y = 9x + 8$$?
5. Which of the points (2, 15), (−8, −74), (2, 18), and (5, 40) is a solution of the equation $$y = 9x − 3$$?
6. Which of the points (−9, −52), (−8, −44), (−7, −37), and (8, 35) is a solution of the equation $$y = 5x − 5$$?
7. Which of the points (−2, 12), (−1, 12), (3, −10), and (−2, 14) is a solution of the equation $$y = −5x + 4$$?
8. Which of the points (6, 25), (−8, −14), (8, 33), and (−7, −9) is a solution of the equation $$y = 3x + 9$$?
9. Determine k so that the point (9, k) is a solution of $$y = −6x + 1$$.
10. Determine k so that the point (−9, k) is a solution of $$y = 2x + 3$$.
11. Determine k so that the point (k, 7) is a solution of $$y = −4x + 1$$.
12. Determine k so that the point (k, −4) is a solution of $$y = 8x + 3$$.
13. Determine k so that the point (k, 1) is a solution of $$y = 4x + 8$$.
14. Determine k so that the point (k, −7) is a solution of $$y = −7x + 5$$.
15. Determine k so that the point (−1, k) is a solution of $$y = −5x + 3$$.
16. Determine k so that the point (−3, k) is a solution of $$y = 3x + 3$$.
In Exercises 17-24, which of the given equations is a linear equation?
17. $$y = 6x^2 + 4, ~ y = x^2 + 6x + 4, ~ y = 6x + 4, ~ y = \sqrt{6x + 4}$$
18. $$y = −2x + 1, ~ y = x^2 − 2x + 1, ~ y = \sqrt{−2x + 1}, ~ y = −2x^2 + 1$$
19. $$y = x + 7, ~ y = \sqrt{x + 7}, ~ y = x^2 + 7, ~ y = x^2 + x + 7$$
20. $$y = x^2 + 5x + 1, ~ y = 5x^2 + 1, ~ y = \sqrt{5x + 1}, ~ y = 5x + 1$$
21. $$y = x^2 − 2x − 2, ~ y = −2x^2 − 2, ~ y = \sqrt{−2x − 2}, ~ y = −2x − 2$$
22. $$y = x^2 + 5x − 8, ~ y = 5x^2 − 8, ~ y = \sqrt{5x − 8}, ~ y = 5x − 8$$
23. $$y = x^2 + 7x − 3, ~ y = 7x^2 − 3, ~ y = 7x − 3, ~ y = \sqrt{7x − 3}$$
24. $$y = \sqrt{−4x − 3}, ~ y = x^2 − 4x − 3, ~ y = −4x − 3, ~ y = −4x^2 − 3$$
In Exercises 25-28, which of the given equations has the given graph?
25. $$y = − \frac{3}{2}x + 2, ~ y = \frac{3}{2} x − 3, ~ y = −3x + 1, ~ y = −2x + 1$$
26. $$y = −3x − 2, ~ y = \frac{3}{2} x + 1, ~ y = −2x − 1, ~ y = \frac{5}{2}x +$$2
27. $$y = \frac{5}{2} x − 2, ~ y = 3x + 3, ~ y = \frac{3}{2} x + 1, ~ y = \frac{1}{2} x + 1$$
28. $$y = 3x + 1, ~ y = \frac{5}{2} x − 1, ~ y = − \frac{5}{2} x − 3, ~ y = \frac{3}{2} x − 2$$
29. $$y = 3x − 2$$
30. $$y = \frac{5}{2}x + 1$$
31. $$y = −2x − 1$$
32. $$y = \frac{5}{2} x + 2$$
33. $$y = −2x + 2$$
34. $$y = − \frac{5}{2}x − 2$$
35. $$y = −2x − 2$$
36. $$y = − \frac{5}{2}x + 1$$
37. $$y = 2x − 2$$
38. $$y = \frac{5}{2} x − 1$$
39. $$y = \frac{3}{2}x + 1$$
40. $$y = 2x + 2$$
41. $$y = 2x − 3$$
42. $$y = − \frac{5}{2}x − 1$$
43. $$y = \frac{3}{2}x + 3$$
44. $$y = 3x + 1$$
45. Sketch the lines $$y = \frac{1}{2} x−1$$ and $$y = \frac{5}{2} x−2$$ on graph paper. As you sweep your eyes from left to right, which line rises more quickly?
46. Sketch the lines $$y = \frac{5}{2} x+ 1$$ and $$y = 3x+ 1$$ on graph paper. As you sweep your eyes from left to right, which line rises more quickly?
47. Sketch the line $$y = − \frac{1}{2} x + 1$$ and $$y = −3x+ 3$$. As you sweep your eyes from left to right, which line falls more quickly?
48. Sketch the line $$y = −3x − 1$$ and $$y = − \frac{5}{2} x−2$$. As you sweep your eyes from left to right, which line falls more quickly?
49. Sketch the line $$y = −3x − 1$$ and $$y = − \frac{1}{2} x−2$$. As you sweep your eyes from left to right, which line falls more quickly?
50. Sketch the line $$y = −3x − 1$$ and $$y = − \frac{1}{2}x+1$$. As you sweep your eyes from left to right, which line falls more quickly?
51. Sketch the lines $$y = \frac{3}{2} x−2$$ and $$y = 3x+ 1$$ on graph paper. As you sweep your eyes from left to right, which line rises more quickly?
52. Sketch the lines $$y = \frac{1}{2} x+ 3$$ and $$y = \frac{5}{2}x+ 1$$ on graph paper. As you sweep your eyes from left to right, which line rises more quickly?
1. (−1, −6)
3. (−4, 31)
5. (2, 15)
7. (−2, 14)
9. k = −53
11. k = $$− \frac{3}{2}$$
13. k = $$− \frac{7}{4}$$
15. k = 8
17. $$y = 6x + 4$$
19. $$y = x + 7$$
21. $$y = −2x − 2$$
23. $$y = 7x − 3$$
25. $$y = −3x + 1$$
27. $$y = \frac{3}{2} x + 1$$
29. $$y = 3x − 2$$
31. $$y = −2x − 1$$
33. $$y = −2x + 2$$
35. $$y = −2x -2$$
37. $$y = 2x-2$$
39. $$y = \frac{3}{2} + 1$$
41. $$y = 2x − 3$$
43. $$y = \frac{3}{2} x + 3$$
45. The graph of $$y = \frac{5}{2} x−2$$ rises more quickly.
47. The graph of $$y = −3x + 3$$ falls more quickly.
49. The graph of $$y = −3x − 1$$ falls more quickly.
51. The graph of $$y = 3x+1$$ rises more quickly.
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Home : Support : Online Help : Mathematics : Algebra : Skew Polynomials : Ore Algebra : Ore_algebra/Weyl_algebra
Overview of Weyl Algebras
Description
• Weyl algebras are algebras of linear differential operators with polynomial coefficients. They are particular cases of Ore algebras.
• A Weyl algebra is an algebra of noncommutative polynomials in the indeterminates ${x}_{1},...,{x}_{n},{\mathrm{D}}_{1},...,{\mathrm{D}}_{n}$ ruled by the following commutation relations:
${\mathrm{D}}_{i}{x}_{i}={x}_{i}{\mathrm{D}}_{i}+1,\mathrm{for}i=1,...,n$
Any other pair of indeterminates commute.
• In the previous equation, x_i and D_i represent multiplication by x_i and differentiation with respect to x_i respectively. The (noncommutative) inner product in the Ore algebra represents the composition of operators. Therefore, the identity reduces to the Leibniz rule:
$\mathrm{diff}\left({x}_{i}f\left({x}_{1},\mathrm{...},{x}_{n}\right),{x}_{i}\right)={x}_{i}\mathrm{diff}\left(f\left({x}_{1},\mathrm{...},{x}_{n}\right),{x}_{i}\right)+f\left({x}_{1},\mathrm{...},{x}_{n}\right)$
• Since Weyl algebras are particular cases of Ore algebras, you can use most commands of the Ore_algebra package on Weyl algebras without knowing the definition of Ore algebras. For details, see Ore_algebra.
• More specifically, Weyl algebras are defined as operators with polynomial coefficients.
• The commands available for Weyl algebras are most of those of the Ore_algebra package, namely the following.
Building an algebra
Calculations in an algebra
Action on Maple objects
Converters
• The skew_algebra and diff_algebra commands declare new algebras to work with. They return a table needed by other Ore_algebra procedures. The diff_algebra command creates a Weyl algebra. The skew_algebra command creates a general Ore algebra, but can also be used to create a Weyl algebra. (The latter alternative is in fact more convenient in the case of Weyl algebras with numerous commutative parameters.)
• The skew_product and skew_power commands implement the arithmetic of Weyl algebras. Skew polynomials in a Weyl algebra are represented by commutative polynomials of Maple. The sum of skew polynomials is performed using the Maple + command. Their product, however, is performed using the skew_product command. Correspondingly, powers of skew polynomials are computed using the skew_power command.
• The rand_skew_poly command generates a random element of a Weyl algebra.
• The applyopr command applies an operator of a Weyl algebra to a function.
• The annihilators, skew_pdiv, skew_prem, skew_gcdex, and skew_elim commands implement a skew Euclidean algorithm in Weyl algebras and provide with related functionalities, such as computing remainders, gcds, (limited) elimination. The annihilators command makes it possible to compute a lcm of two skew polynomials. The skew_pdiv command computes pseudo-divisions in a Weyl algebra, while skew_prem simply computes corresponding pseudo-remainders. The skew_gcdex command performs extended gcd computation in a Weyl algebra. When possible, the skew_elim command eliminates an indeterminate between two skew polynomials.
Examples
> $\mathrm{with}\left(\mathrm{Ore_algebra}\right):$
> $A≔\mathrm{diff_algebra}\left(\left[\mathrm{Dx},x\right],\left[\mathrm{Dy},y\right],\left[\mathrm{Dz},z\right]\right):$
> $\mathrm{skew_product}\left(\mathrm{Dx},x,A\right)$
${\mathrm{Dx}}{}{x}{+}{1}$ (1)
> $\mathrm{skew_product}\left(\mathrm{Dy},y,A\right)$
${\mathrm{Dy}}{}{y}{+}{1}$ (2)
> $\mathrm{skew_product}\left(\mathrm{Dz},z,A\right)$
${\mathrm{Dz}}{}{z}{+}{1}$ (3)
> $\mathrm{skew_product}\left(\mathrm{Dx}\mathrm{Dy}\mathrm{Dz},xyz,A\right)$
${\mathrm{Dx}}{}{\mathrm{Dy}}{}{\mathrm{Dz}}{}{x}{}{y}{}{z}{+}{\mathrm{Dx}}{}{\mathrm{Dy}}{}{x}{}{y}{+}{\mathrm{Dx}}{}{\mathrm{Dz}}{}{x}{}{z}{+}{\mathrm{Dy}}{}{\mathrm{Dz}}{}{y}{}{z}{+}{\mathrm{Dx}}{}{x}{+}{\mathrm{Dy}}{}{y}{+}{\mathrm{Dz}}{}{z}{+}{1}$ (4)
> $\mathrm{skew_product}\left({\mathrm{Dx}}^{3},{x}^{5},A\right)$
${{\mathrm{Dx}}}^{{3}}{}{{x}}^{{5}}{+}{15}{}{{\mathrm{Dx}}}^{{2}}{}{{x}}^{{4}}{+}{60}{}{\mathrm{Dx}}{}{{x}}^{{3}}{+}{60}{}{{x}}^{{2}}$ (5) | 1,325 | 4,229 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "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} | 3.53125 | 4 | CC-MAIN-2021-21 | longest | en | 0.766173 |
https://www.electronicspoint.com/forums/threads/can-someone-idiot-proof-this-simple-circuit.244123/ | 1,618,650,311,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00380.warc.gz | 842,478,060 | 14,183 | # can someone idiot proof this simple circuit?
Discussion in 'General Electronics Discussion' started by maark6000, Feb 8, 2012.
1. ### maark6000
23
0
Feb 8, 2012
I'm trying to build a device that will slowly turn a wheel, but I'd like it to be variable speed.
So I bought a simple turntable motor (45RPM @ 24 Vdc, 60mA - operates at 12Vdc at 1/2 speed. Drive motor - Mabuchi RS-380SH. 1.1" diameter x 1.67". Gearbox - Molon #CHM-2435-1. 2.75" x 3.00" x 0.57" Threaded mounting holes in four corners. 5/16" diameter shaft, flatted and splined.)
and I want to power it with a duracell 12V battery...
but I'm thinking that in order to control the speed of the motor, I should do it using this circuit kit:
http://www.electronickits.com/kit/complete/motor/ck1400.htm
Is that right? Am I missing anything? doing anything dumb?
Thanks!!!
2. ### jackorocko
1,284
1
Apr 4, 2010
You could use a potential divider (potentiometer) and drop part of the voltage before it gets to the motor. That is by far the easiest route.
Even if you dropped the full voltage across the resistor that is 12V * .06A = .72W
That means a 1/2W linear potentiometer is all you would need and they come pretty cheap.
Or you could use PWM as the module in your link does, which is more efficient. But that one in the link is overkill at 7A. You need 1/100th of that current.
#### Attached Files:
• ###### pot_divider.GIF
File size:
3.8 KB
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Last edited: Feb 8, 2012
3. ### maark6000
23
0
Feb 8, 2012
Got it. Guess I was over concerned about what stresses would be put on the motor with just a potentiometer. So the circuit would simply be from battery to pot, pot to motor, and back to battery? Nothing else required?
4. ### Resqueline
2,848
2
Jul 31, 2009
I don't quite understand what's going on there.
A Mabuchi RS-380SH is usually a 3-6V motor that has a mechanical shaft output of up to 40W. It also has a no-load electrical consumption of almost 5W.
That doesn't quite rhyme with 24V & 60mA (= 1.44W).
23
0
Feb 8, 2012
6. ### Resqueline
2,848
2
Jul 31, 2009
Ok, it says the same as you, but it's hard to read the motor label on the picture. It could be a special hi-volt wound version. Still strange about the wattage though.
jacko's suggestion is cheap enough to try no matter what the outcome, but the PWM circuit should be bullet-proof no matter what.
57
1
Jan 28, 2010
8. ### maark6000
23
0
Feb 8, 2012
The label is even hard to read while holding it two feet in front of you.
RS-380SH
RD542/15 (although that slash could be a Y) | 774 | 2,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-17 | latest | en | 0.931425 |
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# contiguity of arrays
P: n/a Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); --Steve Nov 14 '05 #1
197 Replies
P: n/a Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); --Steve Yes it is legal and will print 4. This array describes a contiguously allocated nonempty set of objects of type int. b = a[0] positions the int *b to the beginning of the array object as would b = &a[0][0]. #include int main(void) { int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("a[2][2] = ((1,2},{3,4}};\nint *b;\n" "Using b = a[0]. *(b + 3) = %d\n", *(b + 3)); b = &a[0][0]; printf("Using b = &a[0][0]. *(b + 3) = %d\n",*(b+3)); printf("The pointer values &a[0][0] and a[0] are%sequal\n", (&a[0][0] == a[0])?" ":" not "); return 0; } -- Al Bowers Tampa, Fl USA mailto: xa******@myrapidsys.com (remove the x to send email) http://www.geocities.com/abowers822/ Nov 14 '05 #2
P: n/a Chris Torek wrote:Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); In article Al Bowers wrote:Yes it is legal and will print 4. I believe there are those in comp.std.c, at least, who will disagree with you. If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ then I believe you could access b[3]. -- pete Nov 14 '05 #4
P: n/a On 25 Sep 2004 17:21:37 GMT, Chris Torek wrote: Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3));In article Al Bowers wrote:Yes it is legal and will print 4.I believe there are those in comp.std.c, at least, who will disagreewith you. I could not follow your discussion which followed so I don't know exactly which point I'm about to disagree with. Instead I will try to present an argument that I believe shows the original question about legality must be answered in the affirmative. For an array q (not declared as a function parameter), sizeof q / sizeof *q must evaluate to the number of elements in b. For a pointer p to type T, the expression p+1 must evaluate to the same value as (T*)((char*)p + sizeof(T)) Therefore sizeof a must evaluate to the same value as 2*sizeof a[0] sizeof a[0] and sizeof a[1] both must evaluate to the same value as 2*sizeof a[0][0]. sizeof a[0][0] must evaluate to the same value as sizeof(int) sizeof a must evaluate to the same value as 4*sizeof(int) Since a contains 4 int and is exactly large enough to contain 4 int, these 4 int must be located at b, b+1, b+2, and b+3. snip <> Nov 14 '05 #5
P: n/a On 25 Sep 2004 22:48:34 GMT, in comp.lang.c , Barry Schwarz wrote: On 25 Sep 2004 17:21:37 GMT, Chris Torek wrote:Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3));In article Al Bowers wrote:Yes it is legal and will print 4.I believe there are those in comp.std.c, at least, who will disagreewith you.I could not follow your discussion which followed so I don't knowexactly which point I'm about to disagree with. Instead I will try topresent an argument that I believe shows the original question aboutlegality must be answered in the affirmative. (snip argument which is based on the size of the objects, and 1-d array arithmetic. ) The size argument is spurious. The compiler is allowed to tell you the object is size 4, even if all 4 members were in separate memory arrays in different solar systems. Theres no limit to the internal magic the compiler can perform. -- Mark McIntyre CLC FAQ CLC readme: ----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- Nov 14 '05 #6
P: n/a Barry Schwarz wrote: On 25 Sep 2004 17:21:37 GMT, Chris Torek wrote:Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3));In article Al Bowers wrote:Yes it is legal and will print 4.I believe there are those in comp.std.c, at least, who will disagreewith you. I could not follow your discussion which followed so I don't know exactly which point I'm about to disagree with. Instead I will try to present an argument that I believe shows the original question about legality must be answered in the affirmative. For an array q (not declared as a function parameter), sizeof q / sizeof *q must evaluate to the number of elements in b. For a pointer p to type T, the expression p+1 must evaluate to the same value as (T*)((char*)p + sizeof(T)) Therefore sizeof a must evaluate to the same value as 2*sizeof a[0] sizeof a[0] and sizeof a[1] both must evaluate to the same value as 2*sizeof a[0][0]. sizeof a[0][0] must evaluate to the same value as sizeof(int) sizeof a must evaluate to the same value as 4*sizeof(int) Since a contains 4 int and is exactly large enough to contain 4 int, these 4 int must be located at b, b+1, b+2, and b+3. The problem is that b, with (b = a[0]) is pointed at the first element of a two element array. If you have int a = 0, b = 0; then, you can't do this if (a + 1 == b) { /* The condition check is valid */ printf("%d\n", a[1]); /* Then printf call isn't valid */ } Whether or not there is an int object at (a + 1) is not the point. Overunning the bounds of the object, is the point. -- pete Nov 14 '05 #7
P: n/a pete wrote: Barry Schwarz wrote: On 25 Sep 2004 17:21:37 GMT, Chris Torek wrote:>Steve Kobes wrote:>> Is this legal? Must it print 4?>>>> int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0];>> printf("%d\n", *(b + 3));In article Al Bowers wrote:>Yes it is legal and will print 4.I believe there are those in comp.std.c, at least, who will disagreewith you. I could not follow your discussion which followed so I don't know exactly which point I'm about to disagree with. Instead I will try to present an argument that I believe shows the original question about legality must be answered in the affirmative. For an array q (not declared as a function parameter), sizeof q / sizeof *q must evaluate to the number of elements in b. For a pointer p to type T, the expression p+1 must evaluate to the same value as (T*)((char*)p + sizeof(T)) Therefore sizeof a must evaluate to the same value as 2*sizeof a[0] sizeof a[0] and sizeof a[1] both must evaluate to the same value as 2*sizeof a[0][0]. sizeof a[0][0] must evaluate to the same value as sizeof(int) sizeof a must evaluate to the same value as 4*sizeof(int) Since a contains 4 int and is exactly large enough to contain 4 int, these 4 int must be located at b, b+1, b+2, and b+3. The problem is that b, with (b = a[0]) is pointed at the first element of a two element array. If you have int a = 0, b = 0; then, you can't do this if (a + 1 == b) { /* The condition check is valid */ printf("%d\n", a[1]); /* Then printf call isn't valid */ } Major typos there. Should be: if (&a + 1 == &b) { /* The condition check is valid */ printf("%d\n", (&a)[1]); /* Then printf call isn't valid */ } Whether or not there is an int object at (a + 1) is not the point. Overunning the bounds of the object, is the point. -- pete Nov 14 '05 #8
P: n/a Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nov 14 '05 #9
P: n/a Hiho, (snip argument which is based on the size of the objects, and 1-d array arithmetic. ) The size argument is spurious. The compiler is allowed to tell you the object is size 4, even if all 4 members were in separate memory arrays in different solar systems. Theres no limit to the internal magic the compiler can perform. True. But I have yet to see the compiler that will correctly wrap _all_ calls to memcpy() and all other functions getting size_t and pointer arguments for this case. It is much easier to build non-broken programs the other way. Cheers Michael Nov 14 '05 #10
P: n/a E. Robert Tisdale wrote: Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nonsense. b is an int* pointing to an array of four int's. b[2] and b[3] are absolutely legal. -- Joe Wright mailto:jo********@comcast.net "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Nov 14 '05 #11
P: n/a Joe Wright wrote: E. Robert Tisdale wrote: Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nonsense. b is an int* pointing to an array of four int's. b[2] and b[3] are absolutely legal. Nov 14 '05 #12
P: n/a Joe Wright wrote: E. Robert Tisdale wrote: Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nonsense. b is an int* pointing to an array of four int's. b[2] and b[3] are absolutely legal. For int myobject[4]; myobject, is an object of type array of four int. You don't have an object of that type in the previous code. -- pete Nov 14 '05 #13
P: n/a Joe Wright wrote: E. Robert Tisdale wrote: Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nonsense. b is an int* pointing to an array of four int's. b[2] and b[3] are absolutely legal. No, it's not. b is an int * pointing to the first of to subsequent arrays of two ints. E. is quite correct that a. this is not the same thing as your version, b. strictly speaking this means that this code is not correct, and c. this difference is theoretical; you'd be hard put to find an implementation where this fails, perhaps excepting very strict debugging implementations. Richard Nov 14 '05 #14
P: n/a E. Robert Tisdale wrote: Joe Wright wrote: E. Robert Tisdale wrote: Steve Kobes wrote: Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nonsense. b is an int* pointing to an array of four int's. b[2] and b[3] are absolutely legal. How can it point to such an array? None has been defined in this program. C doesn't have multidimensional arrays, it has arrays of arrays, and this is one of the subtle differences between those two concepts. The named array 'a' contains two elements, which are themselves arrays. The unnamed array identified by a[0] is an array of only two elements, which are ints. The rules for the limits of pointer arithmetic are defined in terms of arrays, and according to the rules of the C language, 'a' is not an array of four integers. The elements of the elements of 'a' must be stored in the same way as the elements of an array of four integers. As a result, on most implementations pf C code like this works exactly as you expect. However, because this code has undefined behavior, an implementation is free to implement pointers in such a fashion that it can keep track of the limits beyond which they can't be dereferenced, and abort the problem if those limits are violated. In particular, when a[0] decays to a pointer, it's legal for the compiler to give that pointer dereferencing limits of a[0] and a[0]+1. There is special wording that allows any object, including an array of arrays, to be accessed completely using pointers to unsigned char. This is what makes memcpy() usable. However, for any other type this is an issue. Nov 14 '05 #15
