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https://www.sqlservercentral.com/Forums/1014270/Predict-the-outcome-of-the-SQL-statements | 1,506,341,140,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691476.47/warc/CC-MAIN-20170925111643-20170925131643-00360.warc.gz | 840,614,259 | 34,927 | ## Predict the outcome of the SQL statements
Author Message bazzkar SSC-Enthusiastic Group: General Forum Members Points: 186 Visits: 72 Comments posted to this topic are about the item Predict the outcome of the SQL statements UMG Developer SSCertifiable Group: General Forum Members Points: 5448 Visits: 2204 Nice question, thanks! Though I think the explanation left a little to be desired. It isn't just that the decimal type was used, it was that the decimal type was used and no precision/scale was specified, so the number got rounded down to 1.0 when it was converted.For example this results in the same thing as all of your float examples:`select Ceiling(convert(decimal(2, 1), 1.09))`Because with a scale of one specified it gets rounded to 1.1 before it goes to ceiling. Toreador SSCarpal Tunnel Group: General Forum Members Points: 4701 Visits: 8171 Knew the answer but had a brain fart and clicked the wrong option, oops!UMG Developer is right about the explanation. Mike Is Here SSCrazy Group: General Forum Members Points: 2060 Visits: 513 Good question but I think a better explaination is:the default of decimal without parameters is (18,0) thus making 1.09 -> 1 and then the ceiling of 1 is 1 John.Norcott SSC Journeyman Group: General Forum Members Points: 75 Visits: 49 Thank you Old Hand. I knew the ceiling function brought you up to the next integer, but I couldn't figure out why the decimal was any different than the float on this one! You're explanation was what I was missing. For some reason I thought the default precision was 2, not 0. I'll be sure to remember that the next time I use a decimal data type. Toreador SSCarpal Tunnel Group: General Forum Members Points: 4701 Visits: 8171 I hope one thing that everybody has learned from these questions is that you should never rely on default precision for anything - you will inevitably get it wrong sooner or later :-) Tom Thomson SSC-Dedicated Group: General Forum Members Points: 36177 Visits: 12793 Good question, but as some have already noted the explanation is a bit lacking.Also, the question was made too easy by the absence of the All 2s option as an answer choice - anyone who knows what ceiling means can eliminate all answers but the correct one with knowing anything at all about decimal or its default precision. Tom Cliff Jones SSCertifiable Group: General Forum Members Points: 5875 Visits: 3648 Tom.Thomson (11/2/2010)Good question, but as some have already noted the explanation is a bit lacking.Also, the question was made too easy by the absence of the All 2s option as an answer choice - anyone who knows what ceiling means can eliminate all answers but the correct one with knowing anything at all about decimal or its default precision.Yes, I agree. The best wrong answer was missing. SQLRNNR SSC Guru Group: General Forum Members Points: 100254 Visits: 18616 Thanks for the question. Jason AKA CirqueDeSQLeilI have given a name to my pain...MCM SQL Server, MVPSQL RNNRPosting Performance Based Questions - Gail Shaw Koen Verbeeck SSC Guru Group: General Forum Members Points: 99853 Visits: 13323 Good question, although it was more about convert than about ceiling.A link to the msdn page for convert and decimal:http://msdn.microsoft.com/en-us/library/ms187928(SQL.90).aspx (convert)http://msdn.microsoft.com/en-us/library/ms187746.aspx (decimal/numeric)The last page describes the gotcha of this question in the explanation of the scale. How to post forum questions.Need an answer? No, you need a question.What’s the deal with Excel & SSIS?My blog at SQLKover.MCSE Business Intelligence - Microsoft Data Platform MVP | 857 | 3,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.917456 |
https://www.physicsforums.com/threads/magnetic-field-strength-that-emits-waves.762555/ | 1,508,767,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826049.46/warc/CC-MAIN-20171023130351-20171023150351-00190.warc.gz | 943,491,363 | 17,136 | # Magnetic field strength that emits waves
1. Jul 21, 2014
### jayayo
1. The problem statement, all variables and given/known data
The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at 2.99GHz, and as they do so they emit 2.99GHz electromagnetic waves. What is the magnetic field strength?
2. Relevant equations
ƒ=qβ/(2Pi*m)
Thus, B= f*2Pi*m/q
3. The attempt at a solution
I got the question right. It's just I am having a bit difficult grasping why. I thought that because they stated that the electrons also emit 2.99Hz electromagnetic waves, that this would influence the answer. Could someone please explain to me if electrons emitting waves of the same frequency has any influence on this question and the found B value? And also, if not, why?
Thank you so much!
2. Jul 21, 2014
### rude man
The electrons ARE what generate the e-m waves. The frequency is just the electrons' rotational frequency. So I don't know what you mean by ".. the electrons ALSO emit 2.99 GHz electromagnetic waves ...".
The problem just involves equating the Lorentz force with centripetal force which your formula does.
3. Jul 22, 2014
### jayayo
Hi~
Sorry- pretty stupid of me. I just thought because they purposely added the fact that the electrons also emitted EM waves of the same frequency, maybe this would increase or decrease the value of the magnetic field.
What would change due to the fact that the electrons were emitting EM waves of the same frequency? Energy, right? Anything else?
4. Jul 22, 2014
### rude man
The B field is not affected. Without the electrons there would not be any e-m radiation. If the electrons were present but stationary you would have a static E field on top of the B field but no e-m radiation. Charges have to be accelerated to emit e-m radiation.
E-m radiation carries energy with it. The so-called Poynting vector P indicates the direction of e-m radiation and energy transport. The power flux is P times the area under consideration, expressed as a dot product since both P and area A are vectors, and the units are watts/square meter. So power flow = P*A over any area.
It can be shown that P = E x H.
.
. | 525 | 2,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-43 | longest | en | 0.937756 |
https://www.meritnation.com/cbse/class10/ncert-solutions/math/math/arithmetic-progressions/page112-exercise-5-3-qno3/9_132_2503 | 1,506,287,076,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690211.67/warc/CC-MAIN-20170924205308-20170924225308-00239.warc.gz | 798,770,610 | 20,982 | 011-40705070 or
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# Class 10 - Math - Arithmetic Progressions
## Arithmetic Progressions
Question 3:
In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x)Given l = 28, S = 144 and there are total 9 terms. Find a.
To view the solution to this question please
OR | 279 | 706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-39 | latest | en | 0.765145 |
https://discuss.leetcode.com/topic/27542/share-my-iterative-way-of-bfs | 1,513,546,278,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597585.99/warc/CC-MAIN-20171217210620-20171217232620-00672.warc.gz | 567,758,658 | 9,114 | # Share my iterative way of BFS
• ``````public class Solution {
class Position{
public int row ;
public int col;
public Position(int r, int c){
row = r;
col = c;
}
}
public List<List<String>> solveNQueens(int n) {
//bfs
List<List<String>> res = new ArrayList<>();
for(int i = 0; i < n; i++){ ArrayList<Position> path = new ArrayList<Position>(); path.add(new Position(0,i)); q.add(path);}
while(!q.isEmpty()){
int size = q.size();
for(int i = 0; i < size; i++){
ArrayList<Position> path = q.remove();
if(path.size() == n) {
List<String> result = new ArrayList<>();
for(Position p: path){
StringBuilder row = new StringBuilder();
for(int j = 0; j < n; j++){
if(j == p.col) row.append("Q");
else row.append(".");
}
}
continue;
}
ArrayList<Position> successors = getSuccessors(path,n);
for(Position successor : successors){
path.remove(path.size()-1);
}
}
}
return res;
}
public ArrayList<Position> getSuccessors(ArrayList<Position> path,int n){
ArrayList<Position> successors = new ArrayList<Position> ();
int row = path.size();
ArrayList<Integer> set = new ArrayList<>();
ArrayList<Integer> removed = new ArrayList<>();
for(int i = 0; i < n ;i++){
}
for(Position pre : path)
set.remove((Integer)pre.col);
for(Position pre : path){
for(int c : set){
if(Math.abs(pre.row - row)==Math.abs(pre.col - c)) | 345 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-51 | latest | en | 0.180696 |
https://gmatclub.com/forum/lack-of-fresh-water-is-an-ongoing-problem-in-the-outposts-and-it-i-208789.html | 1,561,633,217,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001089.83/warc/CC-MAIN-20190627095649-20190627121649-00050.warc.gz | 463,007,531 | 151,920 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Lack of fresh water is an ongoing problem in the outposts, and it i
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Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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18 Nov 2015, 15:43
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Lack of fresh water is an ongoing problem in the outposts, and it is expected to continue until reinforcements arrive.
(A) Lack of fresh water is an ongoing problem in the outposts, and it is expected to continue until reinforcements arrive.
(B) Lack of fresh water is an ongoing problem in the outposts, which was expected to continue until reinforcements arrive.
(C) Lack of fresh water is an ongoing problem in the outposts, and they are expected to continue until reinforcements arrive.
(D) The outposts lack fresh water, a problem that is expected to continue until reinforcements arrive.
(E) The outposts have a lack of fresh water, a problem expected to continue until reinforcements arrive.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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11 Dec 2015, 07:00
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I'm replying to a PM on this one. This is certainly tricky and really comes down to recognizing the difference between the appositive and absolute phrase. They are super similar and hard to differentiate. These constructions require an absolute phrase (modifying the preceding clause) and not an appositive (modifying the preceding noun) because "the problem" needs to refer back to the preceding clause and not fresh water (fresh water is not a problem). An absolute phrase is typically formed with a noun + participle and contains no tensed verbs. You can see that option E does have an absolute phrase (noun + participle) but D is not (contains a tensed verb).
(D) The outposts lack fresh water, a problem that IS expected to continue until reinforcements arrive.
(E) The outposts have a lack of fresh water, a problem EXPECTED to continue until reinforcements arrive.
KW
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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Updated on: 20 Nov 2015, 03:25
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Aksh19 wrote:
OA should be D.
The problem refers to the lack of fresh water in the outposts. The structuring of option E is very awkward.
Something I learnt recently and it might help. Have a look at this difference between Appositive and Absolute phrases.
Best explained in this post by Kevin from Magoosh
http://magoosh.com/gmat/2015/gmat-tuesd ... e-phrases/
Applying that here
Lack of fresh water is an ongoing problem in the outposts, and it is expected to continue until reinforcements arrive.
(A) Lack of fresh water is an ongoing problem in the outposts, and it is expected to continue until reinforcements arrive. E is more concise
(B) Lack of fresh water is an ongoing problem in the outposts, which was expected to continue until reinforcements arrive. which is incorrectly modifying 'outposts'
(C) Lack of fresh water is an ongoing problem in the outposts, and they are expected to continue until reinforcements arrive. they is an incorrect pronoun used here
(D) The outposts lack fresh water, a problem that is expected to continue until reinforcements arrive. Modifiers modify the immediate preceding noun. Here it is fresh water. And hence an incorrect modifier reference
(E) The outposts have a lack of fresh water, a problem expected to continue until reinforcements arrive. An Absolute phrase here. Modifying the entire statement before it.
Hope this helps. Please correct me in case I'm wrong.
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Originally posted by hdwnkr on 18 Nov 2015, 23:51.
Last edited by hdwnkr on 20 Nov 2015, 03:25, edited 1 time in total.
##### General Discussion
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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18 Nov 2015, 21:04
1
i think the best option here is D: The outposts lack fresh water, a problem that is expected to continue until reinforcements arrive.
E, on the other hand though not wrong but is more wordy :The outposts have a lack of fresh water, a problem expected to continue until reinforcements arrive.---->why would one use "have a lack" when one can easily convey the meaning in more succinct way: The outposts lack fresh water
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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18 Nov 2015, 23:14
OA should be D.
The problem refers to the lack of fresh water in the outposts. The structuring of option E is very awkward.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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19 Nov 2015, 05:49
Dear All,
Even I went for D but found it strange that E was OA
this is the explanation provided
"(A) The pronoun it has an unclear antecedent
(B) The pronoun which has an unclear antecedent
(C) The plural pronoun they could only refer to outposts, which implies that the outposts are expected to continue rather than the problem
(D) A problem incorrectly refers to fresh water ; it is the lack of fresh water that is the problem
(E) Correct. A problem correct refers to a lack of fresh water; the sentence is concise and correct. It is not necessary to use that, as in choice D. The correct answer is E"
Experts please help us as explanation given doesn't make it clear why E is better choice than D
This question is of Gmat Club Test V08-01
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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19 Nov 2015, 14:33
The answer is clearly either D or E. I can hardly see the difference between the two.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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19 Nov 2015, 23:40
kanigmat011 wrote:
Dear All,
Even I went for D but found it strange that E was OA
this is the explanation provided
"(A) The pronoun it has an unclear antecedent
(B) The pronoun which has an unclear antecedent
(C) The plural pronoun they could only refer to outposts, which implies that the outposts are expected to continue rather than the problem
(D) A problem incorrectly refers to fresh water ; it is the lack of fresh water that is the problem
(E) Correct. A problem correct refers to a lack of fresh water; the sentence is concise and correct. It is not necessary to use that, as in choice D. The correct answer is E"
Experts please help us as explanation given doesn't make it clear why E is better choice than D
This question is of Gmat Club Test V08-01
the EXPLANATION is NOT CORRECT !!
in a construction such as CLAUSE+ COMMA + ABSTRACT NOUN + MODIFIER------> The "ABSTRACT NOUN" refers to the previous clause and NOT TO SOME SPECIFIC NOUN !!
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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Updated on: 20 Nov 2015, 03:24
1
kanigmat011 wrote:
Dear All,
Even I went for D but found it strange that E was OA
this is the explanation provided
"(A) The pronoun it has an unclear antecedent
(B) The pronoun which has an unclear antecedent
(C) The plural pronoun they could only refer to outposts, which implies that the outposts are expected to continue rather than the problem
(D) A problem incorrectly refers to fresh water ; it is the lack of fresh water that is the problem
(E) Correct. A problem correct refers to a lack of fresh water; the sentence is concise and correct. It is not necessary to use that, as in choice D. The correct answer is E"
Experts please help us as explanation given doesn't make it clear why E is better choice than D
This question is of Gmat Club Test V08-01
the EXPLANATION is NOT CORRECT !!
in a construction such as CLAUSE+ COMMA + ABSTRACT NOUN + MODIFIER------> The "ABSTRACT NOUN" refers to the previous clause and NOT TO SOME SPECIFIC NOUN !!
Not an expert but I am confident of my explanation
D) The outposts lack fresh water, a problem that is expected to continue until reinforcements arrive.
Modifiers modify the immediate preceding noun. The outposts (noun phrase) lack (verb) fresh water (noun), a problem(modifier that refers to what is immediately before it)..... Fresh water is not a problem Lack of it is. Hence, D is incorrect
(E) The outposts have a lack of fresh water, a problem expected to continue until reinforcements arrive.
An Absolute phrase here. Modifying the entire statement before it. Hence, E is correct
In my reply above, I have put a link explaining appositive and absolute phrases. That might help!
Hope this helps! Please let me know whether I'm wrong.
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Originally posted by hdwnkr on 20 Nov 2015, 02:14.
Last edited by hdwnkr on 20 Nov 2015, 03:24, edited 1 time in total.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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20 Nov 2015, 02:39
Quote:
Modifiers modify the immediate preceding noun. The outposts (noun phrase) lack (verb) fresh water (noun), a problem(modifier that refers to what is immediately before it)..... Fresh water is not a problem Lack of it is. Hence, D is incorrect
This is not correct!! in fact you need to check the DISTINCTION between "CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER" construction AND "CLAUSE+COMMA+CONCRETE NOUN+MODIFIER" construction.
ALSO note that in both D and E "a problem" is an ABSTRACTION so there is no distinction as drawn by you.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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Updated on: 20 Nov 2015, 03:02
Quote:
Modifiers modify the immediate preceding noun. The outposts (noun phrase) lack (verb) fresh water (noun), a problem(modifier that refers to what is immediately before it)..... Fresh water is not a problem Lack of it is. Hence, D is incorrect
This is not correct!! in fact you need to check the DISTINCTION between "CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER" construction AND "CLAUSE+COMMA+CONCRETE NOUN+MODIFIER" construction.
ALSO note that in both D and E "a problem" is an ABSTRACTION so there is no distinction as drawn by you.
Can you please explain the distinction between the CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER" construction AND "CLAUSE+COMMA+CONCRETE NOUN+MODIFIER" construction? I couldn't find a promising explanation.
And, I hope you know what CAPS LOCK in written language means! With all due respect Mr Director, you might want to tone down a bit buddy..
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Fais de ta vie un rêve et d'un rêve une réalité
Originally posted by hdwnkr on 20 Nov 2015, 02:58.
Last edited by hdwnkr on 20 Nov 2015, 03:02, edited 2 times in total.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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20 Nov 2015, 03:09
Quote:
Can you please explain the distinction between the CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER" construction AND "CLAUSE+COMMA+CONCRETE NOUN+MODIFIER" construction. Couldn't find a promising explanation?
CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER------>in this construction the "ABSTRACTION" can refer to complete clause that comes before the COMMA
CLAUSE+COMMA+CONCRETE NOUN+MODIFIER------>in this construction "CONCRETE" noun refers to the NOUN that comes BEFORE COMMA
Quote:
And, I hope you know what CAPS LOCK in written language means! With all due respect Mr Director, you might want to tone down a bit buddy..
you got me wrong again. i have put those CONSTRUCTIONS in caps just to HIGHLIGHT else those construction will get mixed up in text. it was not to offend you
hope this clears up.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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20 Nov 2015, 03:21
Quote:
Can you please explain the distinction between the CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER" construction AND "CLAUSE+COMMA+CONCRETE NOUN+MODIFIER" construction. Couldn't find a promising explanation?
CLAUSE+COMMA+ABSTRACT NOUN+MODIFIER------>in this construction the "ABSTRACTION" can refer to complete clause that comes before the COMMA
CLAUSE+COMMA+CONCRETE NOUN+MODIFIER------>in this construction "CONCRETE" noun refers to the NOUN that comes BEFORE COMMA
Quote:
And, I hope you know what CAPS LOCK in written language means! With all due respect Mr Director, you might want to tone down a bit buddy..
you got me wrong again. i have put those CONSTRUCTIONS in caps just to HIGHLIGHT else those construction will get mixed up in text. it was not to offend you
hope this clears up.
Yes this clears up. And apologies for misunderstanding the intent. Peace!
What you have explained - Is it in the following case too?
Abstract
Fractal art was very popular at the museum, never-ending infinitely complex patterns self similar across different scales.
Concrete
Fractal art, never-ending infinitely complex patterns that are self similar across different scales, was the most popular exhibit at the museum.
Source: Magoosh Video
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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20 Nov 2015, 03:36
Quote:
Yes this clears up. And apologies for misunderstanding the intent. Peace!
Quote:
What you have explained is it in the following case too?
Abstract
Fractal art was very popular at the museum, never-ending infinitely complex patterns self similar across different scales.
Concrete
Fractal art, never-ending infinitely complex patterns that are self similar across different scales, was the most popular exhibit at the museum.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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22 Nov 2015, 06:22
1
GMAT Club verbal questions are pretty shady at times. And lack of proper explanations make it worse.
I wouldn't encourage spending too much time on it.
Thanks!
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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22 Nov 2015, 09:40
D is incorrect as -
The problem is "Lack of Fresh Water" and not "Fresh Water" or "Outposts lack Fresh Water"
"A problem" should refer to "Lack of Fresh Water".
While in D the preceding clause is "The outposts lack fresh water", making "a problem" refer to "Outposts lack Fresh Water" which is not the case.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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04 Dec 2015, 01:11
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Although abstract noun modifies the entire previous clause. I can not think of a better explanation.
Confusing question - "D" or "E". Even I am confused. I thought as below. Present your views.
Q:> Lack of fresh water is an ongoing problem in the outposts, and it is expected to continue until reinforcements arrive.
(A) Lack of fresh water is an ongoing problem in the outposts, anditis expected to continue until reinforcements arrive. antecedent for it ?
(B) Lack of fresh water is an ongoing problem in the outposts, which was expected to continue until reinforcements arrive. illogical.
(C) Lack of fresh water is an ongoing problem in the outposts, andtheyare expected to continue until reinforcements arrive. they refers to outposts
(D) The outposts lack fresh water, a problem that is expected to continue until reinforcements arrive.
---------------------Verb object, modifier (additional info). Incorrect reference for "a problem "
(E) The outposts have a lack of fresh water, a problem expected to continue until reinforcements arrive.
---------------- ----Verb object, modifier (additional info) correct reference for "a problem "
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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05 Dec 2015, 09:40
i have seen this sentence in a book that seems to be manhattan. I chose E because i thought that "a problem" should rể to a lack but the correct answer is D, the book explained that "have a lack" is awkward.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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06 Dec 2015, 03:50
thanhphong01 wrote:
i have seen this sentence in a book that seems to be manhattan. I chose E because i thought that "a problem" should rể to a lack but the correct answer is D, the book explained that "have a lack" is awkward.
Can you plz post the exact explanation given in that book.
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Re: Lack of fresh water is an ongoing problem in the outposts, and it i [#permalink]
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08 Dec 2015, 21:29
Greetings
Confusion b/w choices "D" & "E"
a problem" in option "D" as mentioned is modifying "fresh water" but in "E" it's modifying "a lack of fresh water. How come?
Is my explanation in my previous post correct ?
Why in option "D" "a problem" can not refer to the previous clause as "a problem" is an abstract phrase ?
Anyone who can elaborate?
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Display posts from previous: Sort by | 5,022 | 19,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-26 | latest | en | 0.935119 |
https://stats.stackexchange.com/questions/458337/repeated-measurements-with-different-measurement-methods-and-uneven-sample-size | 1,721,107,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514737.55/warc/CC-MAIN-20240716050314-20240716080314-00607.warc.gz | 467,426,921 | 45,709 | # Repeated measurements with different measurement methods and uneven sample size - which test is suited?
I am struggling to find the appropriate statistical test to analyze my data. I hope that my question will be understandable.
I have the following setup:
• A porcine spine with three vertebral bodies (L1,L2,L3).
• The spine was scanned on three different imaging modalities (Modality A,B,C)
• On each of the modalities, different rings of fat were wrapped around the spine resulting in 5 different simulated sizes (size 1 to 5).
• For each vertebral body of each of the sizes of each modality, I can measure the bone density (BD) as BD.L1, BD.L2, BD.L3
Here the first 10 rows of the table structure with some fictional values for the BD:
my.df <- structure(list(Modality = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("A", "B", "C"), class = "factor"),
Size = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 5L, 5L
), .Label = c("1", "2", "3", "4", "5"), class = "factor"),
Repeat = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L,
2L), .Label = c("1", "2", "3"), class = "factor"), BD.L1 = c(1.3,
1.5, 2.2, 1.2, 1.8, 1.7, 0.7, 2.3, 2.5, 1.3), BD.L2 = c(1.2,
1.7, 1.6, 1.6, 1.1, 1.3, 1, 1.3, 1.2, 1.5), BD.L3 = c(1.6,
1, 1.8, 1.2, 1, 1.1, 1.6, 1.5, 1.6, 1.8)), row.names = c(NA,
10L), class = "data.frame")
I would like to answer the following questions:
1. Are there significant differences in bone density (BD) measurements among the three modalities for each phantom size?
2. Are there significant differences in bone density (BD) measurements among the sizes within each modality?
Here the tricky part: for modality A all sizes were scanned twice (2 repeats) while for modalities B and C all sizes were scanned thrice (3 repeats).
Because the data points are very few, I thought to compare the BD measurements for each size not on a per-vertebra basis, but using the BD measurements of all three vertebra together for each modality and size.
Specific questions:
In regards to Analysis 1.) I was thinking about using the Friedman Test. However, I have unequal sample sizes (2 repeats for modality A) vs. (3 repeats for modality B). Which non-parametric test could I use here with unequal sample sizes?
In regards to Analysis 2.): Are the different sizes paired? If I add additional fat rings to the spine is it still considered the same or an independent sample. If independent is it correct to use Kruskal Wallis with Dunn post-hoc test to make comparisons among the five sizes?
I hope that my question is understandable.
Thank you very much!
Update:
For reproducibility a dataset representing the full data with fictive values has been added:
set.seed(23)
df <- data.frame(
Modality = c(rep("A",30),rep("B",45),rep("C",45)),
Size = factor(c(rep(rep(1:5,each=2),3),rep(rep(1:5,each=3),6)), levels=c(1,2,3,4,5),ordered=TRUE),
Repeat = factor(c(rep(1:2,15),rep(rep(1:3,15),2))),
Level = c(rep(c("L1","L2","L3"),each=10),rep(rep(c("L1","L2","L3"),each=15),2)),
BD = c(runif(30,1,3),runif(45,2,4),runif(45,3,5))
)
str(df)
'data.frame': 120 obs. of 5 variables:
$$Modality: Factor w/ 3 levels "A","B","C": 1 1 1 1 1 1 1 1 1 1 ...$$ Size : Ord.factor w/ 5 levels "1"<"2"<"3"<"4"<..: 1 1 2 2 3 3 4 4 5 5 ...
$$Repeat : Factor w/ 3 levels "1","2","3": 1 2 1 2 1 2 1 2 1 2 ...$$ Level : Factor w/ 3 levels "L1","L2","L3": 1 1 1 1 1 1 1 1 1 1 ...
$BD : num 2.15 1.45 1.66 2.42 2.64 ... • A clarifying question: why are all variables measured categorically? For instance, BD seems like it could be measured continuously. Wouldn't continuous metrics, as appropriate, simplify your model? Commented Apr 10, 2020 at 19:23 • BD is on a continuous scale. The other variables are categorical (Modality and Repeat) and ordinal (Size). Commented Apr 11, 2020 at 8:51 • Ok, my view is that a much more powerful model is possible by retaining quantitative information as quantitative, rather than collapsing them into categories. Potentially valuable information is lost with discretization. Commented Apr 18, 2020 at 18:08 • yes, but it had to be ensured that size was really quantitative in a physical context. Commented Apr 21, 2020 at 10:52 ## 2 Answers I am currently exploring the extent of this "aggregation problem" in my own work, for lack of a better term. One answer here suggests that you average the measurements in each modality. This is a form of aggregation and is an easily overlooked violation of assumptions, i.e. that the unobservable error terms of the model are IID, so I believe you are correct in being concerned with it. By averaging them, you are washing out the larger variance of the modality that has fewer measurements. You are also losing degrees of freedom/data points. It is better to include all measurements in the model to account for this difference in variance. One concern I am noticing is whether phantom size should be continuous or ordinal. The answer to the question is in intro statistics books. Does phantom size have a zero? Considering they are fat layers, then I would think no fat layers could be a choice, so yes. Is twice a phantom size of 1 equivalent to a phantom size of 2? You'll have to answer that, but as far as radii of fat go, the answer is yes. These two yeses indicate a continuous variable. You'd also have to determine what's more reasonable, whether, 1) the area spanned by the fat around the spine affects bone density, or 2) the fat's distance from the center of the spine affects bone density. 1) implies a squared function of radius, whereas 2) implies a linear function. It depends on your research question. If both are true then you can include the linear and squared phantom sizes. Note if the only reason the phantom sizes are ordinal is because you have five phantoms to choose from, this is not a justification for ordinality given your (assumed) research question. You'll simply need to get a ruler and measure the radius of fat from the center of the spine minus the radius of the vertebrum for each phantom. If this paragraph is entirely unreasonable and phantom size should be ordinal, (I really believe it isn't), then I'm wrong and no need to read further, since I assume continuous phantom size. Though if I'm right, at this point I'd suggest changing "phantom size" to "fat layer radius". Overall I agree with others that you can use a regression model. The response variable is bone density. Three indicator explanatory columns are used for each vertebrum, three more for modality, and one or two more for phantom size depending on if you include a squared term. Both 1) and 2) would be answered by including an interaction term of phantom size*modality (or two if squared term included). If you do not include the squared term, then you have a total of 3+3+1+3*1+3*1 = 13, or 3+3+2+3*2+3*2 = 20 terms if you include the squared term. If you can assume that the measurements from each vertebrum are IID (seems reasonable), then you don't need the 3 indicator variables for vertebra. That would reduce to 3+1+3*1 = 7, or 3+2+3*2 = 11 terms. You'll have to pick the appropriate model considering the IID vertebra question, linear vs squared question, and the number of data points you have. The rule of thumb I've learned is 10 to 1 data points to predictors. Fewer data points could still work depending on the effect size. Note the above does not require any non-parametric methods. Non-parametric methods should only be used if your residuals do not end up following a normal distribution. The other answers seem to give good code for generating the model, just make sure phantom size is continuous. I see no reason to perform partial f tests, though. Significant interaction terms in the full model suffice in answering your two (equivalent) questions. • Thank you for your helpful comment. Do I understand correctly (I am a physician) that the model with interaction terms (lm(formula = BMD ~ Modality * Size + Vertebra) would be appropriate to answer both questions? Commented Apr 16, 2020 at 9:32 • Yes. (Modality x Size) is equivalent to (Size x Modality), which appears to be your 1) and 2) questions. Again, this assumes Size is continuous, am I correct that this is a reasonable assumption? Since Modality is categorical, it would be split into 3 indicator columns, then each would be multiplied by Size to create 3 new columns. The coefficients from each of these columns should approximately match the slopes that you'd get by fitting a line in the 3 plots provided in your follow-up post. Let me know if they don't match. Commented Apr 17, 2020 at 4:20 • Is there a way to calculate effect size of the (significant) factors of the multiple linear regression model? Commented Apr 21, 2020 at 10:52 • I am actually still in doubt if the sample size of this experiment is enough to perform multiple linear regression because I think that a linear model is based on parametric assumptions. Would a bootstrap hypothesis testing be an alternative or is that not possible because we are looking at 3 groups? Thank you Commented May 5, 2020 at 6:59 ### Answer about analysis 1 If you want to perform a Friedman test, you have to use 15 groups of 3 observations (one group for each combination of spine and fat size, and one observation for modality), so that the samples are paired. Repeated measurements must be reduced to one single observation by averaging them. The alternative is to use not a non-parametric test, but a linear model, which seems a confortable solution, as it helps with both your problems at one time. ### Answer about analysis 2 If you average repeated measurements, you can create five paired samples of 9 observations (same averages as before, but this time divided in samples according to the size). Also, it seems to me that you should avoid Kruskal Wallis test, as it assumes independence between the samples, so it is less powerful than tests that take pairing into account. On the other hand, Kruskal Wallis doesn't requires pairing, so you can use it without averaging repeated measurements; I wouldn't recommend going down this way though, blocking for nuisance factors is generally more effective than having a not too larger sample. So, you could use a Friedman test again. However, I don't understand why you would even be interested in a post-hoc test between sizes, how comes that a specific size of fat around the bone would be specially interesting? Wouldn't you be satisfied by learning that fat wraps have an effect overall? In fact, as the size is clearly an ordinal variable (so, basically a numeric one), I would not go for a Friedman test either, but for a linear model. This seems by any mean the best option: it lets you use both blocking factors and repeated measurements, and it allows better insight into data. If you are worried about scientific correctness, just check the diagnostic plots before looking at the tests results: if the effect of the size is not linear, add non-linear effects; if the residuals have variance depending on factors... well, let's hope not, because in that case you'll have to use a different model. ### Code for the linear model Following the discussion on the comments here there is the code you could use for doing analysis n. 2: # developing the data frame to use it for linear modeling BD= c(as.matrix(my.df[, 4:6])) Vertebra= paste('L', gl(3, nrow(my.df)), sep= '') df= cbind(my.df[, 1:3], Vertebra, BD) df$$Size= as.numeric(as.character(df$$Size)) # linear models m_wo_mod= lm(DB~Size+Vertebra, data= df) m_wo_size= lm(DB~Modality+Vertebra, data= df) m_complete= lm(DB~Modality*Size+Vertebra, data= df) # diagnostic plots plot(m_wo_mod) plot(m_wo_size) plot(m_complete) plot(df$$Size, m_wo_mod$$residuals, col= df$Modality)
# test question 1
anova(m_wo_mod, m_complete)
# test question 2
anova(m_wo_size, m_complete)
• In regards to Answer 1: Do I have enough data points to perform a linear regression? I am interested to test if there are significant differences in BD among the three modalities. Would it be appropriate to take the average BD values from the repeats for all three vertebrae and 5 sizes together (15 values per modality) and compare the modalities using Friedman (then they would be paired)? Or is that inappropriate because I average across different vertebrae? Commented Apr 11, 2020 at 9:40
• In regards to Answer 2: Your idea to assess if size has an impact overall is interesting. However, I think this has to be done for each modality separately. This results in the same question as for Answer 1: Is my sample size large enough for a linear regression? Commented Apr 11, 2020 at 9:42
• linear models have a hard lower limit for sample size, but it is incredibly low. in practice, when we have already collected the sample, we want the most powerful method, which is the one effective with the least data points. the linear model is more powerful than any non-parametric method, and this holds even more true in your case, as it doesn't require any loss of information from data. Commented Apr 11, 2020 at 12:56
• you can definitely test effect of size for each modality separately, but you have to divide your sample, which is not that big. I would recommend that only on a linear model, after assessing overall effect of size, when estimating an interaction between size and modality. ask if you need the code. Commented Apr 11, 2020 at 13:14
• Thank you. Did you see the full dataframe structure I have updated at the end of the question? This is the complete structure of the data with fictive data Commented Apr 12, 2020 at 20:40 | 3,535 | 13,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.870107 |
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Boleslav Březovský
@rebolek
There is actually an easier way, just use sort/compare with a comparatur function:
sort/compare block func [this that] [all [this/1 > that/1 this/2 > that/2]]
Of course you would need to improve the comaparator function, this is just an example how to do it.
Boleslav Březovský
@rebolek
>> block: collect [loop 15 [keep/only reduce [random 9 random 9 random 9]]]
== [[3 5 1] [7 8 5] [4 7 3] [5 7 4] [2 3 3] [7 6 4] [6 4 7] [3 6 5] [9 7 1] [3 7 1] [6 4 9] [2 4 3] [6 9 5] [7 2 8] [2 4 9]]
>> sort/compare block func [this that] [any [this/1 < that/1 this/2 < that/2]]
== [[2 3 3] [2 4 9] [2 4 3] [3 5 1] [3 6 5] [3 7 1] [4 7 3] [5 7 4] [6 4 7] [6 4 9] [6 9 5] [7 2 8] [7 6 4] [7 8 5] [9 7 1]]
hiiamboris
@hiiamboris
In the comparator: use sort!
Galen Ivanov
@GalenIvanov
Will block in /compare => Comparator offset, block (TBD) or function. serve this function - to specify the priority of the fields when sorting?
GiuseppeChillemi
@GiuseppeChillemi
I have looked at sort.r source code and it is really cryptic: no comments and one or 2 letters argument names. I was trying to understand of the comparer function works. @rebolek: you have built a function with two arguments but what is one argument and what is the other? How the whole sorting process works?
@hiiamboris Why and how should be used sort in the comparator?
Also, I see this refinement /all => Compare all fields (used with /skip) which seems near to what is needed. In place of All fields it could compare just few.
Galen Ivanov
@GalenIvanov
@GiuseppeChillemi the anonymous function takes two arguments - each two values that sort compares in order to place at their respective places in the sorted list.
hiiamboris
@hiiamboris
@hiiamboris Why and how should be used sort in the comparator?
I'll leave that for you as an exercise ☻
Tip: sort does not error out if things are not "comparable"
GiuseppeChillemi
@GiuseppeChillemi
@GalenIvanov Thanks, so the whole process consists into scanning the first and next element and running the comparator function. I imagine it does it into multiple steps and creating intermediary blocks until it reaches the "SORTED" condition.
@hiiamboris I need to know more about the sorting process as a whole, otherwise, you know myself, I will end-up considering every possible scenario, even the most complicated.
Galen Ivanov
@GalenIvanov
@GiuseppeChillemi Yes, in principle. I haven't looked at the source and don't know the details - the sorting might be in place.
hiiamboris
@hiiamboris
Of course it's in-place.
Galen Ivanov
@GalenIvanov
:+1:
GiuseppeChillemi
@GiuseppeChillemi
What you mean as "in place"? On the source block with no temporary intermediate blocks?
Gregg Irwin
@greggirwin
@rebolek, of course! Do you know if using a comparator honors /stable. I remember issues under R2 at some point, but confess I haven't tried it in Red.
Boleslav Březovský
@rebolek
@greggirwin I’m not sure. I know two things:
1) this comparison is not stable, sometimes it needs three passes to produce the right result, and doing the fourth pass produces the wrong result again (but sometimes it’s just two so it’s not predictable).
2) with /stable it’s even more unstable as I got hard crash that I wasn’t able to reproduce later, unfortunately, otherwise I would fill a bug.
Gregg Irwin
@greggirwin
Thanks. Under R2 I believe it did finally honor results of [-1 0 1] for stable sorting.
hiiamboris
@hiiamboris
Gregg Irwin
@greggirwin
Almost a year to the day! I know @qtxie is busy, but sort is important.
hiiamboris
@hiiamboris
Issues got neglected after the decision to not do stable releases :)
(or may be because I've found way too many..)
hiiamboris
@hiiamboris
@greggirwin By the way, in your FOR design: do you allow user to modify the index and then continue from that modified index, or do you replace the index no matter what user does with it?
Something to consider.
Right now only forall allows to alter it's offset at iteration time.
Gregg Irwin
@greggirwin
It's TBD. Some cases do and others don't, but not by design...yet. This comes from delegating some loops down to loop, repeat, or foreach internally, but if there's no internal func to do it, the final logic goes through Ladislav's old cfor (for how C does it), which gives you total control.
hiiamboris
@hiiamboris
I'm in favor of total control too. The only cases where I modify the index in repeat are when I'm using it with an offset, in place of missing for. (so if we had for, I'd vote for repeat to support reading index back between iterations)
It becomes a bit trickier though when iterating over a range of pairs..
Ideally, let user be dumb and add 1x0 even if it goes over the current line, then skip the extra number of pixels on the next line.
Gregg Irwin
@greggirwin
The design problem is consistency. We want to leverage natives, but having some calls allow it, and others not, is bad for users. If everything goes through cfor, every loop is slower.