P: n/a James Kuyper wrote: E. Robert Tisdale wrote: Joe Wright wrote: E. Robert Tisdale wrote: Steve Kobes wrote:> Is this legal? Must it print 4?>> int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0];> printf("%d\n", *(b + 3)); There is special wording that allows any object, including an array of arrays, to be accessed completely using pointers to unsigned char. This is what makes memcpy() usable. However, for any other type this is an issue. If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? -- pete Nov 14 '05 #16
P: n/a >>There is special wording that allows any object,including an array of arrays,to be accessed completely using pointers to unsigned char.This is what makes memcpy() usable.However, for any other type this is an issue. If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? There would. The same arguments hold. If you want to be sure that you use the right type of pointer, you have to work with something along the lines of int (*b)[2]. Nov 14 '05 #17
P: n/a James Kuyper wrote: E. Robert Tisdale wrote: Joe Wright wrote: E. Robert Tisdale wrote: Steve Kobes wrote:> Is this legal? Must it print 4?>> int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0];> printf("%d\n", *(b + 3)); Despite the fact that this works everywhere, neither C89 or C99 "guarantees" this. You can reference *(b+0) and *(b+1) and you can compute (b+2) but you can't reference *(b+2). It isn't "legal" to compute (b+3) much less reference *(b+3). If this "equivalence" is important to you, I don't think that you need to be too concerned about what the standard says. Nonsense. b is an int* pointing to an array of four int's. b[2] and b[3] are absolutely legal. How can it point to such an array? None has been defined in this program. C doesn't have multidimensional arrays, it has arrays of arrays, and this is one of the subtle differences between those two concepts. The named array 'a' contains two elements, which are themselves arrays. The unnamed array identified by a[0] is an array of only two elements, which are ints. The rules for the limits of pointer arithmetic are defined in terms of arrays, and according to the rules of the C language, 'a' is not an array of four integers. At a, there are four int objects, one after the other, having values 1, 2, 3 and four respectively. Looks like an array to me, even if undeclared. The elements of the elements of 'a' must be stored in the same way as the elements of an array of four integers. As a result, on most implementations pf C code like this works exactly as you expect. However, because this code has undefined behavior, an implementation is free to implement pointers in such a fashion that it can keep track of the limits beyond which they can't be dereferenced, and abort the problem if those limits are violated. In particular, when a[0] decays to a pointer, it's legal for the compiler to give that pointer dereferencing limits of a[0] and a[0]+1. Name one compiler enforces such limits. There is special wording that allows any object, including an array of arrays, to be accessed completely using pointers to unsigned char. This is what makes memcpy() usable. However, for any other type this is an issue. If I couldn't access a[1][0] as b[2] I would be surprised, and annoyed. -- Joe Wright mailto:jo********@comcast.net "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Nov 14 '05 #18
P: n/a Joe Wright wrote: If I couldn't access a[1][0] as b[2] I would be surprised, and annoyed. But you can. The *de facto* standard is as you describe it. No compiler developer would implement the standard *de jure* unless they were suicidal. The "legal" restrictions of the standard de jure are pure sophistry. Nov 14 '05 #19
P: n/a Michael Mair wrote:There is special wording that allows any object,including an array of arrays,to be accessed completely using pointers to unsigned char.This is what makes memcpy() usable.However, for any other type this is an issue. If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? There would. The same arguments hold. No, the same arguments don't hold. Is this legal? Must it print 4? int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; printf("%d\n", *(b + 3)); The problem originally was that b had the address of the first element of a[0]. a[0] has two elements, a[0][0] and a[0][1]. The address of b[3] is outside of a[0][1], which is to say that b[3] is beyond the boundary of a[0]. For b = (int *)a; b has the address of the first element of a, converted to int *. a has two elements, a[0] and a[1]. The address of b[3] is not outside of a[1], which is to say that b[3] is not beyond the boundary of a. -- pete Nov 14 '05 #20
P: n/a Joe Wright writes: James Kuyper wrote: [...] The elements of the elements of 'a' must be stored in the same way as the elements of an array of four integers. As a result, on most implementations pf C code like this works exactly as you expect. However, because this code has undefined behavior, an implementation is free to implement pointers in such a fashion that it can keep track of the limits beyond which they can't be dereferenced, and abort the problem if those limits are violated. In particular, when a[0] decays to a pointer, it's legal for the compiler to give that pointer dereferencing limits of a[0] and a[0]+1. Name one compiler enforces such limits. There may or may not be such a compiler. The point (especially in comp.std.c) is that any compiler is allowed to enforce such limits. -- Keith Thompson (The_Other_Keith) ks***@mib.org San Diego Supercomputer Center <*> We must do something. This is something. Therefore, we must do this. Nov 14 '05 #21
P: n/a pete wrote in message news:<41***********@mindspring.com>... .... If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? The validity limits of pointer arithmetic are defined in terms of the containing array. There is one and only one only array of int that contains the int object pointed at by 'b'. That array is a[0], and it only contains two integers. 'a' itself is not an array of 'int', but rather is an array of 'int[2]', and therefore is not capable of determining the validity limits for arithmetic on an 'int*'. Nov 14 '05 #22
P: n/a Keith Thompson wrote: Joe Wright writes:James Kuyper wrote: [...]The elements of the elements of 'a' must be stored in the same wayas the elements of an array of four integers. As a result, on mostimplementations pf C code like this works exactly as youexpect. However, because this code has undefined behavior, animplementation is free to implement pointers in such a fashion thatit can keep track of the limits beyond which they can't bedereferenced, and abort the problem if those limits are violated. Inparticular, when a[0] decays to a pointer, it's legal for thecompiler to give that pointer dereferencing limits of a[0] anda[0]+1.Name one compiler enforces such limits. There may or may not be such a compiler. Meaning that there *is* no such compiler. The point (especially in comp.std.c) is that any compiler is allowed to enforce such limits. Correct. The standard de jure allows compiler developers to commit suicide. We expect them to have more sense than that. Nov 14 '05 #23
P: n/a "E. Robert Tisdale" writes: Keith Thompson wrote: Joe Wright writes: [...]Name one compiler enforces such limits. There may or may not be such a compiler. Meaning that there *is* no such compiler. No, meaning that there may or may not be such a compiler. If you're assuming that I'm familiar with all existing C compilers, and that if there were one that does strict bounds checking I would know about it, your confidence is misplaced. -- Keith Thompson (The_Other_Keith) ks***@mib.org San Diego Supercomputer Center <*> We must do something. This is something. Therefore, we must do this. Nov 14 '05 #24
P: n/a "James Kuyper" wrote in message news:8b**************************@posting.google.c om... pete wrote in message news:<41***********@mindspring.com>... ... If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? The validity limits of pointer arithmetic are defined in terms of the containing array. There is one and only one only array of int that contains the int object pointed at by 'b'. That array is a[0], and it I believe the wording is for stand-alone arrays only, and a[0] is a part of continious object. The array above is a single object. Since a[1] + 2 can be pointed to, even b + 4 can be pointed to, and b[3] can certainly be addressed. Nov 14 '05 #25
P: n/a Joe Wright wrote: .... At a, there are four int objects, one after the other, having values 1, 2, 3 and four respectively. Looks like an array to me, even if undeclared. Key word: undeclared. If it is not declared as such in the C program, it doesn't count as such for purposes of determining what the C program is allowed/required to do with it. Name one compiler enforces such limits. I don't know whether are any, though I have vague memories of a compiler that provided such checking in a special debug mode. It would certainly be too expensive for use in the default mode. However, I don't care whether any compiler actually does this; for the purposes of comp.std.c, all I care about is whether compilers are allowed to do this. This is cross-posted to comp.lang.c, where the relevant concerns are different. .... If I couldn't access a[1][0] as b[2] I would be surprised, and annoyed. Your surprise and annoyance wouldn't render such a compiler non-conforming. Nov 14 '05 #26
P: n/a "E. Robert Tisdale" wrote: Joe Wright wrote: If I couldn't access a[1][0] as b[2] I would be surprised, and annoyed. But you can. The *de facto* standard is as you describe it. No compiler developer would implement the standard *de jure* unless they were suicidal. Or designing a _deliberately_ strict implementation, for example a debugging compiler. Richard Nov 14 '05 #28
P: n/a pete wrote: Michael Mair wrote:>There is special wording that allows any object,>including an array of arrays,>to be accessed completely using pointers to unsigned char.>This is what makes memcpy() usable.>However, for any other type this is an issue. If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? There would. The same arguments hold. No, the same arguments don't hold. Actually, yes, they do. Observe - fit the first: int a[2][2] = {{1, 2}, {3, 4}}, You now have an array of two arrays of int. *b = a[0]; You get the address of the first of these int arrays, convert it to an int pointer, and assign it to b. printf("%d\n", *(b + 3)); You invoke undefined behaviour by increasing _that int pointer_ beyond its legal boundary. Fit the second: int a[2][2] = {{1, 2}, {3, 4}}, You now have an array of two arrays of int. *b = a[0]; You get the address of the entire array, convert it to an int pointer, and assign it to b. printf("%d\n", *(b + 3)); You invoke undefined behaviour by increasing _that int pointer_ beyond its legal boundary. Note that: - the address of an array and the address of its first member are identical. - the entire array is properly aligned for ints, so a conversion of its base address (or the address of its first member, which is the same except for type) to int * must succeed and give the address of the first int in the array. - once a pointer has been converted to another pointer type, there is nothing in the Standard that allows you to deduce the original type. IOW, if you have int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a, *c = a[0]; then b==c _must_ evaluate to 1. The two pointers you get from those two conversions have exactly the same value, and exactly the same requirements. Note that this is _not_ true of a and a[0]; but this is because a has a different type than a[0]. Once you convert them both to int *, this distinction is, obviously, lost. Richard Nov 14 '05 #29
P: n/a James Kuyper wrote: Joe Wright wrote: If I couldn't access a[1][0] as b[2] I would be surprised, and annoyed. Your surprise and annoyance wouldn't render such a compiler non-conforming. James has correctly identified the issue: an "array" is more than just a contiguous sequence of similar objects; it also has a declared size. It is entirely possible that a compiler can generate more efficient code if it takes advantage of the declared size; for example, if an architecture has a 64KB limit on segment size and the array is declared smaller than that, then it will not be necessary to generate code that copes with segment boundaries (e.g. loading different segment base addresses for different parts of the array). An array of arrays is guaranteed to have the storage contiguously allocated (without extra padding), but not all elements of each array can be accessed by indexing off a pointer "based on" a pointer to a given element in a particular array. However, on many architectures there is no noticeable speed penalty involved in supporting that usage, due to a uniform, large memory space and wide-enough pointers Thus, some programmers have been getting away with this nonportable practice on the platforms they have used so far. Nov 14 '05 #30
P: n/a "James Kuyper" wrote in message news:41************@saicmodis.com... Ivan A. Kosarev wrote: > ... and a[0] is a part of > continious object. Agreed. If the validity limits of pointer arithmetic cared about contiguity, that would be a relevant argument. They don't. They're defined entirely in terms of the elements of a single array of the pointed-at type. None of types designates objects. :-) Instead, objects are memory areas which are interpreted accordingly to their types. Since that, it's not important how we get a pointer pair to compare it with relational operators; if they point to a single array (that is an *object*, not a *type*), they can be compared with a defined result. > The array above is a single object. Since a[1] + 2 can be pointed to, a[1]+2 is valid pointer value, which can be compared for equality with any valid pointer value, and compared for relative order with any other pointer that points into or one past the end of a[1]. However, it cannot Again, since the array is a single object, a[1] + 2 and b + 4 are values that point to the same object of the same type. Nov 14 '05 #31
P: n/a Douglas A. Gwyn wrote: Thus, some programmers have been getting away with this nonportable practice on the platforms they have used so far. If they are "getting away with" it, it's portable. Nov 14 '05 #32
P: n/a On Tue, 28 Sep 2004 16:11:16 -0700 in comp.std.c, "E. Robert Tisdale" wrote: Douglas A. Gwyn wrote: Thus, some programmers have been getting away with this nonportable practice on the platforms they have used so far.If they are "getting away with" it, it's portable. Not unless the standard says it is! -- Thanks. Take care, Brian Inglis Calgary, Alberta, Canada Br**********@CSi.com (Brian[dot]Inglis{at}SystematicSW[dot]ab[dot]ca) fake address use address above to reply Nov 14 '05 #33
P: n/a In article , E.**************@jpl.nasa.gov says... Douglas A. Gwyn wrote: Thus, some programmers have been getting away with this nonportable practice on the platforms they have used so far. If they are "getting away with" it, it's portable. Not at all. It simply means they haven't tried enough platforms yet. By your usage, anyone using envp as an argument to main is writing portable code, as long as they haven't tried it on a platform where it doesn't work yet. -- Randy Howard (2reply remove FOOBAR) Nov 14 '05 #34
P: n/a Randy Howard wrote: E.Robert.Tisdale wrote:Douglas A. Gwyn wrote:Thus, some programmers have been getting away with thisnonportable practiceon the platforms they have used so far.If they are "getting away with" it, it's portable. Not at all. It simply means they haven't tried enough platforms yet. cat main.c #include #include int main(int argc, char* argv[]) { const int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; const size_t n = sizeof(a)/sizeof(a[0][0]); for (size_t j = 0; j < n; ++j) fprintf(stdout, "b[%u] = %d\t", j, b[j]); fprintf(stdout, "\n"); return EXIT_SUCCESS; } gcc -Wall -std=c99 -pedantic -o main main.c ./main b[0] = 1 b[1] = 2 b[2] = 3 b[3] = 4 This program ports to every platform with a C99 compliant compiler. Nov 14 '05 #35
P: n/a "E. Robert Tisdale" wrote in message news:cj**********@nntp1.jpl.nasa.gov... This program ports to every platform with a C99 compliant compiler. Prove it. Nov 14 '05 #36
P: n/a "E. Robert Tisdale" wrote: Douglas A. Gwyn wrote: Thus, some programmers have been getting away with this nonportable practice on the platforms they have used so far. If they are "getting away with" it, it's portable. Please don't post on subjects you don't understand. Thanks. Nov 14 '05 #37
P: n/a James Kuyper wrote: pete wrote in message news:<41***********@mindspring.com>... ... If it was int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)a; ^^^^^^^ instead, there wouldn't be a problem accessing b[3], would there? The validity limits of pointer arithmetic are defined in terms of the containing array. There is one and only one only array of int that contains the int object pointed at by 'b'. That array is a[0], and it only contains two integers. 'a' itself is not an array of 'int', but rather is an array of 'int[2]', and therefore is not capable of determining the validity limits for arithmetic on an 'int*'. I don't see what difference it makes whether or not object a even contains an int type. As long as object a is as big as an int and also aligned for type int, I can access the object as (*(int*)&a) regardless if a was declared as a structure or an array of floats. With int *b = (int *)&a; b[3] isn't accessing an element of a, b[3] is accessing the memory at (int*)&a + 3, and treating it as an object of type int. -- pete Nov 14 '05 #38
P: n/a "Douglas A. Gwyn" wrote in message news:41***************@null.net... James Kuyper wrote: Joe Wright wrote: If I couldn't access a[1][0] as b[2] I would be surprised, and annoyed. Your surprise and annoyance wouldn't render such a compiler non-conforming. James has correctly identified the issue: an "array" is more than just a contiguous sequence of similar objects; it also has a declared size. It is entirely possible that a compiler can generate more efficient code if it takes advantage of the declared size; for example, if an Does this mean that pointers that are results of array-to-pointer conversion and any other pointers are somehow differ? If they don't, how the Standard allows such optimizations with the first and forbids with the second ones? (Hopefully, we will keep in mind that an abstract machine cannot refer to any optimization, including any kind of folding and propagation.) Nov 14 '05 #39
P: n/a Hi pete I don't see what difference it makes whether or not object a even contains an int type. As long as object a is as big as an int and also aligned for type int, I can access the object as (*(int*)&a) regardless if a was declared as a structure or an array of floats. With int *b = (int *)&a; b[3] isn't accessing an element of a, b[3] is accessing the memory at (int*)&a + 3, and treating it as an object of type int. Assuming a in this case is not an array of any flavour (or that you would have used the appropriate &a[0]...[0]): That is exactly the point! b[3] or b+3 accesses this address but it is not guaranteed that it may do so! You just might try to access memory which you do not have access to as it does not belong to the object you pointed b to... --Michael Nov 14 '05 #40
P: n/a Wojtek Lerch wrote: "E. Robert Tisdale" wrote:This program ports to every platformwith a C99 compliant compiler. Prove it. That shouldn't be difficult to do by exhaustive search ;-) -- David Hopwood Nov 14 '05 #41
P: n/a Hi E.R.T. > cat main.c #include #include int main(int argc, char* argv[]) { const int a[2][2] = {{1, 2}, {3, 4}}, *b = a[0]; const size_t n = sizeof(a)/sizeof(a[0][0]); for (size_t j = 0; j < n; ++j) fprintf(stdout, "b[%u] = %d\t", j, b[j]); fprintf(stdout, "\n"); return EXIT_SUCCESS; } > gcc -Wall -std=c99 -pedantic -o main main.c > ./main b[0] = 1 b[1] = 2 b[2] = 3 b[3] = 4 This program ports to every platform with a C99 compliant compiler. You are aware that gcc is not C99 compliant and that gcc does not even produce C*89* compliant programs when called with "gcc -Wall -std=c89 -pedantic ...", are you? That said: I also use mainly gcc but think it is unfit to "prove" anything. --Michael Nov 14 '05 #42
P: n/a In comp.lang.c Douglas A. Gwyn wrote: It is entirely possible that a compiler can generate more efficient code if it takes advantage of the declared size; for example, if an architecture has a 64KB limit on segment size and the array is declared smaller than that, then it will not be necessary to generate code that copes with segment boundaries (e.g. loading different segment base addresses for different parts of the array). char a[2][2]; I understand that this would be the wrong way to access a[1][0]: a[0][2]; //wrong An array of arrays is guaranteed to have the storage contiguously allocated (without extra padding), but not all elements of each array can be accessed by indexing off a pointer "based on" a pointer to a given element in a particular array. But I still don't see what should be wrong with this: char *b = a[0]; b[2]; //equiv to *(b+2) If that were not allowed, it would mean we cannot access object `a' through char*. -- Stan Tobias sed 's/[A-Z]//g' to email Nov 14 '05 #43
P: n/a Ivan A. Kosarev wrote: Does this mean that pointers that are results of array-to-pointer conversion and any other pointers are somehow differ? If the compiler can see the array context then it is allowed to take advantage of it, i.e. to generate code that works only for the span of the declared array, as in using a single segment base and not worrying about overflow in the offset field. Nov 14 '05 #44