I'd like for most loops to be safe, as long as there is one way you can be in control.
hiiamboris
@hiiamboris
Meaning?
Gregg Irwin
@greggirwin
Basic iteration (count or series end-test) can never spinlock, as long as you have something like while that puts you in control of the end-test.
hiiamboris
@hiiamboris
Not sure how this relates to for :)
Gregg Irwin
@greggirwin
for [i 1 to 10 step 2][...] doesn't let you change i and potentially spinlock.
while or for [[i: 1] [i <= 10] [i: i + 2]] [...] gives you all the rope you need to shoot yourself in the foot.
hiiamboris
@hiiamboris
So you'd rather not let to change i because of spinlock danger?
Clearly, for will have to be consistent about it.
Gregg Irwin
@greggirwin
Currently the first example "compiles down" to the second example, except when delegating to natives.
I'd like to help the user, and their users.
hiiamboris
@hiiamboris
Help can take different forms :)
Gregg Irwin
@greggirwin
hiiamboris
@hiiamboris
Haha OK :)
Gregg Irwin
@greggirwin
But we agree that consistency is a must.
hiiamboris
@hiiamboris
I asked because I thought I'm going to allow that in foreach. Because that's a powerful feature of forall to skip 1 or more items, and even zero. I'm using it a lot, and I would hate to throw away foreach and fall back to forall when I need that.
But these 2 are positional iterators, not numeric. So consistency with for is less important.
Gregg Irwin
@greggirwin
It is powerful, like goto is powerful. Constraints are a pain sometimes but they also provide a lot of value. In this case, the ability to reason about iteration logic without having to read every line of the body.
hiiamboris
@hiiamboris
Good argument.
Gregg Irwin
@greggirwin
If only there was one right answer. :^\
hiiamboris
@hiiamboris
Indeed.. | 2,091 | 7,089 | {"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} | 2.78125 | 3 | CC-MAIN-2022-05 | latest | en | 0.815024 |
https://math.stackexchange.com/questions/3330085/computational-verification-of-collatz-problem | 1,716,665,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00718.warc.gz | 321,138,858 | 43,398 | # Computational verification of Collatz problem
Every positive integer $$n$$ can be represented as a product \begin{align} n &= a \cdot 2^k \text{,} \end{align} where the $$a$$ is odd integer and $$k$$ is an exponent of two. Let $$\varepsilon$$ be the map $$n \mapsto k$$, and let $$\sigma$$ be the map $$n \mapsto a$$.
Now define two auxiliary functions \begin{align} a(n-1) &= \sigma(n) \cdot 3^{ \varepsilon(n) } \text{,} \\ b(n+1) &= \sigma(n) \cdot 1^{ \varepsilon(n) } \text{,} \end{align} and function composition \begin{align} T(n) &= b(a(n)) \text{.} \end{align}
Collatz problem concerns the question of whether the function iterates reach 1, for all $$n>0$$. This is nothing new. I just formulated the Collatz problem in a very complicated manner. Instead of tracking $$n$$ directly, I now track the pair $$(\sigma(n), \varepsilon(n))$$.
The point of this formulation is that I can verify the convergence of the problem for all $$\sigma(\cdot) < \Sigma$$ and $$\varepsilon(\cdot) < E$$. This is much easier than verifying the problem directly on $$n$$ because pairs of values $$(\sigma(n), \varepsilon(n))$$ take much smaller magnitudes than the $$n$$ itself. For example, I have verified the convergence for all $$(\sigma, \varepsilon)$$ below $$(2^{44}, 14)$$. This was fairly fast on my desktop computer. For convenience, the pair $$(2^{44}, 14)$$ corresponds to approximatelly $$2^{66.2}$$.
Can anyone confirm the mathematical correctness of my verification procedure? Would anyone be willing to implement this effectively (in any programming language) and verify it for higher limits?
As requested in the comment, there is an illustration for $$n = 27$$:
$$\begin{matrix} n & (\sigma, \varepsilon) \text{ in a(n-1)} & T(n) \\ \hline 27 & ( 7, 2) & b( 7 \cdot 3^{ 2}) \\ 31 & ( 1, 5) & b( 1 \cdot 3^{ 5}) \\ 121 & ( 61, 1) & b( 61 \cdot 3^{ 1}) \\ 91 & ( 23, 2) & b( 23 \cdot 3^{ 2}) \\ 103 & ( 13, 3) & b( 13 \cdot 3^{ 3}) \\ 175 & ( 11, 4) & b( 11 \cdot 3^{ 4}) \\ 445 & ( 223, 1) & b( 223 \cdot 3^{ 1}) \\ 167 & ( 21, 3) & b( 21 \cdot 3^{ 3}) \\ 283 & ( 71, 2) & b( 71 \cdot 3^{ 2}) \\ 319 & ( 5, 6) & b( 5 \cdot 3^{ 6}) \\ 911 & ( 57, 4) & b( 57 \cdot 3^{ 4}) \\ 577 & ( 289, 1) & b( 289 \cdot 3^{ 1}) \\ 433 & ( 217, 1) & b( 217 \cdot 3^{ 1}) \\ 325 & ( 163, 1) & b( 163 \cdot 3^{ 1}) \\ 61 & ( 31, 1) & b( 31 \cdot 3^{ 1}) \\ 23 & ( 3, 3) & b( 3 \cdot 3^{ 3}) \\ 5 & ( 3, 1) & b( 3 \cdot 3^{ 1}) \\ 1 & ( 1, 1) & b( 1 \cdot 3^{ 1}) \\ \end{matrix}$$
As requested, there is a pseudocode:
function a(n):
return σ(n+1) × 3^ε(n+1);
function b(n):
return σ(n−1) × 1^ε(n−1);
function test_convergence(n):
while n != 1 do:
n := b( a(n) )
Or alternativelly:
function f(s, t):
n := s × 3^t − 1
return ( σ(n), ε(n) )
function g(s, t):
n := s × 1^t + 1
return ( σ(n), ε(n) )
function test_convergence(s, t):
while (s, t) != (1, 1) do:
(s, t) = f( g(s, t) )
Just for the record, I have verified the convergence of the Collatz problem below the following $$(\sigma, \varepsilon)$$ bounds. I stopped working on it now. So I just share the achieved results for the case someone wants to continue. Verification for higher $$\sigma$$ values is very computationally demanding.
$$\begin{matrix} (\sigma, \varepsilon) \text{ upper bound} \\ \hline ( 2^{49}, 2 ) \\ ( 2^{48}, 6 ) \\ ( 2^{47}, 9 ) \\ ( 2^{46}, 10 ) \\ ( 2^{45}, 12 ) \\ ( 2^{44}, 16 ) \\ ( 2^{43}, 15 ) \\ ( 2^{42}, 22 ) \\ ( 2^{41}, 24 ) \\ ( 2^{40}, 45 ) \\ ( 2^{39}, 46 ) \\ ( 2^{38}, 47 ) \\ ( 2^{37}, 53 ) \\ ( 2^{36}, 78 ) \\ ( 2^{35}, 84 ) \\ ( 2^{34}, 108 ) \\ ( 2^{33}, 132 ) \\ ( 2^{32}, 256 ) \\ \end{matrix}$$
• I don't understand what you have done!! can you illustrate it for example for $n=5$? Aug 21, 2019 at 16:27
• You are exploiting the fact that any odd number can be written as $n=\sigma \cdot 2^\varepsilon−1$ and that $\sigma \cdot 2^\varepsilon−1$ always go straight to $\sigma \cdot 3^\varepsilon−1$ in exactly $\varepsilon$ steps, number which is even and from which you strip any factor of 2 to find the next odd number. Isn't this an optimisation already used in current algorithm? Note: you can optimise further math.stackexchange.com/questions/2428060/… Aug 21, 2019 at 17:55
• Why does $(2^{44}, 14)$ corresponds to $\sim 2^{66.2}$ and not to $2^{44} \cdot 2^{14} = 2^{58}$? Aug 21, 2019 at 18:47
• does this mean \begin{align} a(n) &= \sigma(n+1) \cdot 3^{ \varepsilon(n+1) } \text{,} \\ b(n) &= \sigma(n-1) \cdot 1^{ \varepsilon(n-1) } \text{?} \end{align} If so, why don't you write it down it this way? The value of the expression $1^{ \varepsilon(n-1) }$ is always $1$, so why is it part of the formula? Aug 22, 2019 at 12:31
• Aug 22, 2019 at 13:18
Edit 2019-08-30:
Edit 20190831
added Python code and description and reference to C implementation on codereview
The Collatz function is defined as $$\text{collatz}(n):=\begin{cases} 3n+1,& n\equiv 1 \pmod 2 \\ \frac n 2, & n \equiv 0 \pmod 2 \end{cases}$$
A trajectory of n with respect to a function $$f$$ or an $$f$$-trajectory of $$n$$ is the sequence $$n, \;f(n), \;f(f(n)), \;f^3(n),\;\ldots$$
A subsequence of such a trajectory I will call a subtrajectory.
We are interested if the collatz-trajectory of a positive integer $$n$$ is either unbounded or if it will cycle. At the moment the trajectories of all numbers investigated so far will cycle. The cycle for all these number is the cycle $$4,2,1,4,..$$
If a trajectory cycles then a subtracjectory must contain identical values and vice versa.
We define now the following function that is related to the Collatz function: $$\text{c}(n):=\begin{cases} \frac {3n+1} 2, & n\equiv 1 \pmod 2 \\ \frac n 2, & n \equiv 0 \pmod 2 \end{cases}\tag{1.1}$$ A c-trajectory of $$n$$ will be a Collatz-subtrajectory of $$n$$.
Instead of the $$c$$-trajectory of $$n$$ $$n, c(n), c^2(n),\ldots$$ we can construct a new sequence
$$n+1, c(n)+1, c^2(n)+1, \ldots$$
This is a trajectory with respect to the function $$d$$
$$d(n):=c(n-1)+1\tag{2.1}$$
$$\begin{array} 27&41&62&31&47&71&\ldots\\ 28&42&63&32&48&72\ldots \end{array}\tag{2.2}$$
From $$(2.1)$$ follows
$$c(n)=d(n+1)-1$$ and by induction one can prove $$d^k(n)=c^k(n-1)+1\tag{2.3}$$ $$c^k(n)=d^k(n)-1$$
From $$(1.1)$$ and $$(2.1)$$ we get $$\text{d}(n):=\begin{cases} \frac{n+1} 2,& n\equiv 1 \pmod 2 \\ \frac {3n} 2, & n \equiv 0 \pmod 2 \end{cases}\tag{2.4}$$
From $$c$$ and $$d$$ we can generate new functions
$$c^+(n)=\begin{cases} \frac{3n+1}2 , & n\equiv 1 \pmod 2 \\ \frac n {2^k},& n=2^ka, k>0, a\equiv 1\pmod 2 \end{cases}$$
$$d^+(n)=\begin{cases} \frac{n+1}2 , & n\equiv 1 \pmod 2 \\ \left(\frac {3} {2}\right)^kn,& n=2^ka, k>0, a\equiv 1\pmod 2 \end{cases}$$
We can rewrite this definitions as
$$c^+(n)=\begin{cases} c(n) , & n\equiv 1 \pmod 2 \\ c^k(n),& n=2^ka, k>0, a\equiv 1\pmod 2 \end{cases}$$
$$d^+(n)=\begin{cases} d(n) , & n\equiv 1 \pmod 2 \\ d^k(n),& n=2^ka, k>0, a\equiv 1\pmod 2 \end{cases}$$
and we see that $$c^+$$-trajectories are $$c$$-subtrajectories and $$d^+$$-trajectories are $$d$$-subtrajectories.
Finally we define
$$T(n)=\begin{cases} c^+(n) , & n\equiv 1 \pmod 2 \\ c^+(d^+(n+1)-1),& n\equiv 1\pmod 2 \end{cases}$$
An again we have that a trajectory of $$T$$ is a subtrajectory of $$c$$. If $$n$$ is odd this is trivial, if $$n$$ is even then $$T(n)=c^+(d^+(n+1)-1)=c^+(d^{k_1}(n+1)-1)=c^+(c^{k_1}(n))=c^{k_2}(c^{k_1}(n))=c^{k_2+k_1}(n)$$
The function $$T$$ is the function that you use for your calculations.
The following algorithm assumes that $$k$$ is a positive integer and $$u$$ is an odd positive integer. There are two different variables $$n_c$$ and $$n_d$$ instead of one variable to show which values are from the trajectory of $$c$$ and therefore of the Collatz function and which values are from the trajectory of $$d$$ and therefore from the sequence that we get by adding $$1$$ to the trajectory values of the Collatz function. The termination condition depends on the purpose of the algorithm. Step 2 is used to simplify the comments and should not be implemented. $$\begin{array}[lrc]\\ Step&Precondition&Action&Comment &&Comment\\ 1&&n_c\gets n_0&/* n_0 \; \text{is the start value}&*/\\ 2&&&/*x\gets n_c&*/&\\ 3&/*n_c \text{ is odd}*/&n_d\gets n_c+1&/*x+1&*/\\ 4&/*n_d=2^ku*/&n_d\gets 3^ku&/*d^+(x+1)&*/&/*a(x)*/\\ 5&/*n_d \text{ is odd}*/&n_c\gets n_d-1&/*d^+(x+1)-1&*/\\ 6&/*n_c=2^ku*/&n_c\gets u&/*c^+(d^+(x+1)-1)&*/&/*b(a(x))*/\\ 7&&\mathbf{if }\;n_c =1 \; \mathbf{then}&/* \text{or} \; n_c
This algorithm can be easily transformed to a pseudocode/Python3 program.
• % is the modulo operator
• // is integer division
• ** is the power operator
• x += y means x=x+1, similar holds for other operators
Here is the program:
n=n0
while n>1:
n+=1
k=0
while n%2==0:
k+=1
n//=2
n*=3**k
n-=1
while n%2==0:
n//=2
It can be rewritten by using some functions and replacing the variable k by e.
• ctz(n) returns e, where $$n=2^eu$$, $$u$$ is odd
• rsh(n,e) returns $$\frac n{2^e}$$
• lut(e) returns $$3^e$$
the new program:
n=n0
while n>1:
n+=1
e=ctz(n)
n=rsh(n,e)
n*=lut(e)
n-=1
n=rsh(n,ctz(n))
• The function ctz can be implemented by counting how often n can be repeatedly divided by two until the result is odd or by counting the number of trailing $$0$$ of the binary representation of n.
• The function rsh can be implemented by multiplying n n-times by $$2$$ or by shifting the binary representation $$n$$-times to the right.
• The function lut(e) returns $$3^k$$ and can be implemented by a lookup table if the number e will not become too large.
This program now looks like to the C-implementation of the algorithm posted by the OP at codereview.stackexchange.
You can get the $$c^+$$-trajectory from the $$c$$-trajectory in the following way: If you current value on the trajectory is odd, than proceed on the $$c$$-trajectory to the next value. If it is even then proceed to the next odd value (the second branch of the definition of $$c^+$$) The same holds for the construction of $$d^+$$ from $$d$$. This method is shown on the picture. The circled numbers are the values of the $$c^+$$ (first line) and $$d^+$$ (second line) trajectory of 27. The last two lines show how to construct the trajectory of $$T$$ from a trajectory of $$c$$ and $$d$$. If you start from an odd value $$n$$ then got to the opposite even value n+1 of the $$d$$ trajectory. From this go to the next odd value of the $$d$$-trajectory. Then go to the opposite even value of the $$c$$-trajectory by subtracting $$1$$ and from this go to the next odd value of the $$c$$-trajectory.
At the moment I cannot see any advantage in using the function $$T$$ instead of $$c^+$$ or $$d^+$$.
I evaluated the number of function calls one needs using $$c^+$$, $$d^+$$ and $$T$$ until the the trajectory reaches $$1$$. For all odd numbers $$n \in \{3,...,N\}$$ I summed these path lengths up and got the following numbers
N c+ all c+ 2nd d+ all d+ 2nd T all
1000 16506 5469 16267 5461 5452
10000 229650 76314 226297 76302 76275
100000 2848611 949409 2829632 949374 949358
So from this we see that the number of function calls need to reach the value $$1$$ in the trajectory is for the functions $$d$$ and $$c$$ about the same and three times higher than for the function $$T$$. But note that a call of the function $$T$$ contains a call to the second branch of $$c^+$$ and a call to the second branch of $$d^+$$. So all in all one I cannot see that there is any large improvement in using $$T$$
To check if the trajectory of all numbers $$n$$ less than $$N$$ cycles one does not calculate the trajectory values until they reach $$1$$ but only until it reaches a value less than the start value $$n$$. I also calculated the number of iterations for different $$N$$
N c+all c+2nd d+all d+2nd T all
1000 2696 895 2166 637 892
10000 25909 8662 21002 6145 8660
100000 260246 86777 210708 61692 86760
1000000 2612479 871075 2114522 620923 871073
Conclusion
The OP asked if his procedure is correct and I showed here that he uses the function $$T$$ and that a trajectory of $$T$$ is a subtrajectory of the Collatz function. So his procedure is correct. Additionally I showed that he cannot expect a substantial performance gain by using $$T$$ instead of $$c^+$$ because the number of iteration is the same (maybe they differ by a constant factor).
This is the Python 3 program that generates the data of the table
def c(n):
# this is the function c+
if n%2==1:
return (3*n+1)//2
else:
while n%2==0:
n//=2
return n
def d(n):
# this is the function d+
if n%2==1:
return (n+1)//2
else:
m=1
while n%2==0:
n//=2
m*=3
return m*n
def T(n):
# this is the function T
if n%2==1:
return c(d(n+1)-1)
else:
return(c(n))
def statistics(n,f):
if f == d:
i=n+1
else:
i=n
# stop_value=i # stop if trajectory <=n
stop_value=2 # stop if trajectory <=2
cnt=0
even_cnt=0
while i>stop_value:
i=f(i)
cnt+=1
if i%2==0:
even_cnt+=1
return(cnt,even_cnt)
for N in [1000,10000,100000]:
print(N)
for f in (c,d,T):
all_calls=0
even_calls=0
for N in range(3,N,2):
tmp=statistics(N,f)
all_calls+=tmp[0]
even_calls+=tmp[1]
print(f,all_calls,even_calls)
• I think you missed the main point. The improvement is that you need far less iterations of this new algorithm compared to the original Collatz function. The $(3/2)^k$ and $(1/2)^k$ in your equations are just a single CTZ operation. So the convergence test has just changed to iterations of CTZ and multiplication with an entry in a look-up table. Aug 26, 2019 at 9:40
• I tried to prove that conjecture. I did make some progress using Markov Chains, but I am nowhere close to a solution. In any case, I posted my findings on my website at analyticbridge.datasciencecentral.com/forum/topics/…. Could be an interesting read. Aug 26, 2019 at 15:23
• @DaBler I added a "conclusion". The main point is that you asked if your procedure is correct but the switching between c+ an d+ is not an improvement against c+. If using ctz and table lookup will bring a performance improvement is not in scope of this post. But you cannot expect more than improvement by a constant factor. The simplest way to see if your algorithm is an improvement is to implement the standard solution and compare the running times. you will see there is no relevant improvement. Aug 27, 2019 at 6:21
• @DaBler I did some further testing and it seems that length of a trajectory until it cycles of your function is only about 6% of the length of the trajectory of the collatz function. So maybe your function is 17 times faster than the original collatz function. I will post more details later Aug 28, 2019 at 6:38
• @miracle173 Sounds great! Aug 28, 2019 at 10:50 | 5,051 | 14,712 | {"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": 126, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.654807 |
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User: What is the cube of 8? A. 512 B. 24 C. 64 D. 2
Weegy: C. 2
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The cube of 8 is 512.