P: n/a S.Tobias wrote: But I still don't see what should be wrong with this: char *b = a[0]; b[2]; //equiv to *(b+2) If that were not allowed, it would mean we cannot access object `a' through char*. Make that unsigned char *, to be safe. The guarantee that objects can be accessed as byte arrays is a special dispensation, distinct from the question of arrays of other types. So long as a *single object* can be identified, it can indeed be walked using a byte pointer derived from any kind of pointer into the object. An array of arrays is a single object (with identifiable subobjects). For non-byte pointers, the situation is more restricted; partly this is to allow more efficient code on some platforms (as I explained previously) and partly it is to allow type-based nonaliasing assumptions to be made by compilers (another matter of code efficiency). It may be instructive to consider the similar issue that came up some time ago, concerning what guarantees exist when two declared objects are accidentally contiguous: long a[10], b[10], *p = a, *q = b; if (a + 10 == b) stmt_1 else stmt_2 In stmt_1, can b[3] be accessed using a[13], or can a[6] be accessed using b[-4]? Clearly not. Then, can b[3] be accessed using p[13], or can a[6] be accessed using q[-4]? The consensus seems to be that such usage is not strictly conforming, and the compiler is not obliged to make such code work as a naive programmer might expect. It is hard to see what could go wrong with such tiny examples, especially if you're not very familiar with segment addressing, but if you start to think in terms of arrays near, say, 64KB in size the problems may become more evident. Nov 14 '05 #45
P: n/a Michael Mair wrote: Hi pete I don't see what difference it makes whether or not object a even contains an int type. As long as object a is as big as an int and also aligned for type int, I can access the object as (*(int*)&a) regardless if a was declared as a structure or an array of floats. With int *b = (int *)&a; b[3] isn't accessing an element of a, b[3] is accessing the memory at (int*)&a + 3, and treating it as an object of type int. Assuming a in this case is not an array of any flavour (or that you would have used the appropriate &a[0]...[0]): OK. That is exactly the point! b[3] or b+3 accesses this address but it is not guaranteed that it may do so! You just might try to access memory which you do not have access to as it does not belong to the object you pointed b to... I disagree. new.c is a portable program. /* BEGIN new.c */ #include int main(void) { int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)&a; if (b == (int *)&a[1][1] - 3) { puts("There's no chance that this program " "does not own the memory at b[3]"); } return 0; } /* END new.c */ -- pete Nov 14 '05 #46
P: n/a "Douglas A. Gwyn" wrote in message news:tb********************@comcast.com... Ivan A. Kosarev wrote: Does this mean that pointers that are results of array-to-pointer conversion and any other pointers are somehow differ? If the compiler can see the array context then it is allowed to take advantage of it, i.e. to generate Is the "as if" rule still working? Nov 14 '05 #47
P: n/a Hi pete, That is exactly the point! b[3] or b+3 accesses this addressbut it is not guaranteed that it may do so!You just might try to access memory which you do not haveaccess to as it does not belong to the object you pointed bto... I disagree. new.c is a portable program. /* BEGIN new.c */ #include int main(void) { int a[2][2] = {{1, 2}, {3, 4}}, *b = (int *)&a; if (b == (int *)&a[1][1] - 3) { puts("There's no chance that this program " "does not own the memory at b[3]"); } return 0; } /* END new.c */ Well, I am not one of the "chapter and verse" types but in this case it would be nice if you could explain it to me in the words of the standard as I do not understand how you conceive the idea this works portably. "Proof by program" works only for counterexamples. So, if I had a strange compiler or a strange enough platform and we both would agree on the compiler's standard compliantness, I could prove my point by running the above and getting an alternative output (when I provide an "else"...) Personally, I have shown my students the memory layout of a "two dimensional" array using the same means -- and always would, as it demonstrates this very well. However, I never assumed that this will work on some strange machine, too, and gave my students the "don't do this at home, children" warning... Cheers, Michael Nov 14 '05 #48
P: n/a In comp.lang.c Douglas A. Gwyn wrote: and partly it is to allow type-based nonaliasing assumptions to be made by compilers (another matter of code efficiency). It may be instructive to Aha, I think I see what you mean. long a[10], b[10], *p = a, *q = b; if (a + 10 == b) stmt_1 else stmt_2 I think what you mean to say is that compiler is allowed to assume that b[0] will not be changed between first assignment and if() condition, since p does not point within b[]: b[0] = 1; p[10] = 0; if (b[0]) always_executed(); else never_reached(); It is hard to see what could go wrong with such tiny examples, I think my example shows it. especially if you're not very familiar with segment addressing, but if you start to think in terms of arrays near, say, 64KB in size the problems may become more evident. Yes, this is second important problem. And, of course, these are same reasons why similar things will not work for arrays of arrays. One more question: int a[2][2] = {{0}}; void *pa0, *pa; pa0 = a[0]; pa = a; a[1][0] = 1; ((int*)pa0)[2] = 0; // (1) ((int*)pa)[2] = 0; // (2) Is it true, that after (1) compiler *may* assume that a[1][0] is still 1, because pa0 is "based" on a[0] (and implicitly has access only to that sub-object), and after (2) it *has* *to* assume that something changed in the entire a[][] array (and hence has to re-read a[1][0]), because pa is "based" on a and has access to the whole object? [ BTW. In my previous post I forgot to thank you for your explanations. Thanks to them I finally understood why the "struct hack" (as described in the Rationale 6.7.2.1) is not safe. ] -- Stan Tobias sed 's/[A-Z]//g' to email Nov 14 '05 #49
P: n/a "E. Robert Tisdale" wrote: Douglas A. Gwyn wrote: Thus, some programmers have been getting away with this nonportable practice on the platforms they have used so far. If they are "getting away with" it, it's portable. On the platform I currently use most, I can get away with TCHAR string[]=TEXT("This is a text"); and ok=(MessageBox(0, message, TEXT("Question"), MB_YESNO)!=IDYES); Does that mean those are portable, too? Richard Nov 14 '05 #50
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https://planetmath.org/basisfreedefinitionofdeterminant | 1,606,702,226,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00426.warc.gz | 429,027,872 | 6,281 | # basis-free definition of determinant
The definition of determinant as a multilinear mapping on rows can be modified to provide a basis-free definition of determinant. In order to make it clear that we are not using bases. we shall speak in terms of an endomorphism of a vector space over $k$ rather than speaking of a matrix whose entries belong to $k$. We start by recalling some preliminary facts.
Suppose $V$ is a finite-dimensional vector space of dimension $n$ over a field $k$. Recall that a multilinear map $f\colon V^{n}\to k$ is alternating if $f(x)=0$ whenever there exist distinct indices $i,j\in[n]=\{1,\dots,n\}$ such that $x_{i}=x_{j}$. Every alternating map $f\colon V^{n}\to k$ is skew-symmetric, that is, for each permutation $\pi\in\mathfrak{S}_{n}$, we have that $f(x)=\operatorname{sgn}(\pi)f(x^{\pi})$, where $x^{\pi}$ denotes $(x_{\pi(i)})_{i\in[n]}$, the result of $\pi$ permuting the entries of $x$.
Since the trivial map $0\colon V^{n}\to k$ is alternating and any linear combination of alternating maps is alternating, it follows that alternating maps form a subspace of the space of multilinear maps. In the following proposition we show that this subspace is one-dimensional.
###### Theorem.
Suppose $V$ is a finite-dimensional vector space of dimension $n$ over a field $k$. Then the space of alternating maps from $V^{n}$ to $k$ is one-dimensional.
###### Proof.
We use a basis here, but we will throw it away later. We need the basis here because each map we will consider has exactly as many elements as a basis of $V$. So let $B=\{b_{i}\colon i\in[n]\}$ be a basis of $V$.
Suppose $f$ and $g$ are nontrivial alternating maps from $V^{n}$ to $k$. We claim that $f$ and $g$ are linearly dependent. Let $x\in V^{n}$. We may assume that the entries of $x$ are basis vectors, that is, that $X=\{x_{i}\colon i\in[n]\}\subset\{b_{i}\colon i\in[n]\}$. If $X\subsetneq B$, then there exist distinct indices $i,j\in[n]$ such that $x_{i}=x_{j}$. Since $f$ and $g$ are alternating, it follows that $f(x)=g(x)=0$, which implies that $f(b)g(x)=g(b)f(x)$. On the other hand, if $X=B$, then there is a permutation $\pi\in\mathfrak{S}_{n}$ such that $x=b^{\pi}$. Since $f$ and $g$ are skew-symmetric, it follows that
$f(b)g(x)=\operatorname{sgn}(\pi)f(b)g(b)=g(b)f(x).$
In either case we find that $f(b)g(x)=g(b)f(x)$. Since $f(b)$ and $g(b)$ are fixed scalars, it follows that $f$ and $g$ are linearly dependent.
So far we have shown only that the dimension of the space of alternating maps is less than or equal to one. In order to show that the space is one-dimensional we simply need to find a nontrivial alternating form. To do this, let $\{b^{*}_{i}\colon i\in[n]\}$ be the natural basis of $V^{*}$, so that $b^{*}_{i}(b_{j})$ is the Kronecker delta of $i$ and $j$ for any $i,j\in[n]$. Define a map $f\colon V^{n}\to k$ by
$f(x)=\sum_{\pi\in\mathfrak{S}_{n}}\operatorname{sgn}(\pi)\prod_{i\in[n]}b^{*}_% {i}(x_{\pi(i)}).$
One can check that $f$ is multilinear and alternating. Moreover, $f(b)=1$, so it is nontrivial. Hence the space of alternating maps is one-dimensional. ∎
For an alternate view of the above results, we could look instead at linear maps from the exterior product $\bigwedge^{n}V$ into $k$. The proposition above can be viewed as saying that the dimension of $\bigwedge^{n}V$ is $\displaystyle\binom{n}{n}=1$.
We define the determinant of an endomorphism in terms of the action of the endomorphism on alternating maps. Recall that if $M\colon V\to V$ is an endomorphism, its pullback $M^{*}$ is the unique operator such that
$(M^{*}f)(x_{i})_{i\in[n]}=f(M(x_{i}))_{i\in[n]}.$
Since the space of alternating maps is one-dimensional and endomorphisms of a one-dimensional space reduce to scalar multiplication, it follows that $M^{*}f$ is a scalar multiple of $f$. We call this scalar the determinant. It is well-defined because the scalar depends on $M$ but not on $f$.
###### Definition.
Suppose $V$ is a finite-dimensional vector space of dimension $n$ over a field $k$, and let $M\colon V\to V$ be an endomorphism. Then the determinant of $M$ is the unique scalar $\det(M)$ such that
$M^{*}f=\det(M)f$
for all alternating maps $f\colon V^{n}\to k$.
Title basis-free definition of determinant BasisfreeDefinitionOfDeterminant 2013-03-22 16:51:39 2013-03-22 16:51:39 mps (409) mps (409) 9 mps (409) Definition msc 15A15 | 1,331 | 4,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 81, "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} | 3.59375 | 4 | CC-MAIN-2020-50 | longest | en | 0.834446 |
https://bukimyvomuwyle.gabrielgoulddesign.com/analysis-of-high-turnover-rate-31032ve.html | 1,632,532,357,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00230.warc.gz | 188,274,846 | 4,060 | # Analysis of high turnover rate
Someone has to sit there and show him what to do. If the assets financed by debt generate pretax net income sufficient to repay this interest, then any additional net income is profit that goes to the shareholders. As a rule, outstanding receivables should not exceed credit terms by days.
However, results can be distorted by manipulated retained earnings earned surplus data. In this case, the portfolio turnover is 4. If the result is percent or greater, your average inventory is not too high. Also determine the number of employees the company lets go over the same period of time.
This implies that a company is generating "x" number of sales for every dollar of assets on hand. Unless the superior asset selection renders benefits that offset the added transaction costs they cause, a less active trading posture may generate higher fund returns.
Profitability ratio can be determined on the basis of either sales or investment into business. For this example, the smaller amount is the sales. If the result is percent or greater, your average inventory is not too high. A comparison of this ratio with that of similar firms will throw light on the relative profitability and strength of the firm.
The Absolute Liquidity Ratio only tests short-term liquidity in terms of cash and marketable securities. A high working capital turnover ratio shows efficient use of working capital and quick turnover of current assets like stock and debtors. This equation is a basic formula for measuring how efficiently a company is operating.
Earnings per Share Formula Earnings per Share Calculator Use the Earnings per Share Calculator above to calculate the earnings per share from your financial statements. The lower ratio is better from the long-term solvency point of view. The sixth ratio, Cash Flow to Debt, is known as the best single predictor of failure.
The Absolute Liquidity Ratio only tests short-term liquidity in terms of cash and marketable securities. This post will focus on “how to measure inventory performance with analysis, why the analysis are important, How to calculate, what the formula is and what things you should note when you analyze on each“.
There are actually 32 common analysis/measuerements for inventory control purpose, but I will post four of them at first, they are: Inventory Accuracy, Inventory [ ]. You'll hear people talk about the high cost of turnover, but when you try to press for the actual costs they don't really know.
It seems like a mysterious thing that people talk about. Job Openings and Labor Turnover Survey Highlights June Bureau of Labor Statistics August 7, FINANCIAL RATIO ANALYSIS.
Financial ratio analysis involves the calculation and comparison of ratios which are derived from the information given in the company's financial statements. Ratios and Formulas in Customer Financial Analysis. Financial statement analysis is a judgmental process. One of the primary objectives is identification of major changes in trends, and relationships and the investigation of the reasons underlying those changes.
How to Calculate Turnover Rate. Understanding your firm's employee turnover rate is critical. High rates of turnover can damage employee morale and increase company costs significantly. You need to understand each type of employee.
Analysis of high turnover rate
Rated 3/5 based on 83 review | 645 | 3,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-39 | latest | en | 0.931135 |
http://www.cssforum.com.pk/css-past-papers/css-2018-papers/119000-general-science-ability-2018-paper-9.html | 1,680,334,209,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00322.warc.gz | 64,628,785 | 16,792 | Saturday, April 01, 2023
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CSS Forums General Science and Ability 2018 paper
#81
Tuesday, January 29, 2019
Member Join Date: Dec 2012 Posts: 36 Thanks: 6 Thanked 22 Times in 12 Posts
Quote:
Originally Posted by AamirLKR It was very easy. Ans A. all expenditure=100% then, others=100-15-8-10-26-15-10. So its answer was 16% Ans B. utilities=8/100=.08, rent=26/100=.26, food=15/100=.15 Ans C. As expenditure is 8% then, 11600*100/8=145000. Ans D. expenditure on others in 24000 income is 24000*16/100=3840. now others include three things one is saving which ratio is 3/8. So Saving will be 3840*3/8=1440.
I have a lil bit of confusion in this part of question. Can you please help me with it. We have to calculate savings from 'other' expenditure not from the monthly income 24000. I get that, but from the pie chart in the figure we know that 'other' represents 16% of total expenditure. Which is Rs.145000, not 16% of monthly income i.e 24000. Then why are we calculating 'other' expenses as 16/100*24000? Why not 16/100*145000? What if the figure 24000 in the statement is misleading?
If we assume 16% of 24000, that would disturb the whole pie chart and figures won't add up. I need urgent help with this question. Anyone please?
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#82
Wednesday, January 30, 2019
Member Join Date: Dec 2012 Location: Islamabad Posts: 70 Thanks: 34 Thanked 23 Times in 19 Posts
I think we have to treat part C and D separately. It's just the pie chart that's common. Plus, if you see the "others" section consists of "savings", so savings can't be considered as a form of expenditure. Hence we can safely apply the ratios of the chart to the income directly in part D too. and the answers come out 1440.
The Following User Says Thank You to A New Beginning For This Useful Post: Masood Ehsan Bhatti (Wednesday, January 30, 2019)
#83
Wednesday, January 30, 2019
Member Join Date: Dec 2012 Posts: 36 Thanks: 6 Thanked 22 Times in 12 Posts
Quote:
Originally Posted by A New Beginning I think we have to treat part C and D separately. It's just the pie chart that's common. Plus, if you see the "others" section consists of "savings", so savings can't be considered as a form of expenditure. Hence we can safely apply the ratios of the chart to the income directly in part D too. and the answers come out 1440.
But we can't treat C and D separately. Because C & D on their own don't give us the whole information required to calculate the questions asked in C and D part.
For example, in C part we are given the utilities expenditure (11600) only, and are asked to calculate the total expenditure. Total expenditure that represents the whole pie chart. For calculating that we need value 8/100 from the pie chart. So this part is not independent of the pie chart. P.s the amount of total expenditure 145000 (calculated in the C part) matches up with the pie chart. You calculate any expenditure out of 145000, you get the exact same %age value given in the pie chart. This part is fine.
Now for the D part we are only given the breakup of 'other' in ratio 3:4:1. To calculate 'other' we have to go back to the pie chart to get the 16% value of other, so we could calculate savings from 'other'. Hence this part is not independent of the pie chart either. Now if we are taking 16% from the pie chart_which represents 16% of 145000 ((total expenditure) and not 16% of income_ why are we not calculating it accordingly in the D part, that is 16/100* 145000 ? Because in that case figure would match up with the given pie chart.
Can't it be that 24000 is just given as distraction?
#84
Wednesday, January 30, 2019
Member Join Date: Dec 2012 Posts: 36 Thanks: 6 Thanked 22 Times in 12 Posts
I have solved many questions on pie chart. All are asked in 4-5 parts according to the information given in the pie chart. No matter what are you calculating and by what method, at the end all the calculated figures in all parts match up with the information given in the pie chart. This the only exception, i have seen here in the D part of this question. That is, you calculate 'other' out of 24000 and disturb the whole pie chart. The figures don't add up. Hence the confusion. Is there any way to confirm that the answer calculated out of 24000 in this thread is the correct answer?
#85
Wednesday, January 30, 2019
Member Join Date: Dec 2012 Location: Islamabad Posts: 70 Thanks: 34 Thanked 23 Times in 19 Posts
If the income is 24000 how can the expenditure be 1,45,000 (unless IMF is very lenient on this particular family) . If 24000 is the income then the percentages apply as they are. And the only thing the examiner is checking here, in my opinion, is the application of .16 x 3/8 x (whatever is the value regardless). So shouldn't worry much on this part and simply solve it on the face value of 24,000 income, which is expressly mentioned.
#86
Wednesday, January 30, 2019
Member Join Date: Dec 2012 Posts: 36 Thanks: 6 Thanked 22 Times in 12 Posts
Quote:
Originally Posted by A New Beginning If the income is 24000 how can the expenditure be 1,45,000 (unless IMF is very lenient on this particular family) . If 24000 is the income then the percentages apply as they are. And the only thing the examiner is checking here, in my opinion, is the application of .16 x 3/8 x (whatever is the value regardless). So shouldn't worry much on this part and simply solve it on the face value of 24,000 income, which is expressly mentioned.
Exactly my pont. And hence the source of my confusion. If monthly expenditure is 145000, how come the income can be 24000? Expenditure is 145000,we know that for sure. We have calculated that in part C. P.s you can also varify the figure 145000 by this pie chart. Which the statement says represents monthly expenditure.
8% of 145000 gives 11600= utilities
10% of 145000 gives 14500=local tax
26% of 145000 gives 37700= rent
15% of 145000 gives 21750= childcare
10% of 145000 gives 14500= travel
16%of 145000 gives 23200= other
15% of 145000 gives 217500= food
Add them all up; you get 145000, and your expenditure is varified as per this pie chart. So, how they gave the figure of 24000 for monthly income? Which doesn't fit anywhere in this pie chart. That's why i doubted that this figure could be misleading.
If i had been solving it in the exam hall, i also would have done it out of 24000. But that makes no sense. As in part D, they are asking you to calculate savings portion from 'other'. 'other' s value you have to get from the pie chart (16%). Which refers 16% of expenditure. But all of a sudden in part D. You are calculating 16% of income instead of expenditure.