8 * 8 * 8 = 64 * 8 = 512
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Given the formula E = IR what is the formula for R? A. R = EI B. R = IE C. R = I ÷ E D. R = E ÷ I
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Given the formula E = IR the formula for R is:
R = E ÷ I
Confirmed by sujaysen [4/1/2014 1:18:52 PM]
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Post Partisan
## Value from last day (calendar date)
Hello all
I have this table and a date filter from calendar table.
The date filter is from 5 oct till 11 oct . I want that my measure shows the unit stocks for that last day 11/10 which would be 106 for each SKU column.
Thank you all very much
1 ACCEPTED SOLUTION
Solution Specialist
Give this a try:
``````Measure =
VAR MaxDate =
CALCULATE ( MAX ( 'Date'[Date] ), ALLSELECTED ( 'Date'[Date] ) )
VAR Result =
CALCULATE ( MAX ( Sales[Qty Day] ), 'Date'[Date] = MaxDate )
RETURN
Result``````
4 REPLIES 4
Super User
@pedroccamaraDBI ,Try measure liek
calculate(lastnonblankvalue('Date'[Date], sum(Table[Unit stock]) )
Post Partisan
Hello @amitchandak
Let me explain you what is my data:
For each day, i have the product code (SKU) the movement id (ID Mov) and the Qty mov, which is how many products were moved and the Qty day which is the balance. I need to know, according with date filter (from dates) the Qty Day of the MAX date on the filter. Which means in this case, it should be 106. Notice that if i may choose 2 or 3 product codes (SKU).
My measure should show 106 in this case because it's the value from the last date chosen.
I'm not sure and even don't know how to, where to do it, but i believe we should have some kind of this: ALL (Products) in that measure, right?
Thanks a lot for your help
Solution Specialist
Give this a try:
``````Measure =
VAR MaxDate =
CALCULATE ( MAX ( 'Date'[Date] ), ALLSELECTED ( 'Date'[Date] ) )
VAR Result =
CALCULATE ( MAX ( Sales[Qty Day] ), 'Date'[Date] = MaxDate )
RETURN
Result``````
Post Partisan
Hello @SteveHailey
Thank you so much for your help. You're the 7th person that i asked for help.
Best regards
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Wednesday 10-10-12. Today you need: Whiteboard, Marker, Eraser Calculator 1 page handout. Warm-up Need a White Board. 1. Graph the following equation:. Warm-up Need a White Board. 1. Graph the following equation:. Linear Regression. Section 2-6 Pages 95-100. Objectives.
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Wednesday 10-10-12
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### Wednesday 10-10-12
• Today you need:
• Whiteboard, Marker, Eraser
• Calculator
• 1 page handout
### Warm-upNeed a White Board
1. Graph the following equation:
### Warm-upNeed a White Board
1. Graph the following equation:
## Linear Regression
Section 2-6
Pages 95-100
### Objectives
• I can use Linear Regression with a calculator to find linear prediction Equations
• I can find the correlation co-efficient “r” for the data
### Correlation Co-efficient
• The correlation co-efficient “r” tells how linear the data is.
• Values of 1 or –1 indicate perfect linear lines, either positive or negative
• Values closer to zero mean the data has no linear relationship
• Small whiteboard number line with r=1 and r=-1
1.0.85
Sample “r Values
-.57.17
### Plotting Data
• When the data you plot forms a near linear relationship, then we can use a linear equation to approximate the graph.
• We use what’s called a Best-Fit Line. This line is drawn to be as close to the data points as possible, but may not touch them all.
y-axis
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### Using the Calculator (Linear Regression)
• The calculator is a great resource to give us a prediction equation.
• It is more accurate than doing the equation Manually
• We will enter the data into the STAT mode of the calculator
Turn Diagnostics On. 2nd catalog,
arrow to Diagnostic on, enter, enter
Linear Regressions on the calculator:
(you should clear the calculator before beginning)
2nd, +, 7, 1, 2
#1.
### Linear Regression
• Finding the equation of your “Best Fit Line”
• STAT, then EDIT
• Enter X-Values in L1, Y-Values in L2
• STAT, then CALC
• Choose (4) LIN REG
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The table below shows the years of experience for eight technicians at Lewis Techomatic and the hourly rate of pay each technician earns.
### Prediction Equations
• y = 1.234x + 5.574
• Remember:
• x = Experience in Years
• y = Pay rate in dollars
• We can use this to predict other values
### When Dealing with Years
• Must modify years starting at “0”
• If you don’t you get a really negative y-intercept value that won’t match the graph
• Example on next slide
If the Independent variable is Years and these are your values
1901
1903
1905
1910
1913
1920
Then these are the values we will actually enter for L1
0
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4
9
12
19
### Homework
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# Agency_Prob - Staples’ revenue depends upon Barney’s...
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1 UBC COMM 295 Solution to Moral Hazard/Agency Problem 1. Your firm needs to hire an accountant. There are three types of accountants (1000 of each type): good, medium and bad. The value of the hourly contribution of these accountants towards your firm’s profits equals \$500, \$300 and \$100 respectively. The reservation hourly wage (i.e., opportunity cost) of these accountants equals \$310, \$190 and \$100, respectively. You are unable to distinguish among the three types of accountants. Determine which of the three types of accountants will offer their services to your firm and what you will offer to pay per hour in equilibrium. (Assume you pay the expected value of the accountant’s hourly contribution toward your firm’s profits). What type of problem is this (i.e., moral hazard, adverse selection, signaling, principal-agent, etc.)? 2. Staples recently hired Barney as a store manager. Barney can put in either low effort (a = 0) or high effort (a = 1), but Staples cannot observe Barney’s specific effort choice.
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Unformatted text preview: Staples’ revenue depends upon Barney’s effort as follows: Bad Economy (Prob. =0.5) Strong Economy Low effort (a=0) 150,000 250,000 High effort (a =1) 250,000 500,000 Barney’s utility function is U = w – C(a), where w denotes his wage income, and C(a) = 10,000a is his cost of effort. a) Briefly explain why this employment relationship constitutes an example for the principal-agent problem. b) In the absence of monitoring, can a fixed wage induce Barney to implement high effort (i.e., a = 1)? Briefly explain. c) Suppose that Staples considers offering Barney two different incentive contracts: A. Profit Sharing Contract: Barney receives 5% of the revenue and no fixed wage. B. Bonus Contract: Fixed wage of \$10,000, and a bonus of \$30,000 if revenue equals \$500,000. What level of effort would Barney implement under each contract? d) Given Barney’s respective effort choice you identified in part c), which contract is preferred by Staples?...
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Below is information and knowledge on the topic is a cup or a quart bigger gather and compiled by the monanngon.net team. Along with other related topics like: Cups in a quart, Which is bigger pint or quart, Cups in a pint, Which is larger Quart or ounce, How many pints in a quart, How many ounces in a quart, Pint or quart bigger Chinese food.
ts, Quarts, Gallons Background Information for Teachers, Parents and Caregivers | BrainPOP Educators
This page provides information to support educators and families in teaching K-3 students about cups, pints, quarts and gallons. It is designed to complement the Cups, Pints, Quarts, Gallons topic page on BrainPOP Jr.
Review with your children that capacity describes how much a container can hold. Some children may be familiar with volume, or the amount of space something takes up, which is usually measured in cubic units. Your children should also be familiar with standard units of capacity, including cups, pints, quarts, and gallons. We recommend doing plenty of hands-on activities together, such as cooking, baking, or just measuring a variety of classroom materials to help your children understand how the units are related.
Show your children a glass and a pitcher. Which has the greater capacity? Which can hold more? Guide them to understand that the bigger container holds more and therefore has a greater capacity. Then show two different shaped glasses and ask the question again. You may want to pour water, uncooked rice, beans, small cubes, or other classroom materials from one glass into the other to demonstrate how one has a greater, smaller, or equal capacity to the other. Remind your children that just because two glasses are different sizes, it does not necessarily mean they have different capacities. Tall and skinny glasses may hold the same amount of water as short and wide glasses. Have your children experiment with different containers and compare shapes and capacities using a variety of pourable materials. Ask them to estimate and predict which container has the greater capacity before they begin their experiments.
Show a measuring cup and explain that a cup is a unit of measurement. Show a mug or a plastic cup and remind your children that while they are both cups, they are not the standard size used in measurement. Ask your students to discuss why they think we use a universal measurement called a “cup”. Some cups may hold more than a cup! Fill a measuring cup and model how to write the measurement 1 c. Remind your children that we use the abbreviation “c” to stand for cups. Challenge your students to think of items that come in cup-sized containers such as single servings of yogurt or small school milk cartons.
Show a pint measure and explain that a pint is a unit of measurement that is larger than a cup. Ask a student to pour 2 cups into the pint measure to demonstrate that 2 cups are equal to 1 pint. Explain that we use the abbreviation “pt” to stand for pints. Brainstorm different items that come in pint sizes, such as ice cream, milk, and blueberries.
Show a quart measure and explain that a quart is a unit of measurement that is larger than both a pint and a cup. Have students pour 2 pints into the quart measure to demonstrate that 2 pints are equal to 1 quart. Help your students recognize that since there are 2 cups in a pint, there are 4 cups in a quart. They can pour 4 full measuring cups into a quart measure to demonstrate. Remind children that we use the abbreviation “qt” to stand for quarts. Brainstorm different items that come in quart sizes, such as juice, milk, strawberries (large package), and paint.
A gallon is a unit of measurement that is larger than a quart, pint, and cup. You may want to present to your children with an empty gallon carton of milk or a gallon soup pot. With some assistance they can pour 4 quarts into the gallon container to understand that 4 quarts are equal to 1 gallon. Since there are 2 pints in a quart, there are 8 pints in a gallon. Since there are 2 cups to a pint, there are 16 cups in a gallon. You can demonstrate how the units are related by measuring different materials and pouring them into the carton or pot. We use abbreviation “gal” to stand for gallons. Brainstorm different items that come in gallons, such as juice, milk, and gasoline.
Working with cups, pints, quarts, and gallons can be confusing for some children and we suggest using plenty of hand-on activities to help them understand how the units are related. It is helpful to create a class chart of equivalent amounts, including pictures of the different measurements, to help students visualize and retain the relationships between units. Graphic organizers, mnemonics, and silly songs may also help drive the concepts home.
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## Frequently Asked Questions About is a cup or a quart bigger
If you have questions that need to be answered about the topic is a cup or a quart bigger, then this section may help you solve it.
### A cup: Is it larger than a quart?
Explain that a quart is a measurement that is “larger than both a pint and a cup” by displaying a quart measure.
### Is one quart or eight cups larger?
Answer and explanation: If you have 8 US cups and need to know how many US fluid quarts that is, you would divide 8 by 4, which is 2, and this would tell you that b>8 cups equals 2 quarts/b>.
### Is a quart equivalent to a cup?
One quart is equal to 4 cups; a cup is 8 fluid ounces; a quart is equal to 32 fluid ounces; and a half-quart is equal to two cups or 16 fluid ounces.
### A quart is what volume?
Note that a dry quart is equal to 4.6546 cups, which is important when doing conversions for any dry ingredient. 1 US liquid quart is equal to 14 gallon, 2 pints, 4 cups, and 32 ounces.
### The size of a cup
A US cup is approximately 237 mL; rougher equivalents are 240 mL and 250 mL, where the latter fits well with a US pint of 500 mL and a pound of 500 g. A cup is a unit of volume measurement that is equal to volume equal to 16 tablespoons, 12 pint, 14 quart, or 8 fluid ounces.
### Four quartz equals how many cups?
We’ll teach you: The mathematical formula for converting cups to quarts (spoiler alert: 16 cups are in 4 quarts).
### How much is one cup in units?
If a recipe lists the ingredients in metric measurements, use this conversion: One cup of water is equivalent to 236 grams. If a recipe lists the ingredients in imperial measurements, use this conversion: One cup of water is equivalent to 2 cups of liquid with approximately equal density.
### The smallest cup is which one?
The smallest bra size that is typically available is a 28AA, which has a 28 inch band and a bust that is less than an inch larger than the ribcage.
### Is a quart the same as two cups?
A quart contains 4 cups.
### Two cups weigh how much?
If we recall, 8 ounces equals 1 cup and 2 cups (or 16 ounces) equals 1 pint.
### Is 3 cups the same as 1 quart?
A quart contains 4 cups. | 3,446 | 14,009 | {"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} | 3.8125 | 4 | CC-MAIN-2023-06 | latest | en | 0.942972 |
http://www.kandosuli.hu/docs/e4b1qyo.php?e77238=what-was-the-initial-volume-of-your-copper-sample-lab-report-5 | 1,709,494,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00017.warc.gz | 60,418,217 | 10,235 | what was the initial volume of your copper sample lab report 5
This could be said due to the overwhelming difference in mass and mole of the final and initial moles and mass. The knob that was controlling the heat was constantly turned up and down due to the violent bubbling. Show one sample calculation. There were many sources of error present throughout this experiment. This surplus of water may have resulted in the greater mass and moles of the final solution. 425 0 obj <> endobj
The initial and final moles of Copper should have also been the same.
This information also differs. Mass Of Recovered Copper , 5. This result defies the law of conservation of mass.
The most probable source of error must have been that there were still some solid zinc that did not react with the copper sulfate. The group may have not waited long enough to let all of the product reach to the bottom of the beaker. Make and record your observations on page 5 . Density of post-1982 penny = 7.19 g/mL . After all of the chemical reactions have finished, the final mass of copper was 6.85 grams. Mass Of Evaporating Dish 4. These masses differ quite tremendously.
The goal of this lab was to see if the mass of copper would be same after going through a series of chemical reactions. The time it took to heat did not reach 5 minutes. Following the heating, there were some of the solution that was on the side of the beaker. Purpose: The purpose of this lab is to accurately determine the densities of pre-1982 and post-1982 pennies. The results deviated from the weight of copper by 4.841 grams. Cu + HNO 3 - The solution became blue. Privacy The pH tester was suppose indicate a color that matches the number 13. In this lab, each group had the initial mass of the copper close to 2.00 grams. Show one sample calculation trial 1 trial 2 trial 3 mass copper chloride (c) Calculate the mass of chlorine. 445 0 obj <>/Filter/FlateDecode/ID[<4964B7B108E29346BB75D38CE0516694>]/Index[425 38]/Info 424 0 R/Length 94/Prev 54442/Root 426 0 R/Size 463/Type/XRef/W[1 2 1]>>stream It is believed that the process was known (metallurgy) as early as 4500 BC. 36% 76% 3.7%.76% Not enough information is given to make this calculation. %%EOF The purpose of this lab is demonstrate the use of the conservation of mass through a series of chemical reactions. If you recover 1.8 grams of copper from a reaction where you started with 5.00 grams of copper(II) chloride and an excess of aluminum, what is your percent yield of copper? The initial and final moles of Copper should have also been the same. Show one sample calculation trial 1 trial 2 rial 3 moles of copper moles of chloride mol( chlorine) mol (copper) (e) Report the formula CuCly/, using the average ratio CHM 117 Spring 2018. There was a significant amount of moles in surplus in comparison to the moles of the final product and the initial product. Nitrogen gas was formed. Percent Yield (show Calculations) % Yield-reca Era Mass Cre- \$83 6. The cooling of the solution is also needed to speed up the process of the formation of a solid product. When decanting the solution, it was hard to extract all of the water from the solution. Another error at this time could be that it was not 2 grams. One source of error would be the inaccurate calibration. This may have affected the results due to the lack of acidity. This is a lab report for my General Chemistry class. The initial moles of copper was 0.0315 grams. Question: Post-Lab Report (a) Report The Collected Data With The Correct Uncertainty. One of the error occurred when testing the pH.
Question: REPORT SHEET EXPERIMENT Chemical Reactions Of Copper 6 And Percent Yield - 1. Whether it be miscalculation or careless mistakes, the results were altered in some way that differs from the law of conservation of mass. Many errors would follow.
Show One Sample Calculation Trial 1 Trial 2 Trial 3 Mass Copper Chloride (c) Calculate The Mass Of Chlorine. Density of pre-1982 penny = 8.87 g/mL . The basis behind this experiment is the law of conservation of mass. After putting the copper through many chemical reactions and phase changes, the final mass of the copper was 6.845 grams of copper. The initial mass of copper used in the experiment was 2.004 grams, and the initial moles of copper was 0.0315. Trial Initial Volume Final Volume (mL) Concentration Mass Copper (&/L) 16. Laboratory Report 1 Title: Accurate Measurement of Mass and Volume Part A: The Formula of Hydrated Copper (II) Sulfate Aim: The objective of this experiment is to find out the accurate mass of a solid and to calculate the moles of an unknown. Composition: 95% copper, 5% zinc. & h�bbd``b`�\$V ��@�eH\$�� After the lab was finished, the final mass of the copper was 6.845 g, and the final … h�b```���@r ?�30EY8@܍��+�x�x�� ��Y �I7.���{��P�·)�;:,:88X;\$:;<0t5 Create your own unique website with customizable templates. Freezing Point Depression with Antifreeze Lab. A member of the group would use droplets of water to spray down the solution from the sides of the beaker. Another error would be caused during the decanting process. In our case, my group measured out 2.004 grams of copper. After this process, there was a wait time of 5 minutes. Even this small deviation from the ideal amount could cause a difference in calculations. Initial Mass of Copper 2.004 grams. your copper solution. Percent Yield = (moles of copper recovered.
The initial mass of the copper was 2.004 grams. Lab #6 Chemical Transformations of Copper Introduction: Copper was one of the first metals to be isolated, due to the ease of separating it from its ores. View desktop site, Post-Lab Report (a) Report the collected data with the correct uncertainty.
There mere weight of this substance could be enough to alter the final weight of the product. Due to the Law of Conservation of Mass, one would predict that after any number of chemical reactions and phase changes, the final mass and number of moles of copper would remain the same as the initial amounts. This should have been impossible. Terms The percent yield was calculated at 341.9%. These masses differ quite tremendously. All of the copper in precipitate form has been removed. Most of the group’s measurement deviated from the measurement of 2 grams by less than 0.005 grams. There should have only been little deviation from 2.00 grams. The law of conservation of mass says that matter can neither be created or destroyed. A source of error that could be present at this time could have been that the scale was not properly calibrated, thus throwing off the weighted measurement. Mass Of Copper And Evaporating Dish - 3. The beaker would take a while in order to start heating up.
After all of the chemical reactions have finished, the final mass of copper was 6.85 grams. This could affect the results dramatically. The problem was that there was not a 13 on the scale. Initial Mass Of Copper -2. Post-Lab Report (a) Report The Collected Data With The Correct Uncertainty. 1) Th product of the reaction between copper and nitric acid in step 2 is placed on ice because it is needed to control the temperature of the reaction.
462 0 obj <>stream
trial 1 trial 2 trial 3 mass chlorine (d) Calculate the moles of copper () and chlorine (u) and the ratio (w/x).
endstream endobj startxref
0 During the stage when each groups were to heat their beaker with their mixture in it, the heating was considerably inconsistent. The beaker was to be kept on the heater for 5 minutes, starting when it begins to boil. This could have drastically increased the final weight of the copper. This experiment would involve the use of copper (Cu) in a series of reactions that when finished, should equal the same amount of mass as when first started. © 2003-2020 Chegg Inc. All rights reserved.
| trial initial volume final volume (mL) concentration mass copper () (&/L) 16. ol a O: 95% (b) Calculate the mass of copper chloride in the volume of solution used. Pennies dated 1982-present: Composition: 97.5% zinc, 2.5% copper. %PDF-1.5 %���� The initial mass of the copper was 2.004 grams.
In addition the errors above, another source of error may have came from the final weighing process. 6. endstream endobj 426 0 obj <.
The assignment was to create a formal lab report that expresses data and observations, lab procedure, and a discussion of the data with a conclusion. Doing this added more water to the solution which could have altered the results. Another would be that inside the evaporating dish, there was a crusty, gelatin-like substance that surrounded the dried up solution. The goal of this lab was to see if the mass of copper would be same after going through a series of chemical reactions. b=�: �-��"�n�.��2����b"B�_�j[�1012�%�D�g��` �^ D It is a ductile, malleable metal and is easily pounded and/or drawn into various shapes for use as wire, ornaments and implements of various types. While stirring the solution with a glass rod, slowly add 15 mL of Dil (6 M) NaOH to precipitated Cu(OH)2. This particular lab report shows my ability to work with quantitative data, and analyze the calculations and measurements from the lab.… The most probable reason for this result is the abundance in the sources of error throughout the experiment. It is important to add the NaOH slowly, because you are adding a base to an acid. The final weight should have been 2.004 grams of copper. If the product is not cooled, the reaction would be too violent to deal with. 10�����`+-x�%e�7�\o����M�a��%|[��MJi?bqP��s ����00���Y��\$��.W@T1 �8. At the start of the lab, each group was to measure as close to 2 grams of copper as possible. The result of this copper lab should have been that the mass of the initial copper and the mass of the final copper equals each other. Ol A O: 95% (b) Calculate The Mass Of Copper Chloride In The Volume Of Solution Used. Step II (Conducted at your lab bench) 1. The final moles of copper was .1077 moles.
.
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https://www.compadre.org/precollege/items/detail.cfm?ID=5072 | 1,506,268,747,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690035.53/warc/CC-MAIN-20170924152911-20170924172911-00003.warc.gz | 770,549,893 | 8,761 | Editor selections by Topic and Unit
The Physics Front is a free service provided by the AAPT in partnership with the NSF/NSDL.
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This website details the symmetries and structure of minerals and crystals with pictures. The site provides excellent examples and background information.
Subjects Levels Resource Types
Modern Physics
- Condensed Matter
= Crystal Structure
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- Instructional Material
= Curriculum support
= Tutorial
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- Physics First
- Conceptual Physics
- Algebra-based Physics
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Keywords:
crystal, crystal growth, crystal systems, crystalline structure, properties of matter
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Metadata instance created April 25, 2007 by Christopher Bares
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when Cataloged:
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(1995), WWW Document, (http://www.galleries.com/minerals/symmetry/symmetry.htm).
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Crystal Systems: Symmetry Operations and Crystallographic Axes, (1995), <http://www.galleries.com/minerals/symmetry/symmetry.htm>.
APA Format
Crystal Systems: Symmetry Operations and Crystallographic Axes. (2008, January 1). Retrieved September 24, 2017, from http://www.galleries.com/minerals/symmetry/symmetry.htm
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. Crystal Systems: Symmetry Operations and Crystallographic Axes. January 1, 2008. http://www.galleries.com/minerals/symmetry/symmetry.htm (accessed 24 September 2017).
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Crystal Systems: Symmetry Operations and Crystallographic Axes. 1995. 1 Jan. 2008. 24 Sep. 2017 <http://www.galleries.com/minerals/symmetry/symmetry.htm>.
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@misc{ Title = {Crystal Systems: Symmetry Operations and Crystallographic Axes}, Volume = {2017}, Number = {24 September 2017}, Month = {January 1, 2008}, Year = {1995} }
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%T Crystal Systems: Symmetry Operations and Crystallographic Axes
%D January 1, 2008
%U http://www.galleries.com/minerals/symmetry/symmetry.htm
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Save to my folders | 766 | 3,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.759125 |
https://aboutsabah.com/sightseeing/what-is-the-proper-voltage-for-a-malaysia-electrical-outlet.html | 1,653,357,423,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00404.warc.gz | 130,075,297 | 19,225 | # What is the proper voltage for a Malaysia electrical outlet?
Contents
For Malaysia the associated plug type is G, which is the plug that has three rectangular pins in a triangular pattern. Malaysia operates on a 240V supply voltage and 50Hz.
## Can I use 220V in Malaysia?
Electricity in Malaysia – voltage and frequency
All power sockets in Malaysia provide a standard voltage of 240V with a standard frequency of 50Hz. You can use all your equipment in Malaysia if the outlet voltage in your own country is between 220V-240V.
## Can I use 220V 50Hz in Malaysia?
What voltage and frequency in Malaysia? In Malaysia the standard voltage is 240 V and the frequency is 50 Hz. You can use your electric appliances in Malaysia, if the standard voltage in your country is in between 220 – 240 V (as is in the UK, Europe, Australia and most of Asia and Africa).
## Can I use 120v in Malaysia?
Power converters for Malaysia
Power converters allow travellers to use a 100, 110 or 120 volt electric device with a 240 volt Malaysian power outlet.
## What voltage is standard outlet?
The most common electrical outlet in any home is a 110 volt. Sometimes you may hear 110 volt plugs referred to as 120 volt. Do not be confused by this; think of them as one and the same.
IMPORTANT: You asked: Has Singapore Grip ended?
## Is it OK to plug 220V to 240V?
Yes, any device that is classified in 240 volts and can be used in a 220V socket. If the rated voltage is 240 volts, the boxes can have a power supply of 208V, 220V or 240V.
## Is 240V same as 220V?
In North America, the terms 220V, 230V, and 240V all refer to the same system voltage level. However, 208V refers to a different system voltage level. In North America, the utility companies are required to deliver split phase 240VAC for residential use.
## Can 220V motor run on 240V?
for short periods, 240VAC will not burn out a 220VAC motor. furthermore, switching the polarity of the wires will not blow up the motor. here are the things to check: to begin with, disconnect the 240 VAC power and apply an ohmmeter to the SAME motor terminals to which you had applied the 240VAC power.
## Is EU plug same as Malaysia?
It’s the same ie Type G Electrical Outlet (3 pin). The voltage is a little bit different as UK voltage is 220V – 230V while in Malaysia is 240V. The frequency is the same ie 50Hz.
## How many watts can 13 amps take?
It is important to never overload a plug socket, which is 3000 watts = 13 Amps. Some appliances use more than others. Use the calculator on this page as a guide. Overloading electrical sockets can cause plugs to overheat and result in fire.
## What is US plug 110v?
If a nameplate on an appliance shows that it has a 110 plug, this most likely means that the appliance is designed to operate at 120 volts, but will continue to operate normally if the voltage drops to 110 volts.
IMPORTANT: Where can I exchange old Philippine money?
## What voltage is Japan?
The Kansai Electric Power Company supplies electricity at 100V/60Hz. Although 200V has been implemented for some appliances, basically the voltage in Japan is 100V. Appliances brought from overseas might not be used at the voltage in Japan. Note that the plug sockets for 100V and 200V are different in shapes. | 790 | 3,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | latest | en | 0.871025 |
https://wiki.fysik.dtu.dk/ase/dev/_modules/ase/build/tube.html | 1,586,317,279,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371807538.83/warc/CC-MAIN-20200408010207-20200408040707-00428.warc.gz | 774,943,462 | 4,738 | # Source code for ase.build.tube
from math import sqrt, gcd
import numpy as np
from ase.atoms import Atoms
[docs]def nanotube(n, m, length=1, bond=1.42, symbol='C', verbose=False,
vacuum=None):
if n < m:
m, n = n, m
sign = -1
else:
sign = 1
nk = 6000
sq3 = sqrt(3.0)
a = sq3 * bond
l2 = n * n + m * m + n * m
l = sqrt(l2)
nd = gcd(n, m)
if (n - m) % (3 * nd) == 0:
ndr = 3 * nd
else:
ndr = nd
nr = (2 * m + n) // ndr
ns = -(2 * n + m) // ndr
nn = 2 * l2 // ndr
ichk = 0
if nr == 0:
n60 = 1
else:
n60 = nr * 4
absn = abs(n60)
nnp = []
nnq = []
for i in range(-absn, absn + 1):
for j in range(-absn, absn + 1):
j2 = nr * j - ns * i
if j2 == 1:
j1 = m * i - n * j
if j1 > 0 and j1 < nn:
ichk += 1
nnp.append(i)
nnq.append(j)
if ichk == 0:
if ichk >= 2:
raise RuntimeError('more than 1 pair p, q strange!!')
nnnp = nnp[0]
nnnq = nnq[0]
if verbose:
print('the symmetry vector is', nnnp, nnnq)
lp = nnnp * nnnp + nnnq * nnnq + nnnp * nnnq
r = a * sqrt(lp)
c = a * l
t = sq3 * c / ndr
if 2 * nn > nk:
raise RuntimeError('parameter nk is too small!')
rs = c / (2.0 * np.pi)
if verbose:
q1 = np.arctan((sq3 * m) / (2 * n + m))
q2 = np.arctan((sq3 * nnnq) / (2 * nnnp + nnnq))
q3 = q1 - q2
q4 = 2.0 * np.pi / nn
q5 = bond * np.cos((np.pi / 6.0) - q1) / c * 2.0 * np.pi
h1 = abs(t) / abs(np.sin(q3))
h2 = bond * np.sin((np.pi / 6.0) - q1)
ii = 0
x, y, z = [], [], []
for i in range(nn):
x1, y1, z1 = 0, 0, 0
k = np.floor(i * abs(r) / h1)
x1 = rs * np.cos(i * q4)
y1 = rs * np.sin(i * q4)
z1 = (i * abs(r) - k * h1) * np.sin(q3)
kk2 = abs(np.floor((z1 + 0.0001) / t))
if z1 >= t - 0.0001:
z1 -= t * kk2
elif z1 < 0:
z1 += t * kk2
ii += 1
x.append(x1)
y.append(y1)
z.append(z1)
z3 = (i * abs(r) - k * h1) * np.sin(q3) - h2
ii += 1
if z3 >= 0 and z3 < t:
x2 = rs * np.cos(i * q4 + q5)
y2 = rs * np.sin(i * q4 + q5)
z2 = (i * abs(r) - k * h1) * np.sin(q3) - h2
x.append(x2)
y.append(y2)
z.append(z2)
else:
x2 = rs * np.cos(i * q4 + q5)
y2 = rs * np.sin(i * q4 + q5)
z2 = (i * abs(r) - (k + 1) * h1) * np.sin(q3) - h2
kk = abs(np.floor(z2 / t))
if z2 >= t - 0.0001:
z2 -= t * kk
elif z2 < 0:
z2 += t * kk
x.append(x2)
y.append(y2)
z.append(z2)
ntotal = 2 * nn
X = []
for i in range(ntotal):
X.append([x[i], y[i], sign * z[i]])
if length > 1:
xx = X[:]
for mnp in range(2, length + 1):
for i in range(len(xx)):
X.append(xx[i][:2] + [xx[i][2] + (mnp - 1) * t])
transvec = t
numatom = ntotal * length
diameter = rs * 2
chiralangle = np.arctan((sq3 * n) / (2 * m + n)) / np.pi * 180
cell = [[0, 0, 0], [0, 0, 0], [0, 0, length * t]]
atoms = Atoms(symbol + str(numatom),
positions=X,
cell=cell,
pbc=[False, False, True])
if vacuum:
atoms.center(vacuum, axis=(0, 1))
if verbose:
print('translation vector =', transvec)
print('diameter = ', diameter)
print('chiral angle = ', chiralangle)
return atoms | 1,252 | 2,805 | {"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} | 2.703125 | 3 | CC-MAIN-2020-16 | latest | en | 0.282977 |
https://socratic.org/questions/how-do-you-multiply-2-3i-2-7i-in-trigonometric-form | 1,582,451,588,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00536.warc.gz | 566,838,823 | 6,153 | # How do you multiply (2+3i)(2-7i) in trigonometric form?
##### 1 Answer
Jun 17, 2018
color(blue)((2 + i 3) * (2 - i 7) = 26.25 (0.9524 - i 0.3047)
#### Explanation:
$\left(2 + i 3\right) \cdot \left(2 - i 7\right)$
${z}_{1} \cdot {z}_{2} = \left({r}_{1} {r}_{2}\right) \left(\cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)\right)$
${r}_{1} = \sqrt{{2}^{2} + {3}^{2}} = \sqrt{13}$
${r}_{2} = \sqrt{{2}^{2} + {7}^{2}} = \sqrt{53}$
${\theta}_{1} = \arctan \left(\frac{3}{2}\right) = {56.31}^{\circ}$
${\theta}_{2} = \arctan \left(- \frac{7}{2}\right) = - 74.05 = {285.95}^{\circ} , \text{ IV Quadrant}$
${z}_{1} {z}_{2} = \left(\sqrt{13} \cdot \sqrt{53}\right) \left(\cos \left(56.31 + 285.95\right) + \sin \left(74.05 + 285.95\right)\right)$
$\left(2 + i 3\right) \cdot \left(2 - i 7\right) = 26.25 \left(\cos 342.26 + i \sin 342.26\right)$
$\left(2 + i 3\right) \cdot \left(2 - i 7\right) = 26.25 \left(0.9524 - i 0.3047\right)$ | 469 | 990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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} | 4.46875 | 4 | CC-MAIN-2020-10 | latest | en | 0.284231 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(McQuarrie_and_Simon)/12%3A_Group_Theory_-_The_Exploitation_of_Symmetry/12.07%3A_Characters_of_Irreducible_Representations | 1,632,624,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057796.87/warc/CC-MAIN-20210926022920-20210926052920-00347.warc.gz | 204,964,811 | 23,564 | # 12.7: Characters of Irreducible Representations
The two one-dimensional irreducible representations spanned by $$s_N$$ and $$s_1'$$ are seen to be identical. This means that $$s_N$$ and $$s_1'$$ have the ‘same symmetry’, transforming in the same way under all of the symmetry operations of the point group and forming bases for the same matrix representation. As such, they are said to belong to the same symmetry species. There are a limited number of ways in which an arbitrary function can transform under the symmetry operations of a group, giving rise to a limited number of symmetry species. Any function that forms a basis for a matrix representation of a group must transform as one of the symmetry species of the group. The irreducible representations of a point group are labeled according to their symmetry species as follows:
1. 1D representations are labeled $$A$$ or $$B$$, depending on whether they are symmetric (character $$+1$$) or antisymmetric (character $$-1$$) under rotation about the principal axis.
2. 2D representations are labeled $$E$$, 3D representations are labeled $$T$$.
3. In groups containing a center of inversion, $$g$$ and $$u$$ labels (from the German gerade and ungerade, meaning symmetric and antisymmetric) denote the character of the irreducible representation under inversion ($$+1$$ for $$g$$, $$-1$$ for $$u$$)
4. In groups with a horizontal mirror plane but no center of inversion, the irreducible representations are given prime and double prime labels to denote whether they are symmetric (character $$+1$$ or antisymmetric (character $$-1$$) under reflection in the plane.
5. If further distinction between irreducible representations is required, subscripts $$1$$ and $$2$$ are used to denote the character with respect to a $$C_2$$ rotation perpendicular to the principal axis, or with respect to a vertical reflection if there are no $$C_2$$ rotations.
The 1D irreducible representation in the $$C_{3v}$$ point group is symmetric (has character $$+1$$) under all the symmetry operations of the group. It therefore belongs to the irreducible representation $$A_1$$. The 2D irreducible representation has character $$2$$ under the identity operation, $$-1$$ under rotation, and $$0$$ under reflection, and belongs to the irreducible representation $$E$$.
Sometimes there is confusion over the relationship between a function $$f$$ and its irreducible representation, but it is quite important that you understand the connection. There are several different ways of stating the relationship. For example, the following statements all mean the same thing:
• "$$f$$ has $$A_2$$ symmetry"
• "$$f$$ transforms as $$A_2$$"
• "$$f$$ has the same symmetry | 601 | 2,704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-39 | latest | en | 0.883971 |
https://classroom.thenational.academy/lessons/speed-c5jp4t?activity=intro_quiz&step=1 | 1,653,290,254,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00324.warc.gz | 219,681,834 | 28,965 | Speed
In this lesson we will learn the definition and equation of speed.
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What is the equation that links extension, force and spring constant?
1/7
Q2.What store of energy is increased when an elastic object is stretched or compressed?
2/7
Q3.Whose law refers to the relationship between force and extension of an elastic object?
3/7
Q4.What does elastically deform mean?
4/7
Q5.What name is given to the point where a spring no longer returns to its original shape when stretched?
5/7
Q6.What does Hooke's law state?
6/7
Q7.Calculate the force required to extend a spring by 0.4m when the spring has a spring constant of 5 N/m.
7/7
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What is the equation that links extension, force and spring constant?
1/7
Q2.What store of energy is increased when an elastic object is stretched or compressed?
2/7
Q3.Whose law refers to the relationship between force and extension of an elastic object?
3/7
Q4.What does elastically deform mean?
4/7
Q5.What name is given to the point where a spring no longer returns to its original shape when stretched?
5/7
Q6.What does Hooke's law state?
6/7
Q7.Calculate the force required to extend a spring by 0.4m when the spring has a spring constant of 5 N/m.
7/7
Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
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Worksheet
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Quiz:
Speed
Complete the quiz to see how much you have remembered from the speed lesson.
Q1.What is speed defined as?
1/11
Q2.What is the average speed of sound in air?
2/11
Q3.What is the average speed of a person running?
3/11
Q4.What is the average speed of a person walking?
4/11
Q5.What is the average speed of light in a vacuum?
5/11
Q6.What is the average speed of cycling?
6/11
Q7.What is the equation that links distance, speed and time?
7/11
Q8.A man travels 50 m in 25 seconds. What is his speed?
8/11
Q9.A car travels 1.2 km in 50 seconds. What is its speed?
9/11
Q10.A boy is riding a bike at a speed of 6 m/s. He cycles for 5 minutes. How far has he travelled?
10/11
Q11.A woman runs a 5 km race. Her average speed is 3 m/s. What time did she complete the race in? Give your answer to 2 significant figures
11/11
Quiz:
Speed
Complete the quiz to see how much you have remembered from the speed lesson.
Q1.What is speed defined as?
1/11
Q2.What is the average speed of sound in air?
2/11
Q3.What is the average speed of a person running?
3/11
Q4.What is the average speed of a person walking?
4/11
Q5.What is the average speed of light in a vacuum?