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#87
Wednesday, January 30, 2019
Member Join Date: Dec 2012 Posts: 36 Thanks: 6 Thanked 22 Times in 12 Posts
They have disturbed the beauty of Maths in this question �� which lies in the very fact that whatever way you calculate the figures, they all add up ultimately as a whole without any discrepancy. I guess i'm done protesting here. ��. If in the exam hall i would also calculate it out of 24000, since the examiner is also likely to go with the popular opinion and i would prefer marks over keeping the beauty of maths intact.��
#88
Monday, September 02, 2019
Junior Member Join Date: Mar 2018 Location: Gujranwala Posts: 1 Thanks: 0 Thanked 0 Times in 0 Posts
Question combination
Salamm,
plz confirm that is this compulsory to attempt 2,2 questions from both part or can i attempt a 3, 1 combination??
#89
Sunday, November 24, 2019
Junior Member Join Date: Oct 2018 Location: Outer Space Posts: 10 Thanks: 19 Thanked 2 Times in 1 Post
Quote:
Originally Posted by Daniyal88 Most people at my centre also answered “9,000” for D which I think is wrong. Even the person who was masters in math. 1,440 is the right answer probably. Because you have to calculate ratio from ‘others’’ perspective. Not directly from 24,000. Sent from my iPhone using Tapatalk
I don't think 1440 is the right answer either. 16% was to be calculated of the total expenditure, not the income.
So 16% of 145,000 makes 21200, which when applied in ratio gives an answer of 1200.
__________________
If you don't stand for something, you will fall for anything ~ Malcom X
#90
Monday, July 13, 2020
Junior Member Join Date: Jul 2020 Location: Rawalpindi Posts: 3 Thanks: 1 Thanked 0 Times in 0 Posts
subject selection
i am going to opt ECONOMICS, BUSINESS ADMINISTRATION, GENDER STUDIES, CRIMINOLOY, PUNJABI..Can anyone please guide me that my optional subjects are fine or not beacuse i am beginner i have no idea about this. | 2,372 | 8,836 | {"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-2023-14 | latest | en | 0.946491 |
https://www.coursehero.com/file/6298822/Worksheet-5-Solutions/ | 1,512,954,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948511435.4/warc/CC-MAIN-20171210235516-20171211015516-00529.warc.gz | 727,019,877 | 23,994 | Worksheet 5 Solutions
# Worksheet 5 Solutions - MA 2930 Worksheet 5 1 Stability of...
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Unformatted text preview: MA 2930, Feb 23, 2011 Worksheet 5 1. Stability of critical points For each of the following autonomous equations find its critical points, deter- mine if they are stable/unstable/semi-stable, and draw the phase line as well as qualitative sketches of representative solutions. (1) dy/dt = y 2 ( y- 1)( y- 2) (2) dy/dt = (1 + y 2 ) arctan y (3) dy/dt = e y 2- 2 (4) dy/dt = sin 2 y (1) Critical points are the values of y for which dy/dt = 0. Clearly they are y = 0 , 1 and 2 in this case. To determine their stability we can either look at the sign of dy/dt on either side of them, or calculate f ( y ) and look at its sign at the critical point. Here the first method is easy to apply. If y < 0 (say y =- 1), y > 0. If 0 < y < 1 (say y = 0 . 5), y > 0 as well. So y = 0 is a semi-stable critical point: from below it attracts, but from above it repels. If 1 < y < 2, y < 0. So y = 1 is stable. It attracts the flow from either side. If y > 2, y > 0. So y = 2 is unstable. The system flows away from it on either side. I’ll let you draw the phase line etc. because I don’t have a program for drawing pictures! (2) Critical points are where (1+ y 2 ) arctan y = 0, i.e., where arctan y = 0, i.e., y = 0....
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Ask a homework question - tutors are online | 547 | 1,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-51 | latest | en | 0.833858 |
https://www.geogebra.org/m/fxakjsfg | 1,606,391,779,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00232.warc.gz | 687,425,346 | 9,505 | # Triangle Angle Theorems (V1)
Interact with the applet below for a few minutes. Then, answer the questions that follow. Be sure to change the locations of the triangle's WHITE VERTICES each time before you drag the slider!!!
Questions: 1) What geometric transformations took place in the applet above? 2) When working with the triangle's interior angles, did any of these transformations change the measures of the blue or green angles? 3) From your observations, what is the sum of the measures of the interior angles of any triangle? 4) When working with the triangle's exterior angles, did any of these transformations change the measures of the green or gray angles? 5) From your observations, what is the sum of the measures of the exterior angles of any triangle? | 179 | 789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-50 | latest | en | 0.877647 |
https://brainly.in/question/290602 | 1,484,720,776,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00137-ip-10-171-10-70.ec2.internal.warc.gz | 813,746,846 | 10,167 | # How to study the nature position and relative size of image formed by convex lens
2
2016-02-29T21:33:13+05:30
Fr obtaining position, use lens formula and find v. fr nature use formula m=v/u. if m is negative image is real and inverted and if it i is positive image is virtual and erect. fr finding size use frmula m=hi/ho
2016-03-01T14:25:47+05:30
To find out position-
1÷v=1÷f+1÷u
where, v= image distance
u= object distance
f= focal length
To find out the nature-
m = v÷u
where, m= magnification
If m is negative then nature is real and inverted.
If m is positive then nature is virtual an erect.
To find out size-
when v,u and ho ( object size) is given to find hi (image size)-
hi÷ho= v÷u | 212 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-04 | latest | en | 0.735873 |
http://www.cabot.net/investing-advice/options-trading/put-write-strategy | 1,474,948,295,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660957.45/warc/CC-MAIN-20160924173740-00004-ip-10-143-35-109.ec2.internal.warc.gz | 372,439,144 | 8,643 | # Put-Write Strategy
The S&P 500 lost 0.01% yesterday, which was the fourth straight day of losses for the market. That said, the losses were more significant in the morning but the market staged a rebound midway through the day.
When the market sells off (the S&P 500 2.5% off all-time highs is hardly a large selloff), many traders use the selloffs and the increase in option volatility to their advantage. One way to take advantage of these market conditions is to sell puts, also known as Put-Writes.
A Put-Write strategy is used when a rise in the price of the underlying asset is expected, or a significant decline is not expected.
This strategy is often used by traders who are willing to enter a long position in a stock at a price lower than the stock's current price.
This strategy entails selling a put at a specific strike price. It has the potential for loss until the stock hits zero. The maximum profit on the trade is the amount of premium received.
If I were to sell a put, and the stock went below my put's strike price, I would be assigned 100 shares per put I've sold, thus making me long 100 shares per put sold.
Yesterday, a large trader executed a Put-Write in Horizon Pharma (HZNP) in a massive way. With HZNP trading lower by \$0.85, at 31.50, a trader throughout the course of the day sold 21,000 June 30 Puts for an average price of \$0.30. The trade will expire in eight trading days, on Friday, June 19.
Here are the scenarios on how the trade will, or will not work:
If HZNP closes at 30 or above on June expiration, the trader will collect \$630,000. The math behind this is 21,000 multiplied by \$30 = \$630,000.
The breakeven on this trade is 29.70, as the trader collected \$0.30 with the sale of the puts. The math behind this is 30 minus \$0.30 = \$29.70.
If HZNP closes below 30, the trader, if he does not buy back the puts he sold, could be forced to buy 2,100,000 shares of HZNP, a capital outlay of \$63,000,000.
Here is the profit and loss graph of this trade:
I really like Put-Writes, as they're a great way to create yield in stocks I'm willing to own. But I don't like the risk/reward in this trade. While I think the trader will likely collect his \$630,000 next Friday, as order flow has been very bullish in HZNP for the past several weeks, the downside is too high if the market were to sell off in a big way.
This trade was likely executed by a large hedge fund or trading desk, which is perfectly willing to buy 2.1 million shares of HZNP at 30, and if the stock does not drop below 30, the trader will have created a great yield in a week and a half.
## Stock Picks
This undervalued stock has strong future earnings growth expectations.
Biogen is the market-share leader in treating multiple sclerosis.
One of Paul Godwin’s favorite stocks in his Cabot Emerging Markets Investor portfolio.
#### Tesla Model 3 vs. Chevy Bolt: Which Affordable Electric Car Is Better?
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#### As Stock Market Trends Change, Invest in these New Leaders
By Michael Cintolo on September 22, 2016
History tells us that all stock market trends change, and if you don’t recognize the leaders of that change early, you risk missing out on the next big winners.Read More > | 918 | 3,726 | {"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-2016-40 | longest | en | 0.964451 |
http://www.physicsforums.com/showpost.php?p=1367582&postcount=20 | 1,409,558,979,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535917463.1/warc/CC-MAIN-20140901014517-00418-ip-10-180-136-8.ec2.internal.warc.gz | 1,405,542,104 | 2,978 | View Single Post
Sci Advisor HW Helper P: 9,488 and again the second part is zero because of equality of mixed partials and the fact that dr^dt = -dt^dr. so we get just [df/dt dx/dr] dt^dr + df/dr dx/dt dr^dt] = [df/dr dx/dt - df/dt dx/dr] dr^dt, | 84 | 247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2014-35 | latest | en | 0.840125 |
http://math.stackexchange.com/questions/207005/how-many-numbers-will-i-get-right?answertab=active | 1,461,910,462,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860110764.59/warc/CC-MAIN-20160428161510-00176-ip-10-239-7-51.ec2.internal.warc.gz | 116,161,009 | 18,562 | # How many numbers will I get right?
I play a guessing game. In this game, there are 100 equally-sized, folded-up cards randomly dispersed in a bag. The cards are labeled 1 through 100. I draw out the cards one by one and try to guess the number on the card every time I draw. On every guess, a genie will tell me if I am correct or not (but he won't tell me the actual number on the card).
I start by guessing that the card has 1 on it. I keep guessing that the card has 1 on it until I am correct, after which I will keep guessing 2 until I am correct again, after which I will keep guessing 3, and so on until the bag is empty. How many correct guesses should I expect?
My current hunch is that the answer is 1 since this looks like a binomial distribution, and $52\frac{1}{52} = 1$. However, I feel suspicious since there's that genie that gives me extra information...
-
Are you asking about that particular strategy, or an optimal strategy? Even for the $1,1,\dots,2,\dots$ strategy, the expectation is $\gt 1$, since for sure you will get at least $1$ right, and maybe more. – André Nicolas Oct 4 '12 at 6:35
Just for that particular "strategy." It would be interesting to look at the optimal strategy... but I think that might open up a new can of worms. :) – Minden Petrofsky Oct 4 '12 at 6:37
I think this is the optimal strategy. The most likely next draw is the one you've guessed most often so far but haven't yet seen. – mjqxxxx Oct 4 '12 at 6:44
What happens if you draw 2 before you draw 1? Do you then guess 3 (or whatever number is the next unseen one) once you find 1? If so, that would be optimal. If you don't skip the numbers that you've seen, then it's definitely not optimal. – user22805 Oct 4 '12 at 7:48
@David Wallace - if he is not correct, the genie will not tell him the actual number on the card. So he will only know that a sequence of "not-1"s was followed by a "1". – Viliam Búr Oct 4 '12 at 11:16
You’re certain to get at least one correct guess, and in many cases you’ll get more than one, so the expected number of correct guesses is clearly more than one. You’ll get a second correct guess whenever you draw the $1$ before you draw the $2$; the probability of that is $\frac12$, so the expected number of correct guesses can already be seen to be at least $\frac12\cdot1+\frac12\cdot2=\frac32$.
In general you’ll get at least $n$ correct guesses if the numbers $1,\dots,n$ appear in their correct order, ignoring any larger numbers that may appear between them. The probability of that is $\frac1{n!}$, since all $n!$ orders in which they could appear are equally likely. The probability that $n+1$ is the last of the set $\{1,\dots,n+1\}$ to be drawn is $\frac1{n+1}$, so the probability that it is drawn before you draw all of the smaller numbers is $\frac{n}{n+1}$.
Thus, the probability that you’ll get exactly $n$ correct guesses is $$\frac1{n!}\cdot\frac{n}{n+1}=\frac{n}{(n+1)!}\;,$$ and the expected number of correct guesses is
$$\sum_{n=1}^{100}\frac{n}{(n+1)!}\cdot n=\sum_{n=1}^{100}\frac{n^2}{(n+1)!}\;.$$
Now $$e^x=\sum_{n\ge 0}\frac{x^n}{n!}\;,$$ so $$\frac{e^x-1}x=\sum_{n\ge 1}\frac{x^{n-1}}{n!}=\sum_{n\ge 0}\frac{x^n}{(n+1)!}\;,$$
$$x\frac{d}{dx}\left(\frac{e^x-1}x\right)=\sum_{n\ge 0}\frac{nx^n}{(n+1)!}\;,$$ and
$$\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{e^x-1}x\right)\right)=\sum_{n\ge 0}\frac{n^2x^{n-1}}{(n+1)!}\;.\tag{1}$$
Finally, $$\frac{d}{dx}\left(x\frac{d}{dx}\left(\frac{e^x-1}x\right)\right)=\frac{d}{dx}\left(\frac{xe^x-e^x+1}x\right)=e^x-\frac{(x-1)e^x}{x^2}-\frac1{x^2}\;.\tag{2}$$
Evaluating the righthand sides of $(1)$ and $(2)$ at $x=1$, we see that
$$\sum_{n\ge 1}\frac{n^2}{(n+1)!}=e-1\;,$$ so the expected number of correct guesses is just a hair less than $e-1\approx 1.718281828459045$.
-
Though you don't say so, I assume that cards are discarded once they're removed from the bag. The key observation is that you will get at least $n$ correct guesses if and only if the first $n$ cards (that is, the cards numbered $1,2,\ldots,n$) appear in the correct order among your draws. Since each of the $n!$ orderings of the first $n$ cards within your random list of draws is equally likely, you'll get at least $n$ correct guesses with probability $(1/n!)$. Therefore the probability of getting exactly $n$ correct guesses must be $$\frac{1}{n!}-\frac{1}{(n+1)!}=\frac{1}{n!}\left(1-\frac{1}{n+1}\right)=\frac{n}{(n+1)!},$$ and the expected number of correct guesses is $$\sum_{n=1}^{100}\frac{n^2}{(n+1)!} \approx (e-1) = 1.718281828\ldots$$
- | 1,442 | 4,543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.734375 | 4 | CC-MAIN-2016-18 | latest | en | 0.961855 |
http://paradiseo.gforge.inria.fr/index.php?n=Doc.TutoMOLesson5 | 1,606,734,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141213431.41/warc/CC-MAIN-20201130100208-20201130130208-00260.warc.gz | 73,274,710 | 3,685 | # How to use Iterated Local Search
In this lesson, an Iterated Local Search is presented. The Tabu Search of the Lesson 4 is used with an order neighborhood based on a shift operator, to solve the Queen problem.
1. Iterated Tabu Search on the Queen problem.
2. Exercise
1. Iterated Tabu Search (example on the Queen problem)
As in Lesson 4, you have to define a Solution, the method to initialize and evaluate it. Then you have to define a Tabu Search.
Declaration of the Tabu Search:
``` moTS<shiftNeighbor> ts(orderShiftNH, fullEval, shiftEval, 1, 7);
```
To use a simple Iterated Local Search, a mutation operator is needed. So the swap mutation defined in EO is used:
``` eoSwapMutation<Queen> mut;
```
Now, a simple Iterated Tabu Search can be declared as follow:
``` moILS<shiftNeighbor> localSearch1(ts, fullEval, mut, 3);
```
This constructor has got 4 parameters:
1. a local search (ts)
2. a full evaluation function (fullEval)
3. a mutation operator (mut)
4. a number of iterations (3)
localSearch1 performs the Tabu Search 3 times. The first solution of each iteration(except the first one) is obtained by applying the mutation operator on the last visited solution.
A constructor allows to specify the continuator. Be carefull, the continuator must be templatized by a "moDummyNeighbor":
``` moIterContinuator<moDummyNeighbor<Queen> > cont(4, false);
```
The explorer of the Iterated local search don't use its own neighborhood. Here, the neighborhood of the Tabu Search is used. But to respect the conception, we create a "moDummyNeighbor" using as template for Iterated Local Search.
An Iterated Tabu Search with this continuator can be declared as:
``` moILS<shiftNeighbor> localSearch2(ts, fullEval, mut, cont);
```
A general constructor is available allowing to specify the perturbation operator and the acceptance criteria. First, you have to declare a perturbation operator:
``` moMonOpPerturb<shiftNeighbor> perturb(mut, fullEval);
```
And, the acceptance criteria:
``` moSolComparator<Queen> solComp;
moBetterAcceptCrit<shiftNeighbor> accept(solComp);
```
Finally, the Iterated Local Search can be declared as:
``` moILS<shiftNeighbor> localSearch3(ts, fullEval, cont, perturb, accept);
```
You can test these three algorithms by changing problem sizes(use parameter file or the option --vecSize=X on command line to execute "testILS"). It prints the initial and the final solutions.
2. Exercise
• Try to implement an Iterated Hill Climbing on the Queen problem with these caracteristics:
1. Hill Climbing with a "moShiftNeighborhood" and a "moTrueContinuator"
2. Iterated Local Search using a "moIterContinuator" and a "moNeighborhoodPerturb" with a "moSwapNeighborhood". | 671 | 2,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | latest | en | 0.826608 |
https://www.mathworks.com/matlabcentral/cody/problems/148-factorize-this-buddy/solutions/170874 | 1,506,450,754,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696677.93/warc/CC-MAIN-20170926175208-20170926195208-00667.warc.gz | 798,529,795 | 11,497 | Cody
# Problem 148. Factorize THIS, buddy
Solution 170874
Submitted on 30 Nov 2012 by Takehiko KOBORI
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
%% x = 31; y_correct = 31; assert(isequal(unique_prime_factors(x),y_correct))
``` ```
2 Pass
%% x = 37; y_correct = 37; assert(isequal(unique_prime_factors(x),y_correct))
``` ```
3 Pass
%% x = 39; y_correct = [13 3]; assert(isequal(unique_prime_factors(x),y_correct))
``` ```
4 Pass
%% x = 876; y_correct = [73 3 2]; assert(isequal(unique_prime_factors(x),y_correct))
``` ``` | 200 | 661 | {"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-2017-39 | latest | en | 0.474621 |
https://cs.indstate.edu/wiki/index.php?title=Applying_euclid%27s_algorithm_for_gcd&oldid=3092 | 1,669,610,573,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00380.warc.gz | 230,879,526 | 6,274 | Applying euclid's algorithm for gcd
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Euclid's algorithm for the greatest common divisor is often among the first algorithms seen that has a significantly faster run time than the naive algorithm (exponentially faster, in fact). A traditional assignment is to ask you to work through the application of Euclid's algorithm on a pair of integers that are modestly large.
Example
Consider the integers 12345 and 67890. We let a=67890, b=12345, and apply the Euclidian algorithm in the following steps.
• a = 5*b + 6165, so set a=12345 and b=6165
• a = 2*b + 15, so set a=6165 and b=15
• a = 441*b + 0, so gcd(a, b) = b
We conclude that gcd(67890, 12345) = 15. Note that we could perform the divisions required by hand or used a calculator (or other means), but we have showed how we work through the algorithm until we have the gcd at the end (when we get a 0 remainder).
Assignment
You will be assigned values of a and b and asked to work out the gcd in the same way as above.
Pass rating check You should follow the same steps as above, and include a similar amount of text as above so that you have clearly stated what you are doing, and someone can verify that you have applied Euclid's gcd algorithm correctly. You need to both hand in your work, and also check in with the help lab to demonstrate the algorithm. You can use your work that you hand in, and should talk the GA through the algorithm. The GA will make note if you are not able to do this, or if it seems you do not fully understand the process.
Note - the shared spreadsheet that GAs use for submitting information to Jeff Kinne's courses is this link, which should work only for the current term's GAs. | 428 | 1,752 | {"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} | 3.90625 | 4 | CC-MAIN-2022-49 | latest | en | 0.954453 |
https://your-physicist.com/first-law-of-thermodynamics-examples/ | 1,718,524,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00210.warc.gz | 972,840,412 | 14,024 | # Introduction to the First Law of Thermodynamics
The First Law of Thermodynamics is the fundamental law of energy conservation for a closed system. It states that the total amount of energy in a closed system is constant, and that energy can neither be created nor destroyed, but only transformed from one form to another. This law is essential to the study of thermodynamics, as it helps to understand and quantify the behavior of energy in various systems.
The First Law of Thermodynamics is a statement of energy conservation in a closed system. For example, in a closed system, the energy that is added to the system must be accounted for, either as work done on the system or as an increase in the internal energy of the system, or both. Conversely, the energy that is removed from the system must also be accounted for, either as work done by the system or as a decrease in the internal energy of the system, or both.
In the context of the First Law of Thermodynamics, work is defined as the transfer of energy due to a force acting through a distance, while heat is the transfer of energy due to a temperature difference. Both work and heat are forms of energy transfer, and they can be quantified in terms of the units of energy, such as joules or calories. Now that we have an understanding of the First Law of Thermodynamics, let’s look at some examples.
# Example 1: Heat and Work in a Piston-Cylinder System
Consider a piston-cylinder system containing a gas that is initially at a pressure of P1 and a volume of V1. The system is isothermally compressed to a final volume of V2. According to the First Law of Thermodynamics, the change in the internal energy of the system is equal to the work done on the system plus the heat added to the system.
In this case, the internal energy of the gas remains constant, since the process is isothermal. Therefore, the change in the internal energy is zero. The work done on the system is given by the area under the curve on a P-V diagram. The heat added to the system is equal to the work done on the system, since the internal energy remains constant. Therefore, the total energy of the system is conserved, in accordance with the First Law of Thermodynamics.
# Example 2: Conservation of Energy in a Chemical Reaction
Chemical reactions involve the transformation of energy from one form to another. According to the First Law of Thermodynamics, the total amount of energy in a system is conserved, and this is true for chemical reactions as well.
For example, the combustion of methane can be represented by the equation: CH4 + 2 O2 -> CO2 + 2 H2O. In this reaction, the reactants (methane and oxygen) have a certain amount of energy, while the products (carbon dioxide and water) have a different amount of energy. The energy difference between the reactants and the products is released as heat, which can be used to do work or to transfer energy to another system.
The First Law of Thermodynamics ensures that the total amount of energy in the system is conserved, regardless of how it is transformed or transferred. This means that the energy released in a chemical reaction must be accounted for, either as work done by the system, heat transferred to the surroundings, or an increase in the internal energy of the system.
# Example 3: Energy Transfer in an Electrical Circuit
Electrical circuits involve the transfer of energy from a power source to a load, such as a light bulb or a motor. According to the First Law of Thermodynamics, the total amount of energy in the system is conserved, and this is true for electrical circuits as well.
For example, consider a simple circuit consisting of a battery, a resistor, and a switch. When the switch is closed, the battery supplies a certain amount of energy to the circuit, which is then dissipated as heat in the resistor. The heat generated in the resistor is equal to the power dissipated, which is given by the product of the voltage across the resistor and the current flowing through it.
The First Law of Thermodynamics ensures that the total amount of energy in the system is conserved, regardless of how it is transformed or transferred. This means that the energy supplied by the battery must be accounted for, either as work done by the circuit, heat dissipated in the resistor, or an increase in the internal energy of the circuit.
In conclusion, the First Law of Thermodynamics is a fundamental law of energy conservation for closed systems. It states that the total amount of energy in a closed system is constant, and that energy can neither be created nor destroyed, but only transformed from one form to another. The examples we have examined demonstrate how the First Law of Thermodynamics applies to various systems, including piston-cylinder systems, chemical reactions, and electrical circuits. | 995 | 4,843 | {"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} | 3.828125 | 4 | CC-MAIN-2024-26 | latest | en | 0.957827 |
http://mathematica.stackexchange.com/questions/3854/numerical-underflow-for-a-scaled-error-function/3856 | 1,469,549,271,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00279-ip-10-185-27-174.ec2.internal.warc.gz | 164,606,168 | 23,135 | Numerical underflow for a scaled error function
I calculate scaled error function defined as
f[x_] := Erfc[x]*Exp[x^2]
but it can not calculate f[30000.]. f[20000.] is not very small (0.0000282). I think Mathematica is supposed to switch to high precision instead of machine precision, but it does not. It says:
General::unfl: Underflow occurred in computation. >>
General::ovfl: Overflow occurred in computation. >>
How can I calculate f for large values of x? Even with N[f[30000], 50], it does not use high precision and fails.
-
Mathematica doesn't switch to arbitrary precision like you seem to believe. If you enter a machine precison number like 30000. all further calculations are done in machine precision. You may want to read some of the tutorials on the bottom of this page. – Sjoerd C. de Vries Apr 3 '12 at 9:38
If you have an analytic formula for f[x_] := Erfc[x]*Exp[x^2] not using Erfc[x] you could do what you expect. However it is somewhat problematic to do in this form because Erfc[x] < $MinNumber for x == 27300. $MinNumber
1.887662394852454*10^-323228468
N[Erfc[27280.], 20]
5.680044213569341*10^-323201264
Edit
A very good approximation of your function f[x] for values x > 27280 you can get making use of these bounds ( see e.g. Erfc on Mathworld) :
which hold for x > 0.
Here we find values of the lower and upper bounds with relative errors for various x:
T = Table[
N[#, 15]& @ {2 /(Sqrt[Pi] (x + Sqrt[2 + x^2])),
2 /(Sqrt[Pi] ( x + Sqrt[x^2 + 4/Pi])),
1 - ( x + Sqrt[x^2 + 4/Pi])/(x + Sqrt[2 + x^2]),
{x, 30000 Table[10^k, {k, 0, 5}]}];
Grid[ Array[ InputField[ Dynamic[T[[#1, #2]]], FieldSize -> 13] &, {6, 3}]]
Therefore we propose this definition of the function f (namely the arithetic mean of its bounds for x > 27280 ) :
f[x_]/; x >= 0 := Piecewise[ { { Erfc[x]*Exp[x^2], x < 27280 },
{ 1 /( Sqrt[Pi] ( x + Sqrt[2 + x^2]))
+ 1 /( Sqrt[Pi] ( x + Sqrt[x^2 + 4/Pi])), x >= 27280}}
]
f[x_] /; x < 0 := 2 - f[-x]
I.e. we use the original definition of the function f for 0 < x < 27280, the approximation for x > 27280 and for x < 0 we use the known symmetry of the Erfc function, which is relevant when we'd like to calculate f[x] for x < - 27280. Now we can safely use this new definition for a much larger domain :
{f[300], f[300.], f[30000.], f[-30000.]}
{E^90000 Erfc[300], 0.0018806214974, 0.0000188063, 1.99998}
and now we can make plots of f around of the gluing point ( x = 27280.)
GraphicsRow[{ Plot[ f[x], {x, 2000, 55000},
Epilog -> {PointSize[0.02], Red, Point[{27280., f[27280.]}]},
PlotStyle -> Thick, AxesOrigin -> {0, 0}],
Plot[ f[x], {x, 27270, 27290},
Epilog -> {PointSize[0.02], Red, Point[{27280., f[27280.]}]},
PlotStyle -> Thick]}]
-
But why Mathematica does not switch to arbitrary precision, then? – こじま Apr 3 '12 at 9:15
@こじま From the docs: $MinNumber gives the minimum positive arbitrary-precision number that can be represented on a particular computer system. By the way: could you perhaps choose a user name that one can easily type using a standard keyboard? It's pretty difficult to @-refer to you in this way. – Sjoerd C. de Vries Apr 3 '12 at 9:24 I've had to work with that kind of function (relying on cancellation of large terms) before, and the most practical workaround I could figure out to be able to evaluate the function numerically is to use its power expansion near the point of trouble (here,$+\infty$). So, get a good look at the series expansion and find out how it works (or derive it on paper): Series[f[x], {x, ∞, 50}] So, that means that you can use $$f(x) \approx \sum_{n=0}^N \left(-\frac{1}{2}\right)^n \frac{(2n-1)!!}{x^{2n+1}}$$ for$x$larger than some value$X$. All that remains is to find a couple of values$(X,N)$suitable to your needs. Because we have a series with signs alternating and terms decrease, the error is bounded by the$(n+1)$th term. Assuming we want to choose a value of$X$in the range [20,1000], we plot the relative error after the$N$th term as a function of$x$in this range: LogLogPlot[Table[(2 n - 1)!!/x^(2 n + 1)/f[x], {n, 5, 10}], {x, 20, 1000}] So, say we want to have a relative accuracy of$10^{-30}$(which is better than machine precision), we can for example take$X=100$and$N=10$. This gives us the following definition for your f function: f[x_] := Piecewise[{{Erfc[x]*Exp[x^2], x < 100}}, 1/Sqrt[π]*Sum[(-1/2)^n*(2 n - 1)!!/x^(2 n + 1), {n, 0, 10}]]; LogLogPlot[{f[x], Erfc[x]*Exp[x^2]}, {x, 10, 10^6}, PlotStyle -> {Red, Directive[Blue, Thick, Opacity[0.4]]}] - This is a good idea I also wanted to work on the same direction after the university hours! Nice job... – PlatoManiac Apr 3 '12 at 10:42 There is also the possibility of constructing a Padé approximant from your series. For instance, PadeApproximant[Erfc[x] Exp[x^2], {x, Infinity, {4, 4}}] yields an approximant with absolute relative error$< 10^{-13}$for$x > 50$. – J. M. Apr 17 '12 at 14:09 For numerical evaluation, there is the rapidly-converging continued fraction (due to Jones and Thron): $$\exp(x^2)\mathrm{erfc}(x)=\frac{2x}{\sqrt \pi}\cfrac{1}{2x^2+1-\cfrac{1\cdot2}{2x^2+5-\cfrac{3\cdot4}{2x^2+9-\cdots}}},\qquad x > 0$$ One can use the built-in function ContinuedFractionK[] with a suitable cut-off: With[{x = N[30000], n = 10}, -2 x /(1 + 2 x^2 + ContinuedFractionK[2 j (1 - 2 j), 1 + 4 j + 2 x^2, {j, 1, n}])/ Sqrt[Pi]] 0.0000188063 or, even better, use the Lentz-Thompson-Barnett algorithm for evaluating this continued fraction, avoiding unneeded evaluation effort: f[z_?InexactNumberQ] := Module[{c, d, h, k, u, v, y}, y = v = 2 z^2 + 1; c = y; d = 0; k = 1; While[True, u = k (k + 1); v += 4; c = v - u/c; d = 1/(v - u d); h = c*d; y *= h; If[Abs[h - 1] <= 10^-Precision[z], Break[]]; k += 2]; 2 z/y/Sqrt[Pi]] /; Re[z] > 0 With[{z = N[50]}, {Exp[z^2] Erfc[z], f[z]}] {0.01128153626532, 0.0112815} N[f[30000], 30] 0.0000188063194411439209981315314042 Plot[f[x], {x, 1, 50}] - In the interest of showing that there's more than one way to skin a cat, I present a method suitable for large positive arguments, due to Chiarella and Reichel. The method uses the approximation $$\exp(z^2)\mathrm{erfc}(z)\approx\frac{hz}{\pi}\left(z^{-2}+2\sum_{k \geq 1}\frac{\exp(-h^2 k^2)}{z^2+h^2 k^2}\right)$$ where$h$is a suitably chosen parameter, based on the precision needed. f[z_?InexactNumberQ] := Module[{prec = Precision[z], y = z^2, e, h, j, k, s, t}, h = Pi/Sqrt[(Round[prec] + 1) Log[10]]; e = h^2; s = 0; j = k = 1; While[True, t = Exp[-e k]/(y + e k); s += t; If[Abs[t] <= Abs[s] 10^-prec, Break[]]; j += 2; k += j]; h z (1/y + 2 s)/Pi] /; TrueQ[Quiet[z > 0]] Again, this works best for large positive$z$, which seems to be the arguments of interest for the OP anyway. If evaluation for small$z\$ is needed, a correction term has to be added to the Chiarella-Reichel approximation; see their paper for details.
- | 2,341 | 6,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2016-30 | latest | en | 0.779556 |
https://tolstoy.newcastle.edu.au/R/devel/05/11/3283.html | 1,591,418,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00159.warc.gz | 557,487,618 | 4,354 | # Re: [Rd] x[1,], x[1,,], x[1,,,], ...
From: Tony Plate <tplate_at_acm.org>
Date: Thu 24 Nov 2005 - 02:33:46 GMT
Henrik Bengtsson wrote:
> Hi, thanks everyone.
>
>
> Peter Dalgaard wrote:
>
```>>Henrik Bengtsson <hb@maths.lth.se> writes:
>>
>>
>>
>>>Hi,
>>>
>>>is there a function in R already doing what I try to do below:
>>>
>>># Let 'x' be an array with *any* number of dimensions (>=1).
>>>x <- array(1:24, dim=c(2,2,3,2))
>>>...
>>>x <- array(1:24, dim=c(4,3,2))
>>>
>>>i <- 2:3
>>>
>>>ndim <- length(dim(x))
>>>if (ndim == 1)
>>> y <- x[i]
>>>else if (ndim == 2)
>>> y <- x[i,]
>>>else if (ndim == 3)
>>> y <- x[i,,]
>>>else ...
>>>
>>>and so on. My current solution is
>>>
>>>ndim <- length(dim(x))
>>>args <- rep(",", ndim)
>>>args[1] <- "i"
>>>args <- paste(args, collapse="")
>>>code <- paste("x[", args, "]", sep="")
>>>expr <- parse(text=code)
>>>y <- eval(expr)
>>>
>>>Is there another way I can do this in R that I have overlooked?
>>
>>
>>I think this should work:
>>
>>x <- array(1:24, dim=c(3,2,2,2)) # not c(2,2,3,2)....
>>i <- 2:3
>>ndim <- length(dim(x))
>>ix <- as.list(rep(TRUE, ndim))
>>ix[[1]] <- i
>>do.call("[", c(list(x), ix))
```
>
>
> In my case, 'x' is huge, an I have to be careful with allocating memory.
> Doesn't the 'list(x)' statement enforce an extra copy of 'x'? Or will
> lazy evaluation be able to pull out 'x' from the list again without
> evaluating 'list(x)'? I don't think so, but I'm not sure. There is
> also some overhead in 'ix[[1]] <- i', but 'i' is typically much smaller
> than 'x' so this should be of minor importance.
>
>
> array(x[slice.index(x, 1) == 1], dim(x)[-1])?
>
> There 'slice.index(x, 1)' will create an array of same size as 'x'.
>
> I do not think the 'eval(parse(...))' has such overhead (correct me if
> I'm wrong), but on the other hand, it is a more "ugly" solution. I
> prefer not to use parse(), substitute() and friends in my code, if I
> don't have to.
>
> I just want to bring up this flavor of the problem too, because I often
> find myself having to choose from similar options in other situations.
> If you have further comments, I would appreciate those.
>
Here's the type of manipulation I often do to approach these problems:
``` > x <- array(1:24, dim=c(4,3,2))
> i <- 2:3
> x[i,,]
```
, , 1
[,1] [,2] [,3]
[1,] 2 6 10
[2,] 3 7 11
, , 2
[,1] [,2] [,3]
[1,] 14 18 22
[2,] 15 19 23
> xic <- Quote(x[i,])
> xic
x[i, ]
> length(xic)
[1] 4
``` > # now duplicate the empty index argument the appropriate number of times
> xic <- xic[c(1:3,4,4)]
> xic
```
x[i, , ]
> eval(xic)
, , 1
[,1] [,2] [,3]
[1,] 2 6 10
[2,] 3 7 11
, , 2
[,1] [,2] [,3]
[1,] 14 18 22
[2,] 15 19 23
>
I do this type of manipulation for precisely the reasons you bring up. I do know that in S-PLUS, using do.call() in the most obvious manner can result in unnecessary multiple duplications of data objects (as you suspect). I don't think R is quite as bad, but I haven't done careful the experiments with R.
Do be careful though: this type of manipulation can expose a bug in R, which I don't think has been fixed (PR#7924).