5/11
Q6.What is the average speed of cycling?
6/11
Q7.What is the equation that links distance, speed and time?
7/11
Q8.A man travels 50 m in 25 seconds. What is his speed?
8/11
Q9.A car travels 1.2 km in 50 seconds. What is its speed?
9/11
Q10.A boy is riding a bike at a speed of 6 m/s. He cycles for 5 minutes. How far has he travelled?
10/11
Q11.A woman runs a 5 km race. Her average speed is 3 m/s. What time did she complete the race in? Give your answer to 2 significant figures
11/11
Lesson summary: Speed
Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Dance | 1,047 | 3,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.896918 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-865265-0&chapter=7&headerFile=4&lesson=0&state=va | 1,386,393,269,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053558/warc/CC-MAIN-20131204131733-00056-ip-10-33-133-15.ec2.internal.warc.gz | 362,461,465 | 4,114 | 1. Find 6 × 18.52. A. 108.12 B. 111.02 C. 1111.2 D. 111.12 Hint 2. Find 4.2 Χ 7.5. A. 31.4 B. 31.5 C. 3.15 D. 0.315 Hint 3. What is 21.6 divided by 0.03? A. 72.0 B. 7.20 C. 720 D. 0.720 Hint 4. Find 16.2 ÷ 6. A. 5.4 B. 2.7 C. 27 D. 54 Hint 5. Find . Write the answer in simplest form. A. B. C. D. Hint 6. Use the symbol on a calculator to find the value of · 24. Round to the nearest tenth. A. 25.1 B. 27.1 C. 75.4 D. 37.7 Hint 7. Use the symbol on a calculator to find the value of · 42. Round to the nearest tenth. A. 18.8 B. 25.1 C. 19.1 D. 50.3 Hint 8. Use the symbol on a calculator to find the value of · (16 ÷ 2)2. Round to the nearest tenth. A. 100.5 B. 12.6 C. 50.3 D. 201.1 Hint 9. Classify the polygon according to its number of sides. A. triangle B. pentagon C. hexagon D. quadrilateral Hint 10. Classify the polygon according to its number of sides. A. hexagon B. quadrilateral C. pentagon D. triangle Hint | 384 | 919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2013-48 | latest | en | 0.697832 |
https://www.esaral.com/q/if-a-b-are-square-matrices-of-order-3-a-is-non-singular-and-ab-o-then-b-is-a-58426/ | 1,653,799,802,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00334.warc.gz | 850,334,722 | 23,295 | If A, B are square matrices of order 3, A is non-singular and AB = O, then B is a
Question:
If $A, B$ are square matrices of order $3, A$ is non-singular and $A B=0$, then $B$ is a
(a) null matrix
(b) singular matrix
(c) unit-matrix
(d) non-singular matrix
Solution:
(a) null matrix
Since A is non-singular matrix and the determinant of a non-singular matrix is non-zero, B should be a null matrix. | 123 | 406 | {"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} | 3.015625 | 3 | CC-MAIN-2022-21 | latest | en | 0.748023 |
http://www.webiwip.com/quantitative-aptitude-questions-answers/algebra/703616 | 1,708,505,563,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473401.5/warc/CC-MAIN-20240221070402-20240221100402-00890.warc.gz | 73,601,730 | 10,148 | # Algebra - Aptitude Questions and Answers
If (7 - 12x) - (3x - 7) = 14, then the value of x is ? A) -4 B) 0 C) 5 D) 2 Correct Answer : 0 Explanation :(7 - 12x) - (3x - 7) = 14 7 - 12x - 3x + 7 = 14 - 15x + 14 = 14 - 15x = 0 x = 0 Post/View Answer Post comment Cancel Thanks for your comment.! Write a comment(Click here) ... | 134 | 327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.558786 |
https://www.coursehero.com/file/1655348/IPS6eCh04-5bb/ | 1,516,401,234,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888135.38/warc/CC-MAIN-20180119204427-20180119224427-00083.warc.gz | 880,827,127 | 178,038 | IPS6eCh04_5bb
# IPS6eCh04_5bb - Probability The Study of Randomness General...
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Probability: The Study of Randomness General Probability Rules IPS Chapter 4.5 © 2009 W.H. Freeman and Company
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Objectives (IPS Chapter 4.5) General probability rules General addition rules Conditional probability General multiplication rules Tree diagrams Bayes’s rule
General addition rules General addition rule for any two events A and B: The probability that A occurs, B occurs, or both events occur is: P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ) What is the probability of randomly drawing either an ace or a heart from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts. However, 1 card is both an ace and a heart. Thus: P (ace or heart) = P (ace) + P (heart) – P (ace and heart) = 4/52 + 13/52 - 1/52 = 16/52 ≈ .3
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Conditional probability Conditional probabilities reflect how the probability of an event can change if we know that some other event has occurred/is occurring. Example: The probability that a cloudy day will result in rain is different if you live in Los Angeles than if you live in Seattle. Our brains effortlessly calculate conditional probabilities, updating our
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IPS6eCh04_5bb - Probability The Study of Randomness General...
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Ask a homework question - tutors are online | 443 | 1,789 | {"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} | 3.90625 | 4 | CC-MAIN-2018-05 | longest | en | 0.851332 |
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# Mole calculations
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### Mole calculations
1. 1. Using the mole for… CONVERSIONS
2. 2. CONVERSIONS <ul><li>Must use the molar mass </li></ul><ul><li>Can go from: </li></ul><ul><ul><li># of atoms/molecules to moles </li></ul></ul><ul><ul><li>Moles to # of atoms/molecules </li></ul></ul><ul><ul><li>moles to grams </li></ul></ul><ul><ul><li>grams to moles </li></ul></ul>
3. 3. SHORTCUTS for CONVERSIONS <ul><li>Moles to atom/molecules </li></ul><ul><ul><li>moles times 6.022 x 10 23 </li></ul></ul><ul><li>Atom/molecules to moles </li></ul><ul><ul><li># of atoms or molecules divided by 6.022 x 10 23 </li></ul></ul><ul><li>Moles to grams </li></ul><ul><ul><li>moles times molar mass </li></ul></ul><ul><li>Grams to moles </li></ul><ul><ul><li>grams divided by molar mass </li></ul></ul>
4. 4. mole island atom island gram island liter island Divide by 22.4L Mult by 22.4L Divide by molar mass Mult by molar mass divide by 6.022 x 10 23 Mult by 6.022 x 10 23
5. 5. Mole GRAMS LITERS # of atoms Divide by 6.022 x 10 23 X by 6.022 x 10 23 X by 22.4 liters Divide by 22.4 liters Divide by molar mass X by molar mass
6. 6. Sample problems <ul><li>How many atoms of copper are found in 1.34 moles of copper? </li></ul><ul><li>We are on mole island, need to get to atoms island. </li></ul><ul><li>Multiply moles by 6.022 x 10 23 </li></ul><ul><li>1.34 x 6.022 x 10 23 = </li></ul><ul><li>8.07 x 10 23 atoms of copper </li></ul>
7. 7. <ul><li>How many moles does 1.23 x 10 23 molecules of CaCl 2 represent? </li></ul><ul><li>we are on atom/molecule island, need to get to mole island. </li></ul><ul><li>divide # of molecules by 6.022 x 10 23 </li></ul><ul><li>1.23 x 10 23 / 6.022 x 10 23 = </li></ul><ul><li>.20 moles of CaCl 2 </li></ul>
8. 8. <ul><li>How many grams are in 3.4 moles of I 2 ? </li></ul><ul><li>We are on mole island, need to get to grams island. </li></ul><ul><li>multiply # of moles by the molar mass </li></ul><ul><li>3.4 x 253.8 = </li></ul><ul><li>862.92 grams of I 2 </li></ul>
9. 9. <ul><li>How many moles does 134.7 grams of H 2 SO 4 equal? </li></ul><ul><li>We are on gram island, need to get to mole island. </li></ul><ul><li>divide # of grams by the molar mass </li></ul><ul><li>134.7 / 98.08 = </li></ul><ul><li>1.37 moles of H 2 SO 4 </li></ul>
10. 10. <ul><li>How much does 7.12 x 10 24 molecules of H 2 O weigh in grams? </li></ul><ul><li>We are on atoms/molecules island, need to get to gram island. </li></ul><ul><li>divide # of molecules by 6.022 x 10 23 </li></ul><ul><li>then multiply by the molar mass </li></ul><ul><li>(7.12 x 10 24 /6.022 x 10 23 ) x 18.02 = </li></ul><ul><li>213.06 grams of water </li></ul>
11. 11. HOMEWORK <ul><li>1. How many moles are found in a sample containing 6.02 x 10 24 atoms? </li></ul><ul><li>2. If you have 1.28 x 10 22 molecules of HCl, how many moles do you have? </li></ul><ul><li>3. 3.5 moles of gold would have how many atoms? </li></ul><ul><li>4. If you had 12.3 moles of baking soda, how many molecules would it contain? </li></ul><ul><li>5. A sample of NaCl weighing 1.25 grams contains how many molecules? </li></ul>
12. 12. MORE HOMEWORK <ul><li>6. Calculate the number of moles in a 15.3 gram sample of barium hydroxide. </li></ul><ul><li>7. How many grams does 0.14 moles of silver nitrate equal? </li></ul><ul><li>8. If you had 2.3 grams of aluminum, how many atoms would it contain? </li></ul><ul><li>9. If you had 3.4 x 10 34 atoms of tungsten(V) arsenate, how many moles would you have? How many grams? </li></ul> | 1,438 | 3,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-22 | latest | en | 0.581223 |
https://www.jiskha.com/questions/564076/how-many-liters-of-water-should-be-added-to-20-liters-of-40-solution-of-acid-to-obtain | 1,596,929,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738366.27/warc/CC-MAIN-20200808224308-20200809014308-00317.warc.gz | 730,837,517 | 6,020 | # Math
How many liters of water should be added to 20 liters of 40% solution of acid to obtain 30% solution?
1. 👍 0
2. 👎 0
3. 👁 139
1. Acid = 0.4 * 20 = 8 liters.
Water = 0.6 * 20 = 12 lites.
A / (A + W) = 0.30,
8 / (8 + W) = 0.30,
2.4 + 0.3W = 8,
0.3W = 8 - 2.4 = 5.6,
W = 18.7 Liters.
18.7 - 12 = 6.7 Liters of water added.
1. 👍 0
2. 👎 0
2. x = liters of water
.40(20) = .30(x + 20)
8 = .30x + 6
8 - 6 = .30x + 6 - 6
2 = .30x
2 / .30 = .30x / .30
6.666 = x (which is also 6 2/3 as a fraction)
6.7 ~ x
6.7 liters of water is needed.
or
Six and two-third liters of water is need to obtain a 30% acid solution.
1. 👍 0
2. 👎 0
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Assignment 2: Operations Decision
Due Week 6 and worth 300 points
Using the regression results and the other computations from Assignment 1, determine the market structure in which the low-calorie frozen, microwavable food company operates.
Use the Internet to research two (2) of the leading competitors in the low-calorie frozen, microwavable food industry, and take note of their pricing strategies, profitability, and their relationships within the industry (worldwide).
Write a six to eight (6-8) page paper in which you:
1. Outline a plan that will assess the effectiveness of the market structure for the company’s operations. Note: In Assignment 1, the assumption was that the market structure [or selling environment] was perfectly competitive and that the equilibrium price was to be determined by setting QD equal to QS. You are now aware of recent changes in the selling environment that suggest an imperfectly competitive market where your firm now has substantial market power in setting its own “optimal” price.
2. Given that business operations have changed from the market structure specified in the original scenario in Assignment 1, determine two (2) likely factors that might have caused the change. Predict the primary manner in which this change would likely impact business operations in the new market environment.
3. Analyze the major short run and long cost functions for the low-calorie, frozen microwaveable food company given the cost functions below. Suggest substantive ways in which the low-calorie food company may use this information in order to make decisions in both the short-run and the long-run.
TC = 160,000,000 + 100Q + 0.0063212Q2
VC = 100Q + 0.0063212Q2
MC= 100 + 0.0126424Q
4. Determine the possible circumstances under which the company should discontinue operations. Suggest key actions that management should take in order to confront these circumstances. Provide a rationale for your response. (Hint: Your firm’s price must cover average variable costs in the short run and average total costs in the long run to continue operations.)
5. Suggest one (1) pricing policy that will enable your low-calorie, frozen microwavable food company to maximize profits. Provide a rationale for your suggestion.
(Hints:
• In Assignment 1, you determined your firm’s market demand equation. Now you need to find the inverse demand equation. Having found that, find the Total Revenue function for your firm (TR is P x Q). From your firm’s Total Revenue function, then find your Marginal Revenue (MR) function.
• Use the profit maximization rule MR = MC to determine your optimal price and optimal output level now that you have market power. Compare these values with the values you generated in Assignment 1. Determine whether your price higher is or lower.)
6. Outline a plan, based on the information provided in the scenario, which the company could use in order to evaluate its financial performance. Consider all the key drivers of performance, such as company profit or loss for both the short term and long term, and the fundamental manner in which each factor influences managerial decisions.
(Hints:
• Calculate profit in the short run by using the price and output levels you generated in part 5. Optional: You may want to compare this to what profit would have been in Assignment 1 using the cost function provided here.
• Calculate profit in the long run by using the output level you generated in part 5 and cost data in part 3 and assuming that the selling environment will likely be very competitive. Determine why this would be a valid assumption.)
7. Recommend two (2) actions that the company could take in order to improve its profitability and deliver more value to its stakeholders. Outline, in brief, a plan to implement your recommendations.
8. Use at least five (5) quality academic resources in this assignment. Note: Wikipedia does not qualify as an academic resource.
• Be typed, double spaced, using Times New Roman font (size 12), with one-inch margins on all sides; citations and references must follow APA or school-specific format. Check with your professor for any additional instructions.
• Include a cover page containing the title of the assignment, the student’s name, the professor’s name, the course title, and the date. The cover page and the reference page are not included in the required assignment page length.
The specific course learning outcomes associated with this assignment are:
• Analyze short-run and long-run production and cost functions.
• Apply macroeconomic concepts to changes in global and national economies and how they affect economic growth, inflation, interest rates, and wage rates.
• Evaluate the profit-maximizing price and output level for given operating costs for monopolies and firms in competitive industries.
• Use technology and information resources to research issues in managerial economics and globalization.
• Write clearly and concisely about managerial economics and globalization using proper writing mechanics.
Click here to view the grading rubric.
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How it works? | 1,378 | 6,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-14 | latest | en | 0.923744 |
http://www.instructables.com/id/Fix-your-binocularsIt-is-easy/ | 1,495,503,063,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607245.69/warc/CC-MAIN-20170523005639-20170523025639-00576.warc.gz | 545,701,613 | 16,659 | Do you have some primates and see wrong with them?
Do you see the double image and gives you a headache?
Well has a solution.
## Step 1: Deconstructing the Prismatic
If you dismantle the prismatic can see that there is a small screw that is supported in each of the prisms.
Normally these screws are located outside the binoculars.
Not necessary to remove the prismatic, must be located where they are (the screws are very small, and sometimes are hard to find).
=======================================================================
Si desmontamos el prismatico podemos ver que existe un pequeño tornillo que se apoya en cada uno de los prismas.
Normalmente estos tornillos se localizan en el exterior del prismatico. NO hace falta desmontar el prismatico, hay que localizar donde se encuentran (los tornillos son muy pequeños, y a veces son dificiles de encontrar).
## Step 2: Cause of the Problem: NO Parallel Light Rays
We see the double image because light rays (image) that leave for us is not parallel.
=======================================
Vemos la imagen doble porque los rayos de luz (imagen) que salen de los oculares NO es paralela.
## Step 3: Solving the Problem
To solve the problem we must ensure that the light rays go parallel.
For that we turn the screws (move the prisms), so get moving with the light rays.
=================================================
Para solucionar el problema hay que hacer que los rayos de luz salgan paralelos.
Para ello debemos girar los tornillos (mover los prismas), con ello conseguiremos mover los rayos de luz.
## Step 4: Last Tip
To work comfortably:
1 .- Put the binoculars on a tripod and attach it securely.
2 .- Focuses any point with the binoculars (really the double image)
3 .- Move the screw slowly until the two images match.
Once the work can be used (carefully) a little hot glue to prevent movement in the prisms.
If this instructable helps one person I am satisfied.
Thanks for visiting this instructable.
==========================================================
Para trabajar comodamente:
1.- Coloca los prismaticos sobre un tripode y sujetalos fuertemente.
2.- Enfoca cualquier punto con los prismaticos (veras la imagen doble)
3.- Mueve poco a poco los tornillos hasta que los dos imagenes coincidan.
Una vez hecho el trabajo, se pueden usar (con cuidado) un poco de hot glue para evitar movimientos en los prismas.
Si este instructable sirve de ayuda a una sola persona me doy por satisfecho.
Gracias por visitar este instructable.
<p>Many thanks - I was just about to bin my binoculars, but this simple procedure has saved them. Good as new again !!</p>
¿Alguien sabe de algún sitio con instrucciones para reparar el mecanismo de enfoque de los binoculares? tengo unos que se soltaron del tornillo central y tengo que mover los oculares manualmente.<br>GRACIAS
OJO con el HOT GLUE! Comúnmente deja manchas blancas indelebles alrededor de donde se aplicó. (Warning with HOT GLUE! Usually it leaves indelible white spots around where you applied it.) Welcome to the BIC ("Bilingual Instructables Club")!
Gracias por la bienvenida. He buscado mucho un sitio parecido a este en español, pero no lo he encontrado.
Yo lo sigo desde hace 4 años y un día, y gran parte de lo poco que he aprendido de inglés se lo debo a la interacción con esta página, que es la primera que leo luego del correo. Respecto del OJO con el HOT GLUE! mira lo que le pasó a la pobre Ángela:
Quise adjuntar una foto y no anduvo. Acá va el link: <a href="http://funpics.classicfun.ws/index.php/Funpics/merkel-be-careful-with-superglue" rel="nofollow">http://funpics.classicfun.ws/index.php/Funpics/merkel-be-careful-with-superglue</a>
Lo de Angela es un trabajo burdo comparado con los trabajos refinadisimos que se ven en esta presentacion, ahi va el enlace: http://www.megaupload.com/?d=XM1H7SVQ
Muy bueno, seguramente extractado de http://thereifixedit.com/, ¿no?
<br> If you're repairing cheap bins, this is fine, but better ones have water-tight and gas-tight seals, and often have super-dry inert gases filling them, which are lost the instant you open them up.<br> <br> <br> | 1,077 | 4,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.525531 |
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Second Order System Adjustment with first order controller (MIT Rule) | 844 | 3,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.743736 |
https://www.talkstats.com/threads/help-please-with-a-tricky-problem-compare-one-element-to-a-row-in-a-table.58534/ | 1,656,794,092,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00302.warc.gz | 1,070,604,342 | 10,223 | # Help please with a tricky problem: compare one element to a row in a table
#### Lill
##### New Member
Hi
Hi I have a really tricky problem. I tried several approaches and so far I got lost.
I have a table with one column which contains 9 and 2.