• Tony Plate
> Thanks
>
> Henrik
>
> ______________________________________________
> R-devel@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel Received on Thu Nov 24 13:40:08 2005
This archive was generated by hypermail 2.1.8 : Thu 24 Nov 2005 - 08:21:36 GMT | 1,188 | 3,467 | {"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-2020-24 | latest | en | 0.798926 |
https://zh.m.wikipedia.org/wiki/%E5%85%89%E7%9A%84%E8%89%B2%E6%95%A3 | 1,686,433,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646350.59/warc/CC-MAIN-20230610200654-20230610230654-00771.warc.gz | 1,230,144,867 | 48,497 | # 色散 (光學)
（重定向自光的色散
## 光學中的材料色散
${\displaystyle v={\frac {c}{n}}}$
${\displaystyle \lambda _{\rm {r}}>\lambda _{\rm {y}}>\lambda _{\rm {b}}}$
${\displaystyle 1
${\displaystyle {\frac {{\mathrm {d} }n}{{\mathrm {d} }\lambda }}<0}$
${\displaystyle n(\lambda )=B+{\frac {C}{\lambda ^{2}}}+{\frac {D}{\lambda ^{4}}}+\cdots }$
## 群速度色散
${\displaystyle {\rm {v_{g}}}={\frac {\rm {v_{p}}}{1-{\frac {\omega }{\rm {v_{p}}}}{\frac {\rm {dv_{p}}}{d\omega }}}}.}$
${\displaystyle D=-{\frac {\lambda }{c}}\,{\frac {{\rm {d}}^{2}n}{{\rm {d}}\lambda ^{2}}}.}$
${\displaystyle D=-{\frac {2\pi c}{\lambda ^{2}}}\,{\frac {{\rm {d}}^{2}k}{{\rm {d}}\omega ^{2}}}.}$
## 高階色散的廣義公式 – Lah-Laguerre 光學
${\displaystyle {\begin{array}{c}\varphi \mathrm {(} \omega \mathrm {)} =\varphi \left.\ \right|_{\omega _{0}}+\left.\ {\frac {\partial \varphi }{\partial \omega }}\right|_{\omega _{0}}\left(\omega -\omega _{0}\right)+{\frac {1}{2}}\left.\ {\frac {\partial ^{2}\varphi }{\partial \omega ^{2}}}\right|_{\omega _{0}}\left(\omega -\omega _{0}\right)^{2}\ +\ldots +{\frac {1}{p!}}\left.\ {\frac {\partial ^{p}\varphi }{\partial \omega ^{p}}}\right|_{\omega _{0}}\left(\omega -\omega _{0}\right)^{p}+\ldots \end{array}}}$
${\displaystyle {\begin{array}{c}k\mathrm {(} \omega \mathrm {)} =k\left.\ \right|_{\omega _{0}}+\left.\ {\frac {\partial k}{\partial \omega }}\right|_{\omega _{0}}\left(\omega -\omega _{0}\right)+{\frac {1}{2}}\left.\ {\frac {\partial ^{2}k}{\partial \omega ^{2}}}\right|_{\omega _{0}}\left(\omega -\omega _{0}\right)^{2}\ +\ldots +{\frac {1}{p!}}\left.\ {\frac {\partial ^{p}k}{\partial \omega ^{p}}}\right|_{\omega _{0}}\left(\omega -\omega _{0}\right)^{p}+\ldots \end{array}}}$
${\displaystyle {\begin{array}{c}{\frac {{\partial }^{p}}{\partial {\omega }^{p}}}k\mathrm {(} \omega \mathrm {)} ={\frac {1}{c}}\left(p{\frac {{\partial }^{p-1}}{\partial {\omega }^{p-1}}}n\mathrm {(} \omega \mathrm {)} +\omega {\frac {{\partial }^{p}}{\partial {\omega }^{p}}}n\mathrm {(} \omega \mathrm {)} \right)\ \end{array}}}$ , ${\displaystyle {\begin{array}{c}{\frac {{\partial }^{p}}{\partial {\omega }^{p}}}\varphi \mathrm {(} \omega \mathrm {)} ={\frac {1}{c}}\left(p{\frac {{\partial }^{p-1}}{\partial {\omega }^{p-1}}}{\it {OP}}\mathrm {(} \omega \mathrm {)} +\omega {\frac {{\partial }^{p}}{\partial {\omega }^{p}}}{\it {OP}}\mathrm {(} \omega \mathrm {)} \right)\end{array}}(1)}$
${\displaystyle {\begin{array}{l}{\frac {\partial {p}}{\partial {\omega }^{p}}}f\mathrm {(} \omega \mathrm {)} ={}{\left(\mathrm {-} \mathrm {1} \right)}^{p}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{p}\sum \limits _{m={0}}^{p}{{\mathcal {A}}\mathrm {(} p,m\mathrm {)} {\lambda }^{m}{\frac {{\partial }^{m}}{\partial {\lambda }^{m}}}f\mathrm {(} \lambda \mathrm {)} }\end{array}}}$ ${\displaystyle ,}$ ${\displaystyle {\begin{array}{c}{\frac {{\partial }^{p}}{\partial {\lambda }^{p}}}f\mathrm {(} \lambda \mathrm {)} ={}{\left(\mathrm {-} \mathrm {1} \right)}^{p}{\left({\frac {\omega }{\mathrm {2} \pi c}}\right)}^{p}\sum \limits _{m={0}}^{p}{{\mathcal {A}}\mathrm {(} p,m\mathrm {)} {\omega }^{m}{\frac {{\partial }^{m}}{\partial {\omega }^{m}}}f\mathrm {(} \omega \mathrm {)} }\end{array}}(2)}$
${\displaystyle POD={\frac {d^{p}\varphi (\omega )}{d\omega ^{p}}}=(-1)^{p}({\frac {\lambda }{2\pi c}})^{(p-1)}\sum _{m=0}^{p}{\mathcal {B(p,m)}}(\lambda )^{m}{\frac {d^{m}OP(\lambda )}{d\lambda ^{m}}}}$ ${\displaystyle ,}$ ${\displaystyle POD={\frac {d^{p}k(\omega )}{d\omega ^{p}}}=(-1)^{p}({\frac {\lambda }{2\pi c}})^{(p-1)}\sum _{m=0}^{p}{\mathcal {B(p,m)}}(\lambda )^{m}{\frac {d^{m}n(\lambda )}{d\lambda ^{m}}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {GD}}}={\frac {\partial }{\partial \omega }}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(n\mathrm {(} \omega \mathrm {)} +\omega {\frac {\partial n\mathrm {(} \omega \mathrm {)} }{\partial \omega }}\right)={\frac {\mathrm {1} }{c}}\left(n\mathrm {(} \lambda \mathrm {)} -\lambda {\frac {\partial n\mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}\right)=v_{gr}^{\mathrm {-} \mathrm {1} }\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {GDD}}}={\frac {{\partial }^{2}}{\partial {\omega }^{\mathrm {2} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {2} {\frac {\partial n\mathrm {(} \omega \mathrm {)} }{\partial \omega }}+\omega {\frac {{\partial }^{2}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {2} }}}\right)={\frac {\mathrm {1} }{c}}\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)\left({\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}\right)\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {TOD}}}={\frac {{\partial }^{3}}{\partial {\omega }^{\mathrm {3} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {3} {\frac {{\partial }^{2}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {2} }}}+\omega {\frac {{\partial }^{3}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {3} }}}\right)={-}{\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {2} }{\Bigl (}\mathrm {3} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+{\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {FOD}}}={\frac {{\partial }^{4}}{\partial {\omega }^{\mathrm {4} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {4} {\frac {{\partial }^{3}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {3} }}}+\omega {\frac {{\partial }^{4}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {4} }}}\right)={\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {3} }{\Bigl (}\mathrm {12} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {8} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+{\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {FiOD}}}={\frac {{\partial }^{5}}{\partial {\omega }^{\mathrm {5} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {5} {\frac {{\partial }^{4}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {4} }}}+\omega {\frac {{\partial }^{5}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {5} }}}\right)={-}{\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {4} }{\Bigl (}\mathrm {60} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {60} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {15} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+{\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {SiOD}}}={\frac {{\partial }^{6}}{\partial {\omega }^{\mathrm {6} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {6} {\frac {{\partial }^{5}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {5} }}}+\omega {\frac {{\partial }^{6}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {6} }}}\right)={\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {5} }{\Bigl (}\mathrm {360} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {480} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {180} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {24} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+{\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {SeOD}}}={\frac {{\partial }^{7}}{\partial {\omega }^{\mathrm {7} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {7} {\frac {{\partial }^{6}n\mathrm {(} \omega \mathrm {)} }{{\partial \omega }^{\mathrm {6} }}}+\omega {\frac {{\partial }^{7}n\mathrm {(} \omega \mathrm {)} }{{\partial \omega }^{\mathrm {7} }}}\right)={-}{\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {6} }{\Bigl (}\mathrm {2520} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {4200} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {2100} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {420} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {35} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+{\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {EOD}}}={\frac {{\partial }^{8}}{\partial {\omega }^{\mathrm {8} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {8} {\frac {{\partial }^{7}n\mathrm {(} \omega \mathrm {)} }{{\partial \omega }^{\mathrm {7} }}}+\omega {\frac {{\partial }^{8}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {8} }}}\right)={\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {7} }{\Bigl (}\mathrm {20160} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {40320} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {25200} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {6720} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {840} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+\\+\mathrm {48} {\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}+{\lambda }^{\mathrm {8} }{\frac {{\partial }^{8}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {8} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {NOD}}}={\frac {{\partial }^{9}}{\partial {\omega }^{\mathrm {9} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {9} {\frac {{\partial }^{8}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {8} }}}+\omega {\frac {{\partial }^{9}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {9} }}}\right)={-}{\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {8} }{\Bigl (}\mathrm {181440} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {423360} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {317520} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {105840} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {17640} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+\\+\mathrm {1512} {\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}+\mathrm {63} {\lambda }^{\mathrm {8} }{\frac {{\partial }^{8}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {8} }}}+{\lambda }^{\mathrm {9} }{\frac {{\partial }^{9}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {9} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\boldsymbol {\it {TeOD}}}={\frac {{\partial }^{10}}{\partial {\omega }^{\mathrm {10} }}}k\mathrm {(} \omega \mathrm {)} ={\frac {\mathrm {1} }{c}}\left(\mathrm {10} {\frac {{\partial }^{9}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {9} }}}+\omega {\frac {{\partial }^{10}n\mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {10} }}}\right)={\frac {\mathrm {1} }{c}}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {9} }{\Bigl (}\mathrm {1814400} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {4838400} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {4233600} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+{1693440}{\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\\+\mathrm {352800} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+\mathrm {40320} {\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}+\mathrm {2520} {\lambda }^{\mathrm {8} }{\frac {{\partial }^{8}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {8} }}}+\mathrm {80} {\lambda }^{\mathrm {9} }{\frac {{\partial }^{9}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {9} }}}+{\lambda }^{\mathrm {10} }{\frac {{\partial }^{10}n\mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {10} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {\partial {p}}{\partial {\omega }^{p}}}f\mathrm {(} \omega \mathrm {)} ={}{\left(\mathrm {-} \mathrm {1} \right)}^{p}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{p}\sum \limits _{m={0}}^{p}{{\mathcal {A}}\mathrm {(} p,m\mathrm {)} {\lambda }^{m}{\frac {{\partial }^{m}}{\partial {\lambda }^{m}}}f\mathrm {(} \lambda \mathrm {)} }\end{array}}}$ ${\displaystyle ,}$ ${\displaystyle {\begin{array}{c}{\frac {{\partial }^{p}}{\partial {\lambda }^{p}}}f\mathrm {(} \lambda \mathrm {)} ={}{\left(\mathrm {-} \mathrm {1} \right)}^{p}{\left({\frac {\omega }{\mathrm {2} \pi c}}\right)}^{p}\sum \limits _{m={0}}^{p}{{\mathcal {A}}\mathrm {(} p,m\mathrm {)} {\omega }^{m}{\frac {{\partial }^{m}}{\partial {\omega }^{m}}}f\mathrm {(} \omega \mathrm {)} }\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {\partial \varphi \mathrm {(} \omega \mathrm {)} }{\partial \omega }}={-}\left({\frac {\mathrm {2} \pi c}{{\omega }^{\mathrm {2} }}}\right){\frac {\partial \varphi \mathrm {(} \omega \mathrm {)} }{\partial \lambda }}={-}\left({\frac {{\lambda }^{\mathrm {2} }}{\mathrm {2} \pi c}}\right){\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{2}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {2} }}}={\frac {\partial }{\partial \omega }}\left({\frac {\partial \varphi \mathrm {(} \omega \mathrm {)} }{\partial \omega }}\right)={\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {2} }\left(\mathrm {2} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+{\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}\right)\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{3}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {3} }}}={-}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {3} }\left(\mathrm {6} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {6} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+{\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}\right)\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{4}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {4} }}}={\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {4} }{\Bigl (}\mathrm {24} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {36} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {12} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+{\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{\mathrm {5} }\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {5} }}}={-}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {5} }{\Bigl (}\mathrm {120} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {240} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {120} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {20} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+{\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{6}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {6} }}}={\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {6} }{\Bigl (}\mathrm {720} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {1800} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {1200} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {300} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {30} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}\mathrm {\ +} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{7}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {7} }}}={-}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {7} }{\Bigl (}\mathrm {5040} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {15120} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {12600} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {4200} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {630} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {42} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+{\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{8}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {8} }}}={\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {8} }{\Bigl (}\mathrm {40320} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {141120} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {141120} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {58800} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {11760} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {1176} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+\mathrm {56} {\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}+\\+{\lambda }^{\mathrm {8} }{\frac {\partial ^{8}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {8} }}}{\Bigr )}\end{array}}}$ ${\displaystyle {\begin{array}{l}{\frac {{\partial }^{9}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {9} }}}={-}{\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {9} }{\Bigl (}\mathrm {362880} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {1451520} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {1693440} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {846720} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {211680} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {28224} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+\\+\mathrm {2016} {\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}+\mathrm {72} {\lambda }^{\mathrm {8} }{\frac {{\partial }^{8}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {8} }}}+{\lambda }^{\mathrm {9} }{\frac {\partial ^{\mathrm {9} }\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {9} }}}{\Bigr )}\end{array}}}$
${\displaystyle {\begin{array}{l}{\frac {{\partial }^{10}\varphi \mathrm {(} \omega \mathrm {)} }{\partial {\omega }^{\mathrm {10} }}}={\left({\frac {\lambda }{\mathrm {2} \pi c}}\right)}^{\mathrm {10} }{\Bigl (}\mathrm {3628800} \lambda {\frac {\partial \varphi \mathrm {(} \lambda \mathrm {)} }{\partial \lambda }}+\mathrm {16329600} {\lambda }^{\mathrm {2} }{\frac {{\partial }^{2}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {2} }}}+\mathrm {21772800} {\lambda }^{\mathrm {3} }{\frac {{\partial }^{3}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {3} }}}+\mathrm {12700800} {\lambda }^{\mathrm {4} }{\frac {{\partial }^{4}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {4} }}}+\mathrm {3810240} {\lambda }^{\mathrm {5} }{\frac {{\partial }^{5}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {5} }}}+\mathrm {635040} {\lambda }^{\mathrm {6} }{\frac {{\partial }^{6}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {6} }}}+\\+\mathrm {60480} {\lambda }^{\mathrm {7} }{\frac {{\partial }^{7}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {7} }}}+\mathrm {3240} {\lambda }^{\mathrm {8} }{\frac {{\partial }^{8}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {8} }}}+\mathrm {90} {\lambda }^{\mathrm {9} }{\frac {{\partial }^{9}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {9} }}}+{\lambda }^{\mathrm {10} }{\frac {{\partial }^{10}\varphi \mathrm {(} \lambda \mathrm {)} }{\partial {\lambda }^{\mathrm {10} }}}{\Bigr )}\end{array}}}$
## 参考文献
1. ^ 1882-1970., Born, Max,. Principles of optics : electromagnetic theory of propagation, interference and diffraction of light. 7th expanded ed. Cambridge: Cambridge University Press https://web.archive.org/web/20080620012317/http://www.worldcat.org/oclc/40200160. 1999 [2019-01-28]. ISBN 0521642221. OCLC 40200160. （原始内容存档于2008-06-20）. 缺少或|title=为空 (帮助)
2. ^ Dispersion Compensation页面存档备份，存于互联网档案馆） Retrieved 25-08-2015.
3. ^ Born, M. and Wolf, E. (1980) "Principles of Optics, 6th ed." pg. 93. Pergamon Press.
4. ^ Saleh, B.E.A. and Teich, M.C. Fundamentals of Photonics (2nd Edition) Wiley, 2007.
5. ^ Popmintchev, Dimitar; Wang, Siyang; Xiaoshi, Zhang; Stoev, Ventzislav; Popmintchev, Tenio. Analytical Lah-Laguerre optical formalism for perturbative chromatic dispersion. Optics Express. 2022-10-24, 30 (22): 40779–40808. Bibcode:2022OExpr..3040779P. PMID 36299007. （英语）.
6. ^ Popmintchev, Dimitar; Wang, Siyang; Xiaoshi, Zhang; Stoev, Ventzislav; Popmintchev, Tenio. Theory of the Chromatic Dispersion, Revisited. 2020-08-30. [physics.optics] （英语）. | 9,687 | 26,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 54, "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} | 3.640625 | 4 | CC-MAIN-2023-23 | latest | en | 0.288871 |
https://discusstest.codechef.com/t/enigma09-editorial/11936 | 1,627,144,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00291.warc.gz | 218,264,945 | 3,958 | # ENIGMA09 - Editorial
Bots
Author and Editorialist: Shubham Chauhan
Medium
### PREREQUISITES:
Modulo Arithmetic , Tree
### EXPLANATION:
Problem Naturally can be transformed to a more formal way : How many vertices will a trie contain if we add all possible strings with length 2 * N with equal number of zeroes and ones to it.
So first of all, it is obvious that upper half of this tree would be a full binary tree. Lets take a look on N = 3: level (0 - 1) vertex, level (1 - 2) vertices , level (2 - 4) vertices, level (3 - 8) vertices.
Starting from Nth level not every vertex will duplicate: only those that haven’t spent their 0’s or 1’s will.
So here is how to calculate how may vertices will be there on level i + 1 :
1. Lets assign Number of non duplicating vertices from level to PD(i) .
2. Count(i + 1) = PD(i) + (Count(i) - PD(i))*2.
3. And PD can be calculated pretty easily with binomial coefficients: PD(i) = 2 * C(i , N)
4. Everything else is implementation techniques: inverse modulo arithmetics + some fast way to calculate C(i , N) and sum counts
### SOLUTIONS:
To be Update Soon
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No he doesn't. What he's actually asking for is a random shuffle. There are well known O(n) time and memory algorithms for that sort of thing. Look up Fischer-Yates.
update: of course a shuffle won't work so well if the range is large
• Comment on Re^2: How to generate different random numbers?
Replies are listed 'Best First'.
Re^3: How to generate different random numbers?
by thor (Priest) on Sep 05, 2004 at 09:10 UTC
No he doesn't.
I'm going to go ahead and disagree with you there, cowboy. Any time you impose a restriction on random, you've made it not random. A random shuffle of a known interval isn't a set of random numbers. Woe unto he who uses the "random" numbers for cryptographic purposes.
thor
Feel the white light, the light within
Be your own disciple, fan the sparks of will
For all of us waiting, your kingdom will come
Well since we're just arguing semantics, indeed, a random shuffle of a known interval isn't the same thing as a set of random numbers. But I never said they were the same thing. I merely said that it is not any "less random", and indeed it isn't.
A random shuffle is really a random selection from a uniform distribution of the set of all possible permutations on that range of numbers (or N!).
And who said the numbers had to be used for cryptographic purposes? You're adopting a very narrow view of things to make your case, pilgrim.
Any time you impose a restriction on random, you've made it not random.
Well this is just plain false (eg. if I restrict myself to a set of random even numbers, the numbers I've chosen are still random, and indeed, no less random). I think what you meant to say is that for cryptographic purposes, random numbers should only be chosen from a uniform distribution. This is an arguable proposition, especially since many distributions can be translated into one another, and thus any cryptographic function can transform a non-uniform distribution into a uniform one.
A random shuffle is really a random selection from a uniform distribution of the set of all possible permutations on that range of numbers (or N!).
This assumes that you're chosing one member from the interval. However, even this has a gotcha or two. If you do this on a large scale, an attacker will surely notice that all of your "random" numbers come from the same interval and exploit this fact.
Moreover, in terms of randomness, a random shuffle and just choosing a random number are vastly different. One is chosing a member from an finite set without replacement, the other is chosing a member from an infinite set with replacement. That is to say that in the true random case, you are just as likely to pick the number that you just picked as you are any other number. In the random shuffle situation, you are guaranteed that this is not the case. An attacker can exploit this, too. For instance, let's say that the interval was [1,4]. As the attacker, you have this knowledge. Furthermore, you've observed, the following numbers go by: 2, 4, 1. What's the next number in the sequence?
And who said the numbers had to be used for cryptographic purposes?
Is there any other use? ;)
thor
Feel the white light, the light within
Be your own disciple, fan the sparks of will
For all of us waiting, your kingdom will come | 746 | 3,319 | {"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-2024-18 | latest | en | 0.949086 |
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MIT OpenCourseWare http://ocw.mit.edu 6.096 Introduction to C++ January (IAP) 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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Massachusetts Institute of Technology 6.096 Lab 4 Solutions 1. (15 points) #include <iostream> using namespace std; int main() { int grades[6][5] = { {97, 75, 87, 56, 88}, {76, 84, 88, 59, 99}, {85, 86, 82, 81, 88}, {95, 92, 97, 97, 44}, {66, 74, 82, 60, 85}, {82, 73, 96, 32, 77} }; double averages[6]; for (int i = 0; i < 6; i++) { int sum = 0; for (int j = 0; j < 5; j++) { sum += grades[i][j]; } averages[i] = sum/5.0; } for (int k = 0; k < 6; k++) cout << "Student " << k << " has an average of " << averages[k] << endl; return 0; } 2. (5 points) void average( int grades[][5], const int rows, const int columns, double averages[] ) { for (int i = 0; i < rows; i++) { int sum = 0; for (int j = 0; j < columns; j++) { sum += grades[i][j]; } average [i] = sum / static_cast<double>(columns); } } 3. (10 points) void printTTTBoard(int array[][3]) {
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Ask a homework question - tutors are online | 564 | 1,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-17 | longest | en | 0.713925 |
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# 5TH GRADE COMMON CORE DECIMAL LESSONS
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5th Grade Math Curriculum - Common Core Lessons
In Grade 5 Unit 1, Place Value with Decimals, students extend this understanding in two ways. First, they see that a similar pattern emerges with place values to the left of a digit; namely, they are 1/10 of the value. Secondly, students extend these understandings of a digit and the places to the left and to the right to decimal numbers.
5th grade Math Worksheets: Decimal place value (to the ten
Parenting » Worksheets » Decimal place value (to the ten thousandths) Math Decimal place value (to the ten thousandths) MATH | GRADE: 5th . Print full size. Print full size. Skills Understanding place value. Common Core Standards: Grade 5 Number & Operations in Base Ten. CCSSent.5.A.1, CCSSent.5.A.3
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Fifth Grade : Free Common Core Math Worksheets. What you will learn: Fifth graders start writing and solving numerical expressions and equations. Go deep into place value system and practice all four operations with whole numbers and decimals to the hundredths. Master like and unlike fractions, and their operations with whole numbers and vice versa.
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Common Core Learning Standards. 5.3.a Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e..
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Fifth Grade Decimals Worksheets and Printables. But when it comes to learning how to multiply decimals or convert them to fractions or percents, things can get a bit tricky. Our fifth grade decimals worksheets are designed to reduce the confusion with colorful pages, interesting lessons, and easy-to-follow directions so your student can quickly grasp..
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5 Worksheets - Common Core Sheets | 808 | 3,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-51 | latest | en | 0.841152 |
https://www.mathworks.com/matlabcentral/answers/483202-derivative-at-various-points-in-symbolic-math-toolbox?s_tid=prof_contriblnk | 1,597,276,439,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00228.warc.gz | 745,053,543 | 22,005 | # derivative at various points in symbolic math toolbox
1 view (last 30 days)
Mesbahose Salekeen on 2 Oct 2019
Answered: Stephan on 2 Oct 2019
let say i have a vector.
A=[1 2 4 6]
now i am declaring a function using symbolic toolbox
syms x
f=x^2
how do you evaluate f at every elements of A?
Stephan on 2 Oct 2019
A = [1 2 4 6];
syms x
f(x) = x^2
sol = f(A)
gives:
f(x) =
x^2
sol =
[ 1, 4, 16, 36] | 156 | 399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-34 | latest | en | 0.79817 |
https://sg.answers.yahoo.com/question/index?qid=20131023172128AAhF0Jj | 1,600,494,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00383.warc.gz | 653,508,391 | 24,730 | Write an equation of the line that passes through Point P and is perpendicular to the equation:
3x-5y=6
Point P= (-8,0)
I know how to solve the majority of these, but for some reason, this one is stumping me! Thank you!
Relevance
• 7 years ago
For perpendicular lines using standard form, switch the coefficients and change the middle sign:
5x + 3y = ... then plug in the point's coordinates to see what it equals.
5(-8) + 3(0) = -40
So 5x + 3y = -40
OR: get y by itself (slope-intercept form)
3x = 6 + 5y
3x - 6 = 5y
3/5 x - 6/5 = y. This line has slope 3/5, and perpendicular lines have opposite reciprocal slopes, so your new line looks like y = -5/3 x + b. Plugging in (-8,0) you get
0 = -5/3 (-8) + b
0 = 40/3 + b
-40/3 = b so the equation is y = -5/3 x - 40/3
In standard form, multiply by 3: 3y = -5x - 40 then add 5x to both sides: 5x + 3y = -40 | 302 | 869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-40 | latest | en | 0.894837 |
https://www.studyhawks.com/question/which-terms-could-be-used-as-the-first-term-of-the-expression-below-to-create-a-polynomial-written-in-standard-form-check-all-that-apply-8r2s4-3-854192/ | 1,586,164,993,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00108.warc.gz | 1,003,041,391 | 7,838 | # Which terms could be used as the first term of the expression below to create a polynomial written in standard form? Check all that apply. + 8r2s4 – 3r3s3 s5 3r4s5 –r4s6 –6rs5
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What is the quotient of c^2-c-2/c^2-1 / c^2+3c +2/ c^2-3c+2 | 198 | 620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-16 | latest | en | 0.915452 |
www.bwilden.com | 1,726,143,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00172.warc.gz | 645,074,076 | 16,797 | # Does Height Matter When Running for President?