This is the second table:
Code:
> my_crazy_data
V1 V2 V3 V4 V5 V6 V7
1 1 a 2.0 3.0 4.0 9.0 6.0
2 1 b 0.2 0.1 0.8 0.9 0.2
3 1 c 0.1 0.3 0.4 0.1 0.8
4 1 d 0.1 0.1 0.3 0.1 0.4
5 2 a 1.0 2.0 3.0 4.0 5.0
6 2 b 0.1 0.4 0.3 0.5 0.7
7 2 c 0.1 0.4 0.2 0.5 0.8
8 2 d 0.2 0.5 0.6 0.7 0.9
9 3 a 8.0 2.0 3.0 4.0 5.0
10 3 b 0.1 0.4 0.3 0.5 0.7
11 3 c 0.1 0.4 0.2 0.5 0.8
12 3 d 0.2 0.5 0.6 0.7 0.9
Now I want to take the first element from the first table (in this case 9) and find it in the first row V3-V7 as V1 is the index. Once I found 9 (in this case V6) I want to return all V6 for the index V1=1 with V1 and V2 so a result would look like:
Code:
1 a 9.0
1 b 0.9
1 c 0.1
1 d 0.1
then proceed the second element from table1 (in this case 2) and same procedure but only search in the first row that has the first V1=2 so I would search for 2 in line 5 and once I found it only output the columns for index V1=2 so a result would look like:
Code:
2 a 2.0
2 b 0.4
2 c 0.4
2 d 0.5
Maybe doing this will work:
# Your vector of values to look up
Code:
v <- c(9, 2, 4)
# Split the second table by V1
Code:
d.split <- split(d, d\$V1)
Code:
do.call(rbind,
mapply(function(x, y) {
setNames(x[, c(1:2, 2 + match(y, x[1, -(1:2)]))],
c('V1', 'V2', 'val'))
}, d.split, v, SIMPLIFY=FALSE)
)
but the do.call breaks if it can not find an element? How can I search in the row and then return 0 if there is no element that matches and carry on?
Thank you so much!
xx
#### JesperHP
##### TS Contributor
What do you mean with do.call breaks?If it is somekind of error occuring perhaps you could use tryCatch?
#### Lill
##### New Member
What do you mean with do.call breaks?If it is somekind of error occuring perhaps you could use tryCatch?
Hi,
Yes sorry the code works for
Code:
v <- c(9, 2, 4)
but I acutually have 270 elements in there and It cant compare them anymore. Or if I have an element in v that is not in table2 so it cant find it.
I always get the following error:
Code:
Error in [.data.frame(x, , c(1:2, 2 + match(y, x[1, -(1:2)], nomatch = NA_integer_, : undefined columns selected | 961 | 2,372 | {"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} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.858585 |
http://mathoverflow.net/questions/44801/a-5-extension-of-number-fields-unramified-everywhere/44809 | 1,467,106,630,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396872.10/warc/CC-MAIN-20160624154956-00138-ip-10-164-35-72.ec2.internal.warc.gz | 203,334,620 | 20,194 | # $A_5$-extension of number fields unramified everywhere
So I was having tea with a colleague immensely more talented than myself and we were discussing his teaching algebraic number theory. He told me that he had given a few examples of abelian and solvable extensions unramified everywhere for his students to play with and that he had find this easy to construct with class field theory in the back of his head. But then he asked me if I knew how to construct an extension of number fields with Galois group $A_{5}$ and unramified everywhere. All I could say at the time (and now) is:
1. There are Hilbert modular forms unramified everywhere.
2. There are Hilbert modular forms whose residual $G_{{F}_{v}}$-representation mod $p$ is trivial for all $v|p$.
3. There are Hilbert modular forms whose residual $G_{F}$-representation mod $p$ has image $A_{5}$ inside $\operatorname{GL}_{2}(\mathbb F_{p})$.
Suppose there is a Hilbert modular form satisfying all three conditions. Then the Galois extension through which its residual $G_{F}$-representation factors would have Galois group $A_{5}$ and would be unramified everywhere.
Can this be made to work?
Regardless of the validity of this circle of idea, can you construct an extension of number fields unramified everywhere and with Galois group $A_{5}$?
-
I've forgotten the standard example, but I think there's a standard example. You write down some explicit polynomial of degree 5 with rational coefficients (it's something like $x^5+x+1$ but it's not exactly that) and its Galois group over $\mathbf{Q}$ is $S_5$, and then you let the base be the quadratic subextension. – Kevin Buzzard Nov 4 '10 at 10:47
The problem with using Hilbert modular forms is that it is very common that the associated mod $p$ representation is ramified at $p$, making the problem computationally very hard. For example if you want to compute in parallel weight 2 then even the determinant of the mod $p$ representation will be ramified at $p$ unless $p=2$. This rules out using, say, standard tables of elliptic curves of conductor 1. – Kevin Buzzard Nov 4 '10 at 10:50
I don't know whether this is relevant here, but since $A_5$ is isomorphic to $\mathrm{PSL}_2(\mathbf{F}_5)$, one could try to start with an elliptic curve over some number field with good reduction everywhere, and look at the Galois representation on the 5-torsion points. – François Brunault Nov 4 '10 at 11:00
You'll need the base to contain $Q(\zeta_5)$ or something, to stop it being ramified at 5. – Kevin Buzzard Nov 4 '10 at 11:04
If you take the splitting field of $x^5+ax+b$ and consider it as an extension of its quadratic subfield, then it will be unramified with Galois group contained in $A_5$ whenever $4a$ and $5b$ are relatively prime. This is a result of Yamamoto. For almost all $a$ and $b$ (specifically, on the complement of a thin set), the group is $A_5$.
You might also enjoy this preprint of Kedlaya, which I found very readable. A note on Kedlaya's webpage, dated May 2003, says that he will not be publishing this because it has been superseded by a recent result of Ellenberg and Venkatesh. I assume he is referring to this paper, but I can't figure out why that one supersedes his.
-
Surely one needs the hypothesis that the splitting field of $x^5+ax+b$ has Galois group $S_{5}$. – Olivier Nov 4 '10 at 13:08
You're right, I'll edit. – David Speyer Nov 4 '10 at 13:29
Here's the standard example. I found it in Lang's Algebraic Number Theory where he attributes it to Artin. Let $K$ be the splitting field of $X^5-X+1$ over $\mathbb{Q}$. Then $K$ has Galois group $S_5$ over $\mathbb{Q}$ and $A_5$ over $L=\mathbb{Q}(\sqrt{2869})$. Also $K$ is unramified over $L$.
-
Thanks Robin. I knew I was close :-) – Kevin Buzzard Nov 4 '10 at 11:37
Is $2869$ something like the discriminant of $X^5-X+1$ ? – Denis Serre Nov 4 '10 at 11:57
@Denis Serre: is there no computer algebra package on your computer? :-) Yes, 2869 is the discriminant. – Kevin Buzzard Nov 4 '10 at 12:04
@François: I think any quintic $f$ over ${\mathbb Q}$ with Galois group $S_5$ and square-free discriminant $D$ defines an unramified $A_5$-extension of $Q(\sqrt D)$. This is because exactly two roots of $f$ coincide modulo any prime divisor of $D$, so the inertia at $p$ must be exactly $C_2$. E.g. $f(x)=x^5-x^4-x^3+x^2-1$ has $D=1609$, but I don't know whether there are any smaller examples. – Tim Dokchitser Nov 15 '10 at 16:34
Per the philosophy of Bhargava (and indeed it seems plausible that this could be proved via his techniques) the number of S_5-extensions with squarefree discriminant between -N and N should be about (13/120)*(6/pi^2)*2N; i.e. each squarefree discriminant has 13/120 A_5-extensions on average. – JSE Dec 15 '10 at 2:08
Oh, I know how I would try and build examples. First I would write down a random $A_5$ extension $K$ of $\mathbf{Q}$, ramified at some primes (in fact I would look in a table, e.g. in Buhler's thesis or the Frey et al LNM on Artin's conjecture, to find an example ramified at only one prime, probably). I would then try to kill this ramification by making an extension $L$ of $\mathbf{Q}$ which is disjoint from $K$ globally but exactly the same as it locally for the ramified primes (e.g. by going up to a splitting field of a polynomial which is highly $p$-adically congruent to the original poly whose splitting field was the $A_5$ field, for all $p$ where $K$ ramified). I'd then look at the extension $LK/L$ which I think now should be totally split at the primes above the prime where $K/\mathbf{Q}$ ramified and hence unramified everywhere. I think this stands a fair chance of working, and indeed becoming a general machine which turns an extension of number fields with group $G$ into an extension of number fields unramified everywhere with group $G$. Or have I missed something? If not then this is undoubtedly a well-known technique.
-
Here's an explicit worked example in a much easier case. Consider $K=\mathbf{Q}(\sqrt{2})/\mathbf{Q}$. It's the splitting field of $x^2-2$. It's ramified at 2. So let's choose a polynomial highly congruent to $x^2-2$, for example $x^2-34$, and set $L=\mathbf{Q}(\sqrt{34})$. Now $KL/L$ should be unramified everywhere and if my calculations are right, it is. – Kevin Buzzard Nov 4 '10 at 11:42
If I understand correctly, the approach of Yamammoto (in the specific case of $G=A_{n}$) is somewhat dual to what you suggest. Instead of killing the ramification by composing with an auxiliary extension, he realizes $G$ as a normal subgroup of a larger group $G'$ and ensures that the ramification can only occur in the quotient $G'/G$. When $G=A_{n}$, this works nicely because $G'=S_{n}$ and it is easy to build polynomials with at most one double roots modulo $p$ for all $p$. – Olivier Nov 4 '10 at 13:13
@Kevin: this certainly sounds good to me. Indeed, the basic technique (choosing global extensions which have prescribed behavior at finitely many places) is one that I have used many times in my own work... – Pete L. Clark Nov 4 '10 at 19:21
Although the question was only about unramified $\mathfrak{A}_5$-extensions, and has been completely answered, it might not be superfluous to mention the following paper which I happened to come across today:
MR0819826 (87e:11122)
Elstrodt, J.(D-MUNS); Grunewald, F.(D-BONN); Mennicke, J.(D-BLF)
On unramified $A_m$-extensions of quadratic number fields.
Glasgow Math. J. 27 (1985), 31–37.
An explicit description is given of unramified extensions $S/k$ with Galois group equal to the alternating group $A_n$, where $k$ is a quadratic number field. The authors prove that if $f(x)\in {\bf Z}[x]$ is a monic, irreducible polynomial of degree $n$ with square-free discriminant and Galois group $S_n$, then $S/k$ is an unramified $A_n$-extension. Here $S$ denotes the splitting field for $f(x)$ over ${\bf Q}$ and $k={\bf Q}(\sqrt{\Delta})$, where $\Delta$ is the determinant of $f$. The proof involves a series of calculations which show that $S/k$ has relative different 1.
In the final section, 84 examples of unramified $A_5$-extensions of quadratic fields are given. In 15 of the cases the quadratic field is real and in 69 cases it is imaginary. This list contains an example (with real quadratic field) due to E. Artin, which was mentioned by S. Lang [Algebraic number theory, see p. 121, Addison-Wesley, Reading, Mass., 1970].
Reviewed by Charles J. Parry
Addendum (2011/03/30) Kedlaya's preprint mentioned by Speyer is now available on the arXiv. A corollary is that for each $n\geq3$, infinitely many quadratic number fields admit everywhere unramified degree-$n$ extensions whose normal closures have Galois group $\mathfrak{A}_n$.
Addendum Kedlaya's paper has now appeared in the Proceedings of the AMS 140 (2012), 3025--3033
-
That preprint of Kedlaya is exactly the same as the one in Speyer's answer (well, different web links, but the same content). – KConrad Mar 30 '11 at 6:52
I'm really sorry not to have realised this. The Addendum has been edited accordingly. – Chandan Singh Dalawat Mar 30 '11 at 7:21 | 2,553 | 9,093 | {"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} | 2.78125 | 3 | CC-MAIN-2016-26 | latest | en | 0.94866 |
https://physics.stackexchange.com/questions/644919/non-invertible-function-in-the-context-of-legendre-transformations | 1,716,583,940,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00135.warc.gz | 379,596,469 | 40,436 | Non-invertible function in the context of Legendre transformations
I'm in my 4th semester of physics and currently visit a course about Thermodynamics.
We currently deal with Legendre Transformations and my textbook gave the following example:
Given the function $$U(S,V,N)=A\frac{S^2}{N}e^{\frac{S}{k_B N}}\tag{1}$$ try to find $$U(S,V,\mu)$$.
If you do this the standard way, calculating $$(\frac{\partial U}{\partial N})=\mu\tag{2}$$ and then try to find a expression for $$N(\mu)$$, you arrive at a equation which is not invertible in terms of $$N$$. So the just described doesn't really work out.
It further says that the missing invertibility of this special relation is a standard case in Thermodynamics (only in the case of an ideal gas you can find an analytical solution) and I will encounter it again at least once in a course about Statistical Mechanics.
I do not fully understand what is meant by that sentence so I wanted to ask here if anyone knows what the book is talking about and give me some further details about it? I would appreciate any comments, thanks!
• Related: physics.stackexchange.com/q/336818/2451 , physics.stackexchange.com/q/546495/2451 and links therein. Jun 11, 2021 at 17:21
• Thank you for linking these two! So the top answer in the second post talks about how if the mapping between the derivative with respect to velocity and the momentum is neither injective nor surjective we are dealing with so called constrained Hamiltonian theories and to solve such problems we can use "Dirac bracket". Would that be correct so far? Is this procedure and reasoning directly convertible to Thermodynamics aswell? Because all posts that I read so far were dealing with Lagrange/Hamiltonian mechanics. Jun 12, 2021 at 8:00
1. The problem with the standard way (2) for OP's system (1) is that the derivative $$U^{\prime}\equiv\frac{\partial U}{\partial N}<0$$ is negative, so that eq. (2) has no solution for non-negative chemical potential $$\mu\geq 0$$.
However, everything is not lost since one may show that the function (1) is convex $$U^{\prime\prime}\equiv\frac{\partial^2 U}{\partial N^2}>0.$$ Recall that there are at least 2 definitions of Legendre transform, cf. e.g. this related Phys.SE post.
Mathematically, the issue can in principle be solved by using the Legendre-Fenchel transform $$\sup_{N\in\mathbb{R}_+}(\mu N-U(N))~=~\left\{ \begin{array}{rcl} \mu (U^{\prime})^{-1}(\mu) -U((U^{\prime})^{-1}(\mu)) &{\rm for}& \mu<0,\cr 0&{\rm for}& \mu=0, \cr \infty &{\rm for}& \mu>0,\end{array} \right.$$ instead if one allows the Legendre transform to take the value $$\infty$$.
2. Explicit example: Let the Lagrangian be the convex function $$L(v)=e^v>0$$. Then the derivative $$\frac{\partial L}{\partial v}=e^v>0$$ is positive so that the equation $$p=\frac{\partial L}{\partial v}$$ has no solution for non-positive momentum $$p\leq 0$$. Nevertheless, one may show that the Legendre-Fenchel transform exists $$H(p)~:=~\sup_{v\in\mathbb{R}}(pv-L(v))~=~\left\{ \begin{array}{rcl} p(\ln p -1)&{\rm for}& p>0,\cr 0&{\rm for}& p=0, \cr \infty &{\rm for}& p<0.\end{array} \right.$$
• Thanks again for taking the time to answer this! I would have a follow up question reading this: Why is it a problem that there exists no solution for $\mu\geq 0$? Wouldn't a system that only has negative chemical potential also be totally valid? Isn't the real problem that $\frac{\partial U}{\partial N}=\mu$ is not invertible for $N(\mu)$? I guess also with the Legendre Fenchel transformation it's not possible to do a legendre transform in my example because also the toy model example has an invertible derivative. If that's the case, is there even a way to solve it? Jun 13, 2021 at 7:18 | 1,032 | 3,721 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.928677 |
https://aiuta.org/en/it-takes-a-hose-2-minutes-to-fill-a-rectangular-aquarium-8-inches-long-9-inches-wide-and-13-inches.43266.html | 1,558,710,015,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257660.45/warc/CC-MAIN-20190524144504-20190524170504-00491.warc.gz | 380,866,560 | 6,997 | Mathematics
It takes a hose 2 minutes to fill a rectangular aquarium 8 inches long, 9 inches wide, and 13 inches tall. How long will it take the same hose to fill an aquarium measuring 21 inches by 29 inches by 30 inches?
gejoja
3 years ago
To make it easier, you calculate the volume of the first aquarium.
1st aquarium:
V = L x W x H
V = 8 x 9 x 13
V = 72 x 13
V = 936 in.
Rate: 936 in./2 min.
Now that you've got the volume and rate of the first aquarium, you can find how many inches of the aquarium is filled within a minute, which is also known as the unit rate. To do that, you have to divide both the numerator and denominator by their least common multiple, which is 2. 936 divided by 2 is 468 and 2 divided by 2 is 1.
So the unit rate is 468 in./1 min. Now that you've got the unit rate, you can find out how long it'll take to fill the second aquarium up by finding its volume first.
2nd aquarium:
V = L x W x H
V = 21 x 29 x 30
V = 609 x 30
V 18,270 inches
Calculations:
Now, you divide 18,270 by 468 to find how many minutes it will take to fill up the second aquarium. 18,270 divided by 468 is about 39 (the answer wasn't exact, so I said "about").
2nd aquarium's rate:
18,270 in./39 min.
As a result, it'll take about 39 minutes to fill up an aquarium measuring 21 inches by 29 inches by 30 inches using the same hose. I really hope I helped and that you understood my explanation! :) If I didn't, I'm sorry. I tried. :( | 421 | 1,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-22 | latest | en | 0.936776 |
https://discourse.julialang.org/t/how-to-compute-analytical-derivative-and-integral-of-trigonometric-functions-using-symbolic-jl-and-diffrules-jl/89480 | 1,670,298,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00312.warc.gz | 238,339,445 | 6,034 | # How to Compute Analytical Derivative and Integral of trigonometric functions using Symbolic.jl and DiffRules.jl
Hi all,
I have this equation
y = \frac{1}{e^{\frac{t}{2}}} \left[ \int_{t_{0}}^{t} e^{\frac{s}{2}} (10 + 5 \ sin(2s)) \ ds + c \right] \\
I want to ask how to solve this with Symbolics.jl and DiffRules.jl (even if the packages are derivatives, I know they can be used to check whether the integration is correct)
or are there any package that can help for the integration analytically? or both if possible…
From an old 2021 thread on this discourse I try this code (Derivative of a Symbolic Expression as a Symbol Valued Function):
using Symbolics
@variables a3, a2, a1, a0, t
f0(t) = a3*t^3 + a2*t^2 + a1*t + a0
and it does not work:
┌ Info: Precompiling Symbolics [0c5d862f-8b57-4792-8d23-62f2024744c7]
ERROR: LoadError: UndefVarError: similar_type not defined
Stacktrace:
** [1] include**
** @ ./Base.jl:418 [inlined]**
** [3] top-level scope**
** @ none:1**
** [4] eval**
** @ ./boot.jl:373 [inlined]**
** [5] eval(x::Expr)**
** @ Base.MainInclude ./client.jl:453**
** [6] top-level scope**
** @ none:1**
in expression starting at /home/browni/.julia/packages/DiffResults/YLo25/src/DiffResults.jl:1
ERROR: LoadError: Failed to precompile DiffResults [163ba53b-c6d8-5494-b064-1a9d43ac40c5] to /home/browni/.julia/compiled/v1.7/DiffResults/jl_sNyCGj.