Bayes
Author
Bertrand Wilden
Published
August 9, 2022
## Does Height Matter When Running for President?
Height is supposed to confer all sorts of advantages in life. Taller people make more money, have an easier time finding romantic partners, and can reach things off the highest shelves without using a step stool. But does height matter when it comes to politics? The topic has been the subject of extensive debate—so much so that a Wikipedia page was written to provide information on the heights of US presidential candidates. In this post I analyze this debate quantitatively using R and Bayesian regression methods. My results conclusively show that height probably doesn’t matter much when it comes to winning the presidency.
# Loading in the packages used
library(tidyverse)
library(rvest)
library(MetBrewer)
library(ggdist)
library(brms)
library(distributional)
library(geomtextpath)
library(tidybayes)
# Global plotting theme for ggplot
theme_set(theme_ggdist())
# Set global rounding options
options(scipen = 1,
digits = 3)
## Getting the data
The first thing to do is gather the data on presidential candidate heights. The package rvest is a great way to scrape the Wikipedia page above. It’s pretty easy to get data off Wikipedia because the HTML is relatively simple. But tables of data on Wikipedia need a bit of cleaning before they can be used for any statistical analysis. You have to remove things like citation markers, as well as fix column names and make sure columns containing numbers are actually numeric types.
url <- "https://en.wikipedia.org/wiki/Heights_of_presidents_and_presidential_candidates_of_the_United_States"
height_table <- url |>
# Parse the raw html
# Pull out the table elements
html_elements("table") |>
purrr::pluck(5) |>
# Turn the candidate height table into a tibble
html_table()
heights <- height_table |>
# Assign names to all columns to fix duplicate originals
colnames<-(c("election", "winner", "winner_height_in", "winner_height_cm",
"opponent", "opponent_height_in", "opponent_height_cm",
"difference_in", "difference_cm")) |>
# Removing problematic elections
filter(!election %in% c("1912", "1860", "1856", "1836", "1824"),
opponent_height_cm != "") |>
# Cleaning up the citation markers and fixing column types
mutate(across(everything(),
~ str_remove_all(., "\$.*\$")),
across(contains("_cm"),
~ str_remove_all(.x, "\\D") |>
as.numeric()),
# Making a few new variable for the analysis
winner_difference_cm = winner_height_cm - opponent_height_cm,
winner_taller = if_else(winner_difference_cm > 0, 1, 0))
In the process of cleaning the presidential candidate height data I decided to remove all elections in which more than two candidates ran (1824, 1836, 1856, 1860, 1912), all elections in which a candidate’s height was missing from Wikipedia (1816: Rufus King, 1868: Horatio Seymour), and all uncontested elections (1788 and 1792: George Washington, 1820 James Monroe). No information regarding a height advantage can be gleaned from the latter two categories (unless it was Washington’s large stature that helped dissuade any potential challengers) so their exclusion should be uncontroversial. The removal of multi-candidate elections, however, was a choice I made in order to simplify the analysis. The role of height in a multi-candidate election is less straightforward than in a two-candidate election. Should we suppose voters simply gravitate towards the tallest candidate running? Or are they making height comparisons between all three candidates at once? Because political science lacks a good theory to support any of these explanations I dropped multi-candidate elections and moved on.
After these cleaning steps I made a new variable called winner_taller which simply denotes whether that taller candidate won the particular election, 1 or lost, 0. Using mean(heights\$winner_taller) we see that the proportion of elections won by the taller candidate is 0.551. The taller candidate wins more on average! Skeptical readers will object that the sample size is too low for this result to be conclusive. “What is the standard error of the proportion!” they will say, “I want to see a p-value!” These are valid critiques, but as a fervent Bayesian I refuse to calculate any p-values. Let’s move on to some further analysis.
## Presidential candidates compared to the general population
The original candidate height data set was at the election-level, meaning that every row represented a presidential election year. In order to look at candidate heights individually, I transformed the data into “long” format such that each row represents a single candidate. With the data at the candidate-level, we can now investigate how the heights of presidential candidates compare to the overall population.
heights_long <- heights |>
pivot_longer(cols = c("winner", "opponent"),
values_to = "candidate",
names_to = "status") |>
# Creating a single variable for candidate height
mutate(height_cm = case_when(status == "winner" ~ winner_height_cm,
status == "opponent" ~ opponent_height_cm))
The graph below shows the distribution of candidate heights compared to the US adult male population. The variable “height” is often used to illustrate a Normal distribution in action. But technically, the Normal distribution does not accurately reflect height unless we first narrow the population down. Children and adults do not share the same height distribution, and neither do different genders. Each country, or region of the globe, likely also has a distinct height distribution. So unless we clearly define which population we’re talking about, “height” is best characterized as a mixture of normal distributions. Since almost all US presidential candidates have been adult men, however, I overlaid only the distribution for US adult males (mean 178 cm, standard deviation 8 cm).
heights_long |>
select(height_cm, candidate) |>
distinct() |>
ggplot() +
stat_function(geom = "textpath", vjust = 0, hjust = .2,
label = "US Male Population",
fun = function(x) dnorm(x, mean = 178, sd = 8) * 20) +
geom_dots(aes(x = height_cm,
fill = candidate == "Hillary Clinton",
group = NA),
size = .1) +
scale_fill_manual(values = met.brewer("Lakota", 2)) +
xlim(150, 200) +
labs(x = "Height in cm", y = "",
title = "Heights of US Presidential Candidates\nCompared to US Male Population") +
theme(legend.position = "none",
axis.line.y = element_blank(),
axis.text.y = element_blank())
Hillary Clinton (represented by the yellow dot in the candidate distribution) should not be compared to the average US male in terms of height—but interestingly, isn’t the shortest candidate in US history. That honor goes to James Madison at 163 cm (5’ 4”). The graph shows that presidential candidates roughly align with overall male population heights. Perhaps candidates are slightly taller than the average US male, but the difference appears small.
## How much does height contribute to winning the presidency?
Okay, so we discovered that the taller candidate wins slightly more often on average, but how does raw height affect a candidate’s chances of becoming president? To answer this question, we need to add a new dummy variable to our candidate-level data set indicating whether they won or lost.
heights_long <- heights_long |>
mutate(winner = if_else(status == "winner", 1, 0))
Then I fit the following Bayesian logistic regression model to the data:
\begin{equation*} \begin{aligned} \text{Winner}_i &\sim \text{Bernoulli}(p) \\ p &= \text{logit}^{-1}(\alpha + \beta \ \text{Height}_i) \\ \alpha &\sim \text{Normal}(0, 2) \\ \beta &\sim \text{Normal}(0, 2) \end{aligned} \end{equation*}
There’s nothing too fancy going on in this model—just a standard logistic regression with a binary outcome (winning the presidency or not winning the presidency). The Normal(0, 2) priors on the intercept and slope coefficients are weakly informative, meaning they are wide enough to let the data inform our results, but narrow enough to be skeptical of extreme values. Given background knowledge of height in presidential campaigns, it’s unlikely it has a big effect on the outcome.
The code below fits the model using the brms package in R. Because the data only contain 96 candidate observations, the MCMC chains converge extremely quickly. Only 1.2 seconds for 12,000 iterations! Good practice when working with Bayesian models dictates that we look into the diagnostic measures (R-hat, effective sample size, number of divergent transitions, etc) of our fitted model. But since this model is very simple, I hope you will trust me that the fitting process worked reliably well.
height_model <- brm(
winner ~ 1 + height_cm,
prior = prior(normal(0, 2), class = "b") +
prior(normal(0, 2), class = "Intercept"),
data = heights_long,
seed = 111,
refresh = 0,
iter = 12000,
backend = "cmdstanr"
)
Start sampling
Running MCMC with 4 sequential chains...
Chain 1 finished in 0.2 seconds.
Chain 2 finished in 0.2 seconds.
Chain 3 finished in 0.2 seconds.
Chain 4 finished in 0.2 seconds.
All 4 chains finished successfully.
Mean chain execution time: 0.2 seconds.
Total execution time: 1.3 seconds.
Attempting to directly interpret coefficient values from logit models is rarely a good idea. Instead we can graph the results and compare the predicted probabilities of the outcome variable (winning the presidency) against a range of input variable values (candidate height in cm). This is what the (logit dotplot)[https://www.barelysignificant.com/post/glm/] below shows. The dots on the top and bottom of the graph represent candidates that either won or lost, and the line between them shows what our model predicts the winning probability to be at each height value on the x-axis. the weakly upward slope on this prediction line tells us that there is barely any benefit to being an extra cm taller when it comes to winning a presidential election.
# Generate a set of values across the range of the height data
prediction_grid <- with(heights_long,
data.frame(height_cm = seq(min(height_cm), max(height_cm), length.out = 100))
)
prediction_grid |>
# Generate posterior draws
# Collapse down to the height level
group_by(height_cm) |>
summarise(.median = median(.epred),
.sd = sd(.epred)) |>
# Convert log odds into predicted probabilities
mutate(log_odds = dist_normal(.median, .sd),
p_winner = dist_transformed(log_odds, plogis, qlogis)) |>
ggplot(aes(x = height_cm)) +
geom_dots(
aes(y = winner, side = ifelse(winner == 1, "bottom", "top")),
scale = 0.4,
fill = "#931e18",
size = .1,
data = heights_long) +
stat_lineribbon(
aes(ydist = p_winner), alpha = .25, fill = "#931e18", size = .5) +
labs(title = "Predicted Probability of Winning the Presidency\nBased on Candidate Height",
x = "Height cm",
y = "Pr(Winning)")
Apparently height has little effect on the probability that a candidate wins the presidency. But what if the candidate was extremely tall? It is now time to make a confession. The true reason I started this project was for selfish reasons. As someone who is 206 cm tall (6’ 9”), I wanted to know what my chances were of becoming president based only on my height. Plugging my 206 cm into the logistic regression model produces the posterior probability distribution shown in the graph below. While there is considerable uncertainty due to the small sample size of candidates, the model says I have between a 60 and 70% chance to win. Amazing!
prediction_grid <- with(heights_long,
data.frame(height_cm = 206)
)
prediction_grid |>
mutate(p_winner = 1 / (1 + exp(-.epred))) |>
ggplot(aes(x = p_winner)) +
stat_slabinterval(fill = "#04a3bd", trim = FALSE) +
labs(title = "Posterior Probability of Winning\nfor Someone Very Tall",
x = "Pr(Winning | Height = 206 cm)",
y = "") +
theme(axis.line.y = element_blank(),
axis.text.y = element_blank())
As we all know, numbers don’t lie. So keep an eye out for the Bert–2024 campaign coming soon.
## Session info
sessionInfo()
R version 4.1.1 (2021-08-10)
Platform: x86_64-apple-darwin17.0 (64-bit)
Running under: macOS Big Sur 10.16
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/4.1/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/4.1/Resources/lib/libRlapack.dylib
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] tidybayes_3.0.2.9000 geomtextpath_0.1.0 distributional_0.3.0
[4] brms_2.17.4 Rcpp_1.0.8.3 ggdist_3.0.99.9000
[7] MetBrewer_0.2.0 rvest_1.0.2 forcats_0.5.1
[10] stringr_1.4.0 dplyr_1.0.9 purrr_0.3.4
[16] ggplot2_3.3.6 tidyverse_1.3.1
loaded via a namespace (and not attached):
[4] selectr_0.4-2 plyr_1.8.7 igraph_1.3.1
[7] svUnit_1.0.6 splines_4.1.1 crosstalk_1.2.0
[10] rstantools_2.2.0 inline_0.3.19 digest_0.6.29
[13] htmltools_0.5.2 fansi_1.0.3 magrittr_2.0.3
[16] checkmate_2.1.0 tzdb_0.2.0 modelr_0.1.8
[19] RcppParallel_5.1.5 matrixStats_0.62.0 xts_0.12.1
[22] prettyunits_1.1.1 colorspace_2.0-3 textshaping_0.3.6
[25] haven_2.4.3 xfun_0.31 callr_3.7.0
[28] crayon_1.5.1 jsonlite_1.8.0 lme4_1.1-27.1
[31] zoo_1.8-10 glue_1.6.2 gtable_0.3.0
[34] emmeans_1.7.2 V8_4.2.0 pkgbuild_1.3.1
[37] rstan_2.26.11 abind_1.4-5 scales_1.2.0
[40] mvtnorm_1.1-3 DBI_1.1.1 miniUI_0.1.1.1
[43] viridisLite_0.4.0 xtable_1.8-4 stats4_4.1.1
[49] httr_1.4.2 threejs_0.3.3 arrayhelpers_1.1-0
[52] posterior_1.2.1 ellipsis_0.3.2 pkgconfig_2.0.3
[55] loo_2.5.1 farver_2.1.0 dbplyr_2.1.1
[58] utf8_1.2.2 labeling_0.4.2 tidyselect_1.1.2
[61] rlang_1.0.2 reshape2_1.4.4 later_1.3.0
[64] munsell_0.5.0 cellranger_1.1.0 tools_4.1.1
[67] cli_3.3.0 generics_0.1.2 broom_0.8.0
[70] ggridges_0.5.3 evaluate_0.15 fastmap_1.1.0
[73] yaml_2.3.5 processx_3.5.3 knitr_1.39
[76] fs_1.5.2 nlme_3.1-152 mime_0.12
[79] projpred_2.0.2 xml2_1.3.2 compiler_4.1.1
[82] bayesplot_1.9.0 shinythemes_1.2.0 rstudioapi_0.13
[85] curl_4.3.2 gamm4_0.2-6 reprex_2.0.1
[88] stringi_1.7.6 ps_1.7.0 Brobdingnag_1.2-7
[91] lattice_0.20-44 Matrix_1.3-4 nloptr_1.2.2.2
[94] markdown_1.1 shinyjs_2.1.0 tensorA_0.36.2
[97] vctrs_0.4.1 pillar_1.7.0 lifecycle_1.0.1
[100] bridgesampling_1.1-2 estimability_1.3 data.table_1.14.2
[103] httpuv_1.6.5 R6_2.5.1 promises_1.2.0.1
[106] gridExtra_2.3 codetools_0.2-18 boot_1.3-28
[109] colourpicker_1.1.1 MASS_7.3-54 gtools_3.9.2.1
[112] assertthat_0.2.1 withr_2.5.0 shinystan_2.6.0
[115] mgcv_1.8-36 parallel_4.1.1 hms_1.1.1
[118] grid_4.1.1 coda_0.19-4 minqa_1.2.4
[121] cmdstanr_0.4.0 rmarkdown_2.14 shiny_1.7.1
[124] lubridate_1.7.10 base64enc_0.1-3 dygraphs_1.1.1.6 | 4,286 | 15,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.833939 |
https://www.justintools.com/unit-conversion/area.php?k1=square-hectometers&k2=square-dekameters | 1,579,348,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592565.2/warc/CC-MAIN-20200118110141-20200118134141-00510.warc.gz | 964,403,943 | 25,840 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# AREA Units Conversionsquare-hectometers to square-dekameters
1 Square Hectometers
= 100 Square Dekameters
Category: area
Conversion: Square Hectometers to Square Dekameters
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Hectometers] symbol/abbrevation: (hm2, sq hm)
[Square Dekameters] symbol/abbrevation: (dm2, sq dam)
How to convert Square Hectometers to Square Dekameters (hm2, sq hm to dm2, sq dam)?
1 hm2, sq hm = 100 dm2, sq dam.
1 x 100 dm2, sq dam = 100 Square Dekameters.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Square Hectometers (hm2, sq hm) is equal to 10000 square-meters, while 1 Square Dekameters (dm2, sq dam) = 100 square-meters.
1 Square Hectometers to common area units
1 hm2, sq hm = 10000 square meters (m2, sq m)
1 hm2, sq hm = 100000000 square centimeters (cm2, sq cm)
1 hm2, sq hm = 0.01 square kilometers (km2, sq km)
1 hm2, sq hm = 107639.15051182 square feet (ft2, sq ft)
1 hm2, sq hm = 15500031.000062 square inches (in2, sq in)
1 hm2, sq hm = 11959.900463011 square yards (yd2, sq yd)
1 hm2, sq hm = 0.0038610215859254 square miles (mi2, sq mi)
1 hm2, sq hm = 15500031000062 square mils (sq mil)
1 hm2, sq hm = 1 hectares (ha)
1 hm2, sq hm = 2.4710516301528 acres (ac)
Square Hectometersto Square Dekameters (table conversion)
1 hm2, sq hm = 100 dm2, sq dam
2 hm2, sq hm = 200 dm2, sq dam
3 hm2, sq hm = 300 dm2, sq dam
4 hm2, sq hm = 400 dm2, sq dam
5 hm2, sq hm = 500 dm2, sq dam
6 hm2, sq hm = 600 dm2, sq dam
7 hm2, sq hm = 700 dm2, sq dam
8 hm2, sq hm = 800 dm2, sq dam
9 hm2, sq hm = 900 dm2, sq dam
10 hm2, sq hm = 1000 dm2, sq dam
20 hm2, sq hm = 2000 dm2, sq dam
30 hm2, sq hm = 3000 dm2, sq dam
40 hm2, sq hm = 4000 dm2, sq dam
50 hm2, sq hm = 5000 dm2, sq dam
60 hm2, sq hm = 6000 dm2, sq dam
70 hm2, sq hm = 7000 dm2, sq dam
80 hm2, sq hm = 8000 dm2, sq dam
90 hm2, sq hm = 9000 dm2, sq dam
100 hm2, sq hm = 10000 dm2, sq dam
200 hm2, sq hm = 20000 dm2, sq dam
300 hm2, sq hm = 30000 dm2, sq dam
400 hm2, sq hm = 40000 dm2, sq dam
500 hm2, sq hm = 50000 dm2, sq dam
600 hm2, sq hm = 60000 dm2, sq dam
700 hm2, sq hm = 70000 dm2, sq dam
800 hm2, sq hm = 80000 dm2, sq dam
900 hm2, sq hm = 90000 dm2, sq dam
1000 hm2, sq hm = 100000 dm2, sq dam
2000 hm2, sq hm = 200000 dm2, sq dam
4000 hm2, sq hm = 400000 dm2, sq dam
5000 hm2, sq hm = 500000 dm2, sq dam
7500 hm2, sq hm = 750000 dm2, sq dam
10000 hm2, sq hm = 1000000 dm2, sq dam
25000 hm2, sq hm = 2500000 dm2, sq dam
50000 hm2, sq hm = 5000000 dm2, sq dam
100000 hm2, sq hm = 10000000 dm2, sq dam
1000000 hm2, sq hm = 100000000 dm2, sq dam
1000000000 hm2, sq hm = 100000000000 dm2, sq dam
(Square Hectometers) to (Square Dekameters) conversions | 1,174 | 3,107 | {"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-2020-05 | latest | en | 0.801034 |
https://www.esaral.com/q/factorize-94463 | 1,720,809,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00741.warc.gz | 588,821,066 | 11,311 | # Factorize:
Question:
Factorize:
$a(a-2 b-c)+2 b c$
Solution:
We have:
$a(a-2 b-c)+2 b c=a^{2}-2 a b-a c+2 b c$
$=\left(a^{2}-2 a b\right)-(a c-2 b c)$
$=a(a-2 b)-c(a-2 b)$
$=(a-2 b)(a-c)$ | 96 | 198 | {"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} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.312634 |
https://www.clutchprep.com/chemistry/practice-problems/72390/if-an-ideal-gas-has-a-pressure-of-6-07-atm-a-temperature-of-10-42-c-and-has-a-vo | 1,611,398,056,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703537796.45/warc/CC-MAIN-20210123094754-20210123124754-00064.warc.gz | 721,679,612 | 31,613 | # Problem: If an ideal gas has a pressure of 6.07 atm, a temperature of 10.42 °C, and has a volume of 69.83 L how many moles of gas are in the sample?
###### FREE Expert Solution
Using the ideal gas law: PV=nRT
where P is 6.07 atm, V is 69.83 L, T as 10.42oC and R as 0.0821
97% (382 ratings)
###### Problem Details
If an ideal gas has a pressure of 6.07 atm, a temperature of 10.42 °C, and has a volume of 69.83 L how many moles of gas are in the sample? | 153 | 460 | {"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-2021-04 | latest | en | 0.94666 |
https://numbermatics.com/n/5044075/ | 1,685,694,843,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00391.warc.gz | 480,679,149 | 6,206 | # 5044075
## 5,044,075 is an odd composite number composed of three prime numbers multiplied together.
What does the number 5044075 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 12 divisors.
5044075 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twelve divisors.
## Prime factorization of 5044075:
### 52 × 89 × 2267
(5 × 5 × 89 × 2267)
See below for interesting mathematical facts about the number 5044075 from the Numbermatics database.
### Names of 5044075
• Cardinal: 5044075 can be written as Five million, forty-four thousand and seventy-five.
### Scientific notation
• Scientific notation: 5.044075 × 106
### Factors of 5044075
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 2361
### Divisors of 5044075
• Number of divisors d(n): 12
• Complete list of divisors:
• Sum of all divisors σ(n): 6327720
• Sum of proper divisors (its aliquot sum) s(n): 1283645
• 5044075 is a deficient number, because the sum of its proper divisors (1283645) is less than itself. Its deficiency is 3760430
### Bases of 5044075
• Binary: 100110011110111011010112
• Base-36: 30417
### Squares and roots of 5044075
• 5044075 squared (50440752) is 25442692605625
• 5044075 cubed (50440753) is 128334849704717921875
• The square root of 5044075 is 2245.9018233217
• The cube root of 5044075 is 171.4985734285
### Scales and comparisons
How big is 5044075?
• 5,044,075 seconds is equal to 8 weeks, 2 days, 9 hours, 7 minutes, 55 seconds.
• To count from 1 to 5,044,075 would take you about twelve weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 5044075 cubic inches would be around 14.3 feet tall.
### Recreational maths with 5044075
• 5044075 backwards is 5704405
• 5044075 is a Harshad number.
• The number of decimal digits it has is: 7
• The sum of 5044075's digits is 25
• More coming soon!
MLA style:
"Number 5044075 - Facts about the integer". Numbermatics.com. 2023. Web. 2 June 2023.
APA style:
Numbermatics. (2023). Number 5044075 - Facts about the integer. Retrieved 2 June 2023, from https://numbermatics.com/n/5044075/
Chicago style:
Numbermatics. 2023. "Number 5044075 - Facts about the integer". https://numbermatics.com/n/5044075/
The information we have on file for 5044075 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 5044075, math, Factors of 5044075, curriculum, school, college, exams, university, Prime factorization of 5044075, STEM, science, technology, engineering, physics, economics, calculator, five million, forty-four thousand and seventy-five.
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http://www.lookingforananswer.net/what-is-the-code-to-print-odd-numbers-from-20-to-30-in-qbasic.html | 1,481,381,782,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543315.68/warc/CC-MAIN-20161202170903-00302-ip-10-31-129-80.ec2.internal.warc.gz | 568,629,204 | 10,305 | # What is the code to print odd numbers from 20 to 30 in qbasic?
C++ code to print all odd and even numbers in given range ... Java Netbean Code from My Videos (30) Java Swing (11) ... (20) 2012 ... - Read more
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Related Questions
Recent Questions | 1,780 | 7,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-50 | latest | en | 0.816139 |
https://www.grasshopper3d.com/photo/s1 | 1,725,858,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00203.warc.gz | 741,115,812 | 14,601 | Grasshopper
algorithmic modeling for Rhino
s1
strokes of a drawing to vector field
tangent vectors on strokes learnt by resilient propagation algorithm to construct a vector field function from it;
learning time 10 minutes (8 cores)
Views: 820
Comment
You need to be a member of Grasshopper to add comments!
Comment by Shahrokh Kamyab on May 28, 2018 at 1:27pm
Hi!
Can you explain the workflow a bit more? you mentioned it's gonna be part of Octopus?
Best regards
Shahrokh
Comment by Robert Vier on December 15, 2015 at 4:25am
The 'field' is a collection of Bipolar Sigmoid Functions. Their coefficients get altered by the algorithm.
The network has
2 input nodes = x|y of point (a point on the curves)
2 output nodes = x|y of assigned direction (the tangent)
and a number of hidden nodes which are fully connected.
Dividing the curves then gives you a number of samples (pairs of inputs with known desired pairs of output values) to figure out some coefficients which minimize the error between inputs and outputs.
The rest is simple RungeKutta4 integration with the trained 'field' giving directions.
A big drawback is that it's not PI-invariant. I cant figure an efficient way to make it ignore the orientation of the input vectors.
Compared to just interpolating the vectors it's also dead slow and inaccurate. But it's an intuitive way to get a grip on what those networks can and cannot do well.
Comment by David Rutten on December 15, 2015 at 3:09am
How is the field defined? Is it a sampling grid? A collection of polynomials? A collection of trig functions?
Comment by Robert Vier on December 15, 2015 at 2:40am
basically the field tries to fit itself to a curve you've drawn. will be part of the next release of octopus
Comment by Tudor Cosmatu on December 15, 2015 at 2:35am
beautiful
Comment by Kellan Shanahan on December 14, 2015 at 4:47pm
Looks fantastic! Meanwhile I translated Alexander's link into GH using Anemone, one can get a variety of forms using the scale input on the Perlin Component.
Comment by Robert Vier on December 14, 2015 at 8:18am
schiele's standing nude with stockings btw
Comment by Alexander on December 14, 2015 at 7:24am
Oh now I understand. Your explanation flew right by me but makes sense now. Trippy. Interesting.
Comment by Robert Vier on December 14, 2015 at 3:39am
nice link! although I'm not doing noise, at least trying :)
Comment by Alexander on December 13, 2015 at 8:39pm
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© 2024 Created by Scott Davidson. Powered by | 764 | 2,933 | {"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-2024-38 | latest | en | 0.926624 |
https://www.mathsisfun.com/puzzles/breaking-up-a-chocolate-bar-solution.html | 1,529,542,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00284.warc.gz | 837,735,400 | 2,674 | # Breaking Up a Chocolate Bar Puzzle - Solution
## The Puzzle:
How many steps are required to break an m × n sized bar of chocolate into 1 × 1 pieces?
You can break an existing piece of chocolate horizontally or vertically.
You cannot break two or more pieces at once (so no cutting through stacks).
## Our Solution:
You need m×n - 1 steps.
By breaking an existing piece horizontally or vertically, you merely increase the total number of pieces by one.
You already have 1 piece, so need m×n - 1 steps to get to m×n pieces.
See this puzzle without solution
Discuss this puzzle at the Math is Fun Forum | 139 | 609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-26 | latest | en | 0.878438 |
https://www.airmilescalculator.com/distance/dfw-to-cho/ | 1,603,935,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902683.56/warc/CC-MAIN-20201029010437-20201029040437-00564.warc.gz | 615,813,591 | 43,240 | # Distance between Dallas, TX (DFW) and Charlottesville, VA (CHO)
Flight distance from Dallas to Charlottesville (Dallas/Fort Worth International Airport – Charlottesville–Albemarle Airport) is 1106 miles / 1780 kilometers / 961 nautical miles. Estimated flight time is 2 hours 35 minutes.
Driving distance from Dallas (DFW) to Charlottesville (CHO) is 1232 miles / 1983 kilometers and travel time by car is about 21 hours 13 minutes.
## Map of flight path and driving directions from Dallas to Charlottesville.
Shortest flight path between Dallas/Fort Worth International Airport (DFW) and Charlottesville–Albemarle Airport (CHO).
## How far is Charlottesville from Dallas?
There are several ways to calculate distances between Dallas and Charlottesville. Here are two common methods:
Vincenty's formula (applied above)
• 1105.765 miles
• 1779.557 kilometers
• 960.884 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1103.809 miles
• 1776.408 kilometers
• 959.184 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Dallas/Fort Worth International Airport
City: Dallas, TX
Country: United States
IATA Code: DFW
ICAO Code: KDFW
Coordinates: 32°53′48″N, 97°2′16″W
B Charlottesville–Albemarle Airport
City: Charlottesville, VA
Country: United States
IATA Code: CHO
ICAO Code: KCHO
Coordinates: 38°8′18″N, 78°27′10″W
## Time difference and current local times
The time difference between Dallas and Charlottesville is 1 hour. Charlottesville is 1 hour ahead of Dallas.
CDT
EDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 157 kg (346 pounds).
## Frequent Flyer Miles Calculator
Dallas (DFW) → Charlottesville (CHO).
Distance:
1106
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1106
Round trip? | 509 | 2,070 | {"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-2020-45 | latest | en | 0.787747 |
https://www.jiskha.com/display.cgi?id=1274323338 | 1,502,971,780,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103270.12/warc/CC-MAIN-20170817111816-20170817131816-00201.warc.gz | 886,019,832 | 3,670 | # Chemistry
posted by .
How many kilojoules of heat are absorbed when 1.00L of water is heated from 18 degrees C to 85 degrees C?
Did I do this right?
1000g H2O x 4.18 x 67 = 280060 kilojoules
(my book says the specific heat for water in J/g x degrees C = 4.18, and for cal/g x degrees C = 1.00) Would i multiply 1000g x 67 x by 4.18 or 1?
• Chemistry -
Whether you use 1 cal/g*C or 4.18 J/g*C depends upon what unit you want the answer in. The first one gives the answer in calories, the second in joules. Your math is ok but the answer is not kJ, but J. You need to divide you answer by 1000 to convert to kJ.
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More Similar Questions | 605 | 2,087 | {"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-2017-34 | latest | en | 0.91053 |
https://diy.stackexchange.com/questions/152366/is-joist-blocking-safe-on-these-old-warped-joists/152404 | 1,603,455,645,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00333.warc.gz | 286,001,179 | 33,244 | # Is joist blocking safe on these old warped joists?
I purchased a house built in 1960 and like other people have mentioned the terrible bounce, I got the same exact situation.
I'm thinking of installing blocking to strengthen the floor. The 7 joists are spaced 14½ However, since they're all buckled at the lower part of the joist, ranging from 14⅛ - 14⅞ I wanted to know if it's safe to wedge between 14½ blocks forcing them to move back that extra space. (see below picture)
I also so wanted to know if this load sharing would cause to much weight on other parts of floor beams.
• It looks like you have 2x12’s (1 5/8” x 11 5/8”) floor joists at 16” on center. Can you confirm this? AND what is the span of these joists (face of support to face of support). – Lee Sam Dec 10 '18 at 2:15
• It's 2x10 spaced 15 in I don't understand the the rest of what you're asking me to check – Joseph Wit Dec 10 '18 at 3:03
• What is the distance the 2x10’s span from support to support? (I’m guessing it’s about 16 feet or so?) – Lee Sam Dec 10 '18 at 3:11
• Yup it's 16 feet – Joseph Wit Dec 10 '18 at 3:30
• I reconstructed my question and concern – Joseph Wit Dec 17 '18 at 5:28
Bouncy floors are caused from too much deflection in the floor. Too much deflection is caused by “undersized” floor joists for the load and span.
Depending on the species and grade of your floor joists, they should only span about 14’-15’ or so. When your five year old runs across the floor it “pounds” (impacts) the floor supports. Because the floor joists are already overloaded, those little feet can cause the bouncy floor problem.
To correct the situation, you’ll need to 1) increase the size of the joists or 2) decrease the span.
1) You can increase the size of your supports by adding a “sister joist” or adding a joist between each existing joist.
2) You can dresses the span by adding a support beam under the joists with new footings, etc.
It’s complicated and expensive. But the good news: When he turns into a teenager, that bouncy floor will seem trivial.
• Are any of the three options going to work for my floor see link below – Joseph Wit Dec 10 '18 at 18:18
• I’ve never heard of adding bridging to solve deflection problems. I doubt if it works, but I like the third option: add a support to decrease the span. – Lee Sam Dec 10 '18 at 19:56
• Can I use this Simpson Strong-Tie 22-Gauge Tension Bridging – Joseph Wit Dec 13 '18 at 19:27
• Or this 3/4 in. x 10 ft. Galvanized Steel Hanger Strap – Joseph Wit Dec 13 '18 at 19:28 | 700 | 2,528 | {"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-2020-45 | latest | en | 0.951545 |
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Egge September 2, 2018 05:25
Runtime Integration and using the result to correct a field
Hello,
I am calculating a pipe flow with heat transfer (Neumann-BC) using cyclic boundary conditions.
I modified the solver pimplefoam to solve for the "modified Temperature field", so the temperature stays more or less constant over the time. This has to be done due to the cyclic boundary conditions.
But unfortunately the temperature changes a little bit. The definition of the "modified temperature field" implies that the modified bulk temperature is zero, which appears to be not the case according to my post-processing
As a solution I want to substract the "modified bulk temperature" from each time step from the "modified temperature" at each time step.
And here is were my question occurs. I clicked through several threads but did not find the exact answer I was looking for, maybe one of you had a similiar problem in the past.
Def. Bulk Temp.:
So I need to do a surface integral to get the Bulk temperature, which is in my case the pipe cross section.
My algorithm so far would be:
1. average along the axial direction of the pipe (here z-direction)
2. integrate above the cross section and get a scalar value
3. read the bulk temperature in a vector corresponding to the length of uncorrected
temperature field
4. substract the bulk temperature
Step 3 and 4 I at least guess it would not be a problem, the opposites holds for 1 and 2.
My Problems are:
1. How can I average along the axial-direction. Can I tell openfoam to average all elements along the axis?
2. I know there is something like patchIntegrate from swak4foam, BUT I guess it cannot handle the averaged Values, because there is not patch which contains the information. Or can I define one? I mean it could be arbitary, because Only the face integral is important not the axial position anymore. But it must have the same geometric characteristics of the e.g. inlet or outlet or any cross section in the pipe.
3. is there a already existing face integration option in openfoam, because the way over swak4foam could be more complicated than using an existing option. | 557 | 2,520 | {"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-2021-31 | latest | en | 0.922506 |
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# If A = -|B| and A-B =-10, what is B?
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If A = -|B| and A-B =-10, what is B? [#permalink]
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Updated on: 07 Mar 2016, 07:52
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If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
Source: Sravna test prep
Originally posted by vinoo7 on 07 Mar 2016, 07:10.
Last edited by ENGRTOMBA2018 on 07 Mar 2016, 07:52, edited 1 time in total.
Reformatted the question and added the OA.
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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07 Mar 2016, 07:40
1
vinoo7 wrote:
If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
Follow posting guidelines (link in my signatures), especially mention the source of the question and the OA.
Question is not of a good quality.
For this question, let us use the options given to us.
A. B=5 ---> A= -|B| = -5 ---> -5-5 = -10. , True. Keep
B. B=-5 ---> A= -|B| = -5 ---> -5+5 = 0 $$\neq$$ -10. Eliminate.
C. B=0 ---> A= -|B| = 0 ---> -5-0 = -5 $$\neq$$ -10 , False. Eliminate.
D. B=-5 or +5 ---> As mentioned in option B, -5 does not follow the given statements. Eliminate
E. B=-10 ---> A= -|B| = -10 ---> -10+10=0 $$\neq$$ -10. Eliminate.
Thus, A is the correct answer.
Hope this helps.
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If A = -|B| and A-B =-10, what is B? [#permalink]
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Updated on: 16 Sep 2017, 12:49
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Given $$A = -|B|$$
=> if B =-ve then $$A= -|-B|$$ = -ve
=> if B=+ve then $$A=-|B|$$ = -ve
as given$$A-B= -10$$ and A and B have same magnitude then possible values of $$|B|= 5$$=> B is either -5 or 5
if B =5 =>$$A =-|5|=-5$$ => $$A-B = -5-(5) =-10$$ Answer
if B = -5 =>$$A=-|-5| = -5$$=> $$A-B = -5-(-5) =-5+5 =0$$
So for$$A-B =-10$$ => value of B =5
Originally posted by Nikkb on 16 Sep 2017, 11:44.
Last edited by Nikkb on 16 Sep 2017, 12:49, edited 2 times in total.
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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16 Sep 2017, 11:59
Nikkb wrote:
Given A = -|B|
=> if B =-ve then A=+ve B
=> if B=+ve then A=-ve B
as given A-B= -10 and A and B have same magnitude then possible values of |B|= 5 => B is either -5 or 5
if B =5 => A =-5 => A-B = -5-(5) =-10 Answer
if B = -5 => A=-(-5) =5 => A-B = 5-(-5) =5+5 =10
So for A-B =-10 => value of B =5
Hi Nikkb
You have got the right answer but the highlighted portions are wrong.
$$A = -|B|$$. Notice that the negative sign is outside the mod function and mod is always positive. Hence $$A$$ will be always negative irrespective of $$B$$
So if $$B=-5$$, then $$A=-|-5| = -|5| =-5$$
Hence $$A-B = -5-(-5) = 0$$. Hence B cannot be $$-5$$
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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16 Sep 2017, 12:27
niks18 wrote:
Nikkb wrote:
Given A = -|B|
=> if B =-ve then A=+ve B
=> if B=+ve then A=-ve B
as given A-B= -10 and A and B have same magnitude then possible values of |B|= 5 => B is either -5 or 5
if B =5 => A =-5 => A-B = -5-(5) =-10 Answer
if B = -5 => A=-(-5) =5 => A-B = 5-(-5) =5+5 =10
So for A-B =-10 => value of B =5
Hi Nikkb
You have got the right answer but the highlighted portions are wrong.
$$A = -|B|$$. Notice that the negative sign is outside the mod function and mod is always positive. Hence $$A$$ will be always negative irrespective of $$B$$
So if $$B=-5$$, then $$A=-|-5| = -|5| =-5$$
Hence $$A-B = -5-(-5) = 0$$. Hence B cannot be $$-5$$
Hey yah thanks for pointing it out. Totally agree with you. While solving question i made correct assumptions.
while posting solution was in between 2 things so made mistake. Will update the solution.
Thanks
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If A = -|B| and A-B =-10, what is B? [#permalink]
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17 Sep 2017, 00:16
Option: A
Since the absolute value of both A and B is same, all we have to do is plug values from option in the equation A-B = -10.
Only A suffices. Answer can be reached within a minute.
+1 Kudos if it helped
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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19 Sep 2017, 00:45
1
A=-IBI
A-B=-10
-IBI-B=-10
-B-B=-10
-2B=-10
2B=10
B=10/2=5=A
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If A = -|B| and A-B =-10, what is B? [#permalink]
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01 Oct 2017, 09:24
If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
-|B|-B=-10
-B-B=-10
-2B=-10
B=10/2
B=5 ANS.
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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03 Oct 2017, 00:18
1
Since A=-|B|, there are two possible relations
A=-B (or) A=B
If we substitute both of them in the equation,
If A=-B ---> A-B = -2B => -2B=-10 => B=5
If A=B ---> A-B = 0 (This is not possible since (A-b=-10))
Hence, A is the correct ans
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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07 Mar 2018, 05:29
I understand the solutions above and now understand why the answer must be 5.
But I think what tripped me up is that I think that because A=-|B| that A has to be negative. Because if we make B=5, then A=-5. If we make B=-5, then A=-5.
I understand that because A-B=-10, that B has to be 5, but using the first equation in the questions I feel like 5 doesn't work because that makes it ==> A=-|B| ==>5=-|5|==>5=-5, which is not true. Can someone explain? Why are we allowed to make A a positive number when the equation makes it seem like A has to be negative.
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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07 Mar 2018, 05:39
msurls wrote:
If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
I understand the solutions above and now understand why the answer must be 5.
But I think what tripped me up is that I think that because A=-|B| that A has to be negative. Because if we make B=5, then A=-5. If we make B=-5, then A=-5.
I understand that because A-B=-10, that B has to be 5, but using the first equation in the questions I feel like 5 doesn't work because that makes it ==> A=-|B| ==>5=-|5|==>5=-5, which is not true. Can someone explain? Why are we allowed to make A a positive number when the equation makes it seem like A has to be negative.
A = -|B| = -(non-negative value) = (non-positive value). So, A = -|B| means that A is positive or 0.
Now, if B = 5, then A = -|B| = -|5| = -5 = negative.
If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
A = -|B| and A - B = -10, thus:
-|B| - B = -10;
B + |B| = 10.
If B is negative or 0, then we'd have B - B = 10 --> 0 = 10, which is not true.
If B is positive, then we'd have B + B = 10 --> B = 5.
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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08 Mar 2018, 17:56
1
vinoo7 wrote:
If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
Sine A = B - 10, we have:
B - 10 = -|B|
When B is positive we have:
B - 10 = -B
2B = 10
B = 5
When B is negative we cannot get a solution for B. So B must be 5.
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Re: If A = -|B| and A-B =-10, what is B? [#permalink]
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21 Mar 2018, 23:24
vinoo7 wrote:
If A = -|B| and A-B =-10, what is B?
A. 5
B. -5
C. 0
D. -5 or +5
E. -10
Source: Sravna test prep
A = - |B|
means A = B or A = -B
A - B = -10 --------(I)
Replace A with B,
B - B = -10
0 = -10 (Not true)
Replace A with -B,
-B -B = -10
B = 5
Hence (A)
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Re: If A = -|B| and A-B =-10, what is B? [#permalink] 21 Mar 2018, 23:24
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