I also try this:
using Symbolics
@variables x, y, z
f(x,y,z) = x^2 + sin(x+y) - z
Symbolics.derivative(::typeof(my_function), args::NTuple{N,Any}, ::Val{i})
still not working:
Info: Precompiling Symbolics [0c5d862f-8b57-4792-8d23-62f2024744c7]
ERROR: LoadError: UndefVarError: similar_type not defined
Stacktrace:
** [1] include**
** @ ./Base.jl:418 [inlined]**
** [3] top-level scope**
** @ none:1**
** [4] eval**
** @ ./boot.jl:373 [inlined]**
** [5] eval(x::Expr)**
** @ Base.MainInclude ./client.jl:453**
** [6] top-level scope**
** @ none:1**
in expression starting at /home/browni/.julia/packages/DiffResults/YLo25/src/DiffResults.jl:1
ERROR: LoadError: Failed to precompile DiffResults [163ba53b-c6d8-5494-b064-1a9d43ac40c5] to /home/browni/.julia/compiled/v1.7/DiffResults/jl_ixnUlq.
Stacktrace:
Thanks
This has nothing to do with your own code, you have something broken in your package installation. What version of Julia are you using? What version of Symbolics (type status at the pkg> prompt).
I am using IJulia with kernel Julia 1.7.3
This is the status of packages in my environment:
Status ~/LasthrimProjection/JupyterLab/DynamicalSystems/Project.toml
[91a5bcdd] Plots v1.35.5
[24249f21] SymPy v1.1.7
[0c5d862f] Symbolics v4.13.0
Works fine for me with Julia 1.8.2 and Symbolics 4.13.0. | 912 | 2,699 | {"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} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.588299 |
https://denizen.io/en/times-table-worksheets-printable.html | 1,713,709,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00788.warc.gz | 189,266,175 | 8,799 | # Times Table Worksheets Printable
Times table worksheets printable - Even and odd numbers chart worksheets. These multiplication charts for kids feature traditional multiplication equation facts and the common core technique of using counters to show and explain. Web this section includes math worksheets for practicing multiplication facts to from 0 to 49. I recommend making sure that your students are comfortable with the traditional 12x12 chart and the corresponding. Download our free printable worksheets today! Click here for our free online interactive times table tester, here. Timestablebug.com is for schools, tutors or personal use. All others either contain all the possible questions plus some repeats or a. Multiplying whole tens by whole tens (including missing factors) Web in this free printable times tables worksheet students have to learn and practice 3 times table multiplication.
Web the times tables are the basis of more elaborate calculations and one's general maths comprehension. This makes it easy to. 3 digits times 2 digits. Learning through a multiplication table enables faster calculation in kids and increases mental arithmetic skills. Click here for our new printable pdf multiplication wheel worksheet generator.
In the second exercise you have to enter the missing number to complete. Web printable individual multiplication tables 1 to 12. Web times tables worksheet (randomly generated) welcome to our times tables worksheets area. Best of all, many worksheets across a variety of subjects feature. Using the 9x table worksheets can help to keep youngsters interested in learning. The 49 question worksheet with no zeros and the 64 question worksheet with zeros. It is not news that a solid grasp of multiplication is essential for success in daily activities. The use of worksheets allows kids to learn independently. Web 12 times tables 12 times tables. On the first attempt, let students use their multiplication tables to find the products;
Web in this free printable times tables worksheet students have to learn and practice 3 times table multiplication. If no blue box is displayed on hover, you can manually fill or add. I recommend making sure that your students are comfortable with the traditional 12x12 chart and the corresponding. Type your text to fill the field. Practice multiplication with times tables worksheets; You can also make a. Web click on one of the worksheets to view and print the table practice worksheets, then of course you can choose another worksheet. Click here for our free online interactive times table tester, here. Web grade 4 mental multiplication worksheets. 8 and 9 times table worksheets 15 download print.
Web practise with tables worksheets. Web if times tables worksheets printable contains any form fields, the form fields are detected automatically. 3 digits times 2 digits. Tables times table printable charts math nine chart 9s worksheets grade counting multiplication 2nd sheet. Children can practise times tables with our popular free times tables web app. A great addition to practising your tables online is learning them with the assistance of worksheets. This makes it easy to. Timestablebug.com is for schools, tutors or personal use. 9, 10, 11 and 12 times tables. We also have the following two worksheet generators which enable you to put together your own worksheets and quizzes to meet your. | 641 | 3,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.841367 |
https://brilliant.org/problems/row-over-the-river/ | 1,601,063,992,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228707.44/warc/CC-MAIN-20200925182046-20200925212046-00124.warc.gz | 326,160,306 | 8,349 | # Row Across The River
Logic Level 2
Alex, Brook, Chris, and Dusty need to cross a river in a small canoe. The canoe can carry only 100 kg. Alex weighs 90 kg, Brook weighs 80 kg, Chris weighs 60 kg, Dusty weighs 40 kg, and they have 20 kg of supplies.
Find the minimum number of times they have to row across the river so that everyone and the supplies are safely across, assuming that at least one person must be in the rowboat during the trips.
× | 114 | 452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-40 | latest | en | 0.941212 |
https://www.studymode.com/essays/Final-Project-48775155.html | 1,604,056,191,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00716.warc.gz | 909,564,271 | 26,139 | # FINAL PROJECT
Topics: Series and parallel circuits, Inductor, Incandescent light bulb Pages: 3 (851 words) Published: March 14, 2014
Home Bulbs
ECET105
FINAL PROJECT ASSIGNMENT
7-Jan-2014
2013-2014
Objectives:
We are explaining in this project houses parallel connections for the lights (the bulbs). We choose this subject because parallel connection is the best for houses. We will mention the equipment and the procedure of the circuit so it could be clearer. Also the energy that comes out is a renewable energy system.
Introduction/Theory:
parallel circuit: a closed circuit in which the current divides into two or more paths before recombining to complete the circuit. Conventional house wiring systems are made up of parallel circuits—it’s the normal wiring mode for most electrical systems. The main energy supply can be the utility grid or a renewable energy system. The energy comes into a main circuit panel, where the wiring divides into multiple circuits. Each circuit leads to a group of loads, which are also wired in parallel within the circuit. Our project shows how chemical energy turns out to electrical energy. A parallel circuit has more than one resistor (anything that uses electricity to do work) and gets its name from having multiple (parallel) paths move along. Charges can move through any of several paths. If one of the items in the circuit is broken then no charge will move through the path, but other paths will continue to have charges flow through them. Parallel circuits are found in most household electric wiring. This is done so the light doesn’t stop working just because you turned off the TV for example.
Apparatus/Equipment:
Wires
Bulbs
Batteries
Switch
Procedure:
At first we started doing the home to know how where can we put the light, and how can we connect the bulb. After we finished the home we connect the circuit at the beginning on the board, but all the LED light were burned when we connected with resisters. So we change the LED with the light bulb and we use the battery (9v). We noticed that when one light was burned it was not... | 446 | 2,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-45 | longest | en | 0.935661 |
http://www.greylabyrinth.com/discussion/viewtopic.php?p=36925 | 1,369,164,142,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700477029/warc/CC-MAIN-20130516103437-00046-ip-10-60-113-184.ec2.internal.warc.gz | 509,141,914 | 9,058 | # The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.
Author Message
Ghost Post
Icarian Member
Posted: Thu Mar 01, 2001 3:48 am Post subject: 1 All you do is get 1 bar from the first, then 2, 4, 8, 16, 32, 64, etc.
Ghost Post
Icarian Member
Posted: Mon Mar 05, 2001 10:43 pm Post subject: 2 The thing is that they asked for the smallest amount of bars. So i think that 1,2,4,8,15,22 is the best combination. You don't need the last one because you can figure that out just by eliminating the rest.
Coyote
Posted: Tue Mar 06, 2001 12:19 am Post subject: 3 No, both those answers fail, as they're both higher than the current best guess, 1, 2, 4, 7,13, (24). Which you would have known if you'd bothered to read this! I'm sorry this seems a little rude, but I'm amazed at how many people go through the trouble of registering just to post a solution, yet won't go through the trouble of reading what's already been posted about the puzzle. ------------------ Gravity is a harsh mistress
Zarriar
Daedalian Member
Posted: Tue Mar 06, 2001 1:15 am Post subject: 4 touché Coyote. You haven't checked out my answer yet. I claim to be able to do it with less bars than you.
Coyote
Posted: Tue Mar 06, 2001 3:52 am Post subject: 5 Oh, a wiiise guy, eh? Geez, I hate wise guys! Actually, I did read your answer, but either I misunderstood it, or it doesn't work. Let's say you got a reading of 147. Sure, you've proven vaults 1 and 2 hold counterfeits, but how can you tell which of vaults 5, 6, or 7 holds the final set?
Zarriar
Daedalian Member
Posted: Tue Mar 06, 2001 4:21 am Post subject: 6 No you understood it, it doesn't work. It serves me right posting answers before I think about them.
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You cannot vote in polls in this forum | 749 | 2,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | latest | en | 0.933217 |
http://www.cs.utep.edu/vladik/cs2401.11/assign08.html | 1,539,679,520,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510415.29/warc/CC-MAIN-20181016072114-20181016093614-00113.warc.gz | 451,215,056 | 1,220 | ## CS 2401 Assignment #8
Due Date: Monday, April 18 or Tuesday, April 19, depending on the day of your lab.
Objective: The goal of this assignment is to practice the use of binary search trees.
Assignment: Write several recursive methods related to trees:
• a method that, given a binary search tree, returns its smallest element;
• a method that, given a binary search tree, returns its largest element;
• a method that, given a binary search tree, returns the total number of elements in the tree;
• a method that, given a binary search tree, returns its height;
• a method that, given an element and a binary search tree, checks whether this element belongs to this tree;
• a method that, given an element and a binary search tree, adds this element to the tree (no balancing is needed).
Main idea. For example, to find a given element in a given tree, we do the following. First, we check whether the tree is empty. If the tree is empty, then the given element is not there, otherwise we compare the given element with the root of the tree:
• if the element is equal to the root, we return "true";
• if the element is smaller than the root, we search in the left subtree;
• if the element is larger than the root, we search in the right subtree.
For extra credit: write similar recursive methods that compute other characteristics of the tree: e.g., total number of elements which are positive or which are even, etc. | 317 | 1,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.87452 |
https://programming.vip/docs/analysis-of-recursion-recursion-and-dynamic-programming.html | 1,718,484,819,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00360.warc.gz | 415,378,402 | 6,609 | # Concept introduction and understanding
Recurrence: deduce the following results from the previous calculation results. Mathematical language description can deduce f(n) from f(0),f(1)... f(n-1), so recurrence formula is the key.
Recursion: push forward from the back, start with the result, and push backward. The mathematical language description is that if I ask f(n), I have to find f(n-1) first. If I ask f(n-1), I have to know f(n-2) first until f(1) or f(0) is an easy result, and then push back to calculate f(n). It can be seen that the key point is also the recursive formula.
Dynamic programming: it is a branch of operations research, which decomposes a complex problem into multiple simple subproblems and divides them in the same way until it can be solved in the simplest way.
For the time being, I will not give a lecture in theory. From several practical problems, I will see how I use recursion or recursion to solve them.
# Total permutation problem
Problem Description: given A sequence, such as (A, B, C...), output the number of fully arranged groups and sequences of this sequence.
Problem analysis:
Assuming that there are N characters that need to be arranged in full, the simplest idea is to consider first selecting one from N, then selecting one from the remaining N-1, and so on until the last one, so the quantity is N * (N-1) *... * 1, that is, N!; This is a mathematical method. Is it feasible to let the computer follow this idea? If it is three characters, it can be realized as follows:
```a=[0,1,2]#Consider a, B and C as 0,1,2
s=[]
for i in a:
temp1=a[:]
temp1.remove(i)
for j in temp1:
temp2=temp1[:]
temp2.remove(j)
el=[i,j,temp2[0]]#Group the three selected elements together
s.append(el)
print(s,len(s))
```
The idea is very simple, that is, one element is removed from each layer, and then the selected elements are combined at last; But if I follow this idea, if I have a lot of elements to arrange, there will be a lot of cycle layers. Is there any way to reduce the number of cycle layers, at least write it down a little less? The answer is OK, because we see that the processing process in each layer of the cycle is not much different. In fact, we can extract this process by recursion, The following code is implemented as follows:
```a=[0,1,2]
s=[]
def swap(arr,p):
#arr, array to be arranged; p. Sequence of selected elements
if arr:
for i in arr:
temp=arr[:]
temp.remove(i)
p1=p[:]
p1.append(i)
swap(temp,p1)#temp is the array after element i is deleted, and p1 is the sequence with element i selected
else:
s.append(p)#arr is empty, indicating that the element has been selected
return
swap(a,[])
print(s,len(s))
```
The problem comes again. If the number of full rows is relatively large, the recursive stack may overflow, but it is not realized by recursion. After exploration, let's start with a simple case and push forward:
1. Suppose there is one element. Obviously, there is only one arrangement A
2. If there are two elements, there are two kinds of AB and BA. The mathematical representation set is {AB,BA}, and the record S(2)={AB,BA}
3. There are 6 of the three elements, ABC, ACB,BAC,BCA,CAB,CBA, S(3)={ABC,ACB,BAC,BCA,CAB,CBA}
Observe that in S(2), there are three methods for AB element plus C: cab, ACB and ABC; According to this idea, we can push that for each element of S(n-1), when we add a new element X, there is an insertion mode in n. Therefore, according to this method, we can write code recursively as follows:
```# -*- coding=utf-8 -*-
'''
For implementation sequence N The full array and number of the full array are output
'''
N=3
x=[i for i in range(N)]
dps=[[] for i in range(N+1)]#A sequence used to record a fully arranged sequence from 0 to N
for i in range(1,N+1):
if i==1:
dps[i]=[[0]]#Arrangement of one element
if i>1:
for arr in dps[i-1]:
for j in range(len(arr)+1):
temp=arr[:]
temp.insert(j,x[i-1])#There can be n-1 positions to insert
dps[i].append(temp)#The newly formed sequence is included in dps
print(dps[N],len(dps[N]))
```
# Indefinite equation solving problem
Problem Description: given an equation x1+x2+x3 +... + xn=N, where x1,x2... Xn are positive integers, solve all solution sets of the equation.
Problem analysis:
Let's first consider the case of three unknowns, x1+x2+x3=N, obviously n > = 3. When N=3, there is only one set of solutions x1=1,x2=1,x3=1, written as S(3)={[1,1,1]}; When N=4, it is obvious that an unknown number can be added with 1 on the basis of N=3. There are three ways of adding, which are x1,x2 and X3 respectively. Therefore, S(4)={[2,1,1],[1,2,1],[1,1,2]}, and so on. Therefore, we can realize it in a recursive way. The solution set of S(n) can be realized on the solution set of S(n-1). Suppose that a group of solutions of S(n-1) are x1,x2... x(n-1), A set of solutions of S(n) must be an element plus 1 based on the above solution. According to this idea, we can implement it in code, as follows:
```# -*- coding=utf-8 -*-
'''
Given equation sequence, given sum N，Solve the number of equations and enumerate them
'''
N=4#Number of unknowns
M=6#The sum of unknowns, N in the equation
dps=[[] for i in range(M+1)]#Store each and the corresponding solution set
def solv(n):
for i in range(1,n+1):
if i==N:
dps[i]=[[1 for x in range(N)]]
if i>N:
for arr in dps[i-1]:
for j,el in enumerate(arr):
temp=arr[:]
temp[j]=arr[j]+1#Element + 1 of each solution of traversal i-1
solv(M)
print(dps[M],len(dps[M]))
```
# Money problem
Problem Description: there are currently given amounts, such as 1 yuan, 3 yuan and 11 yuan. How many to the right of each face value? Given an amount, find the minimum quantity to get the quantity of the amount, and give the distribution scheme.
Problem analysis:
Assuming that the given amount is n and the minimum number of coins is f(n), if there is one less, it may be 1 yuan, 3 yuan or 11 yuan. Therefore, f(n) is the minimum value of F (n-1), f (n-3) and f (N-11) plus 1. The code is as follows:
```# -*- coding=utf-8 -*-
'''
There are several RMB with a given amount. Given an amount, the quantity of the amount can be rounded up by finding the minimum quantity,The distribution scheme is given.
'''
coin=[1,3,11]
N=10#Given amount
dp=[0 for i in range(N+5)]#Record the minimum quantity of each amount
dp[0]=0
s=[0 for i in range(N+1)]#Record which face value is selected each time the amount increases
f=[0,0,0]#Used to record the amount of each denomination
def coin_solve():
for i in range(1,N+1):
num=[1000,1000,1000]
if i>=1:
num[0]=dp[i-1]#The previous one had a face value of 1
if i>=3:
num[1]=dp[i-3]#The previous one had a face value of 3
if i>=11:
num[2]=dp[i-11]#The previous one had a face value of 11
minNum=1000
index=0
#Find the minimum value in num
for j,el in enumerate(num):
if el<minNum:
minNum=el
index=j
dp[i]=minNum+1
s[i]=index#Record the face value of the selection
coin_solve()
i=N
count=0
#Push forward from the back in the s sequence. If the last one is 1, the face value is N-1. Find the face value in s(N-1) in the s sequence, and so on
while True:
f[s[i]]=f[s[i]]+1
i=i-coin[s[i]]
count=+1
if i<=0:
break
print(f"The minimum quantity is:{dp[N]},The quantity of each currency is:{f}")
```
# Two dimensional array shortest path problem
Problem Description: given a two-dimensional array (n rows, m columns), with (0,0) as the starting point and (n-1,m-1) as the end point, the movement can only be right or down, and find the shortest path value and walking method between them.
Problem analysis:
For the two-dimensional array shown in the figure, assuming that the shortest distance to the coordinate (i,j) is f(i,j), it is obvious that there can only be two methods to get to (i,j). One is from the left (i,j-1) and the other is from (i-1,j). Only compare f (i,j-1) and f (i-1,j). The smaller value plus (i,j) is the value of f(i,j); If you look at the boundary, there is obviously only one way for the point in the first line, that is, f(0,j), which can only come from f(0,j-1); Similarly, f(i,0) in the first column can only come from f(i-1,0). According to this idea, the code is as follows:
```# -*- codfing=utf-8 -*-
'''
Problem: given a two-dimensional array( n that 's ok, m Column),[0,0]As the starting point,[n-1,m-1]For the end point, find the shortest path
'''
X=3
Y=3
arr=[[1,2,1],[3,1,2],[2,1,1]]#Given two-dimensional array
dp=[[0 for i in range(X)] for j in range(Y)]#Used to record the shortest distance to each point
s=[[[0,0] for i in range(X)] for j in range(Y)]#Used to record the coordinates of the last point to each point
print(s)
def find_path():
for i in range(X):
for j in range(Y):
if i==0 and j==0:
dp[i][j]=arr[i][j]#The end is at the beginning
s[i][j]=[0,0]
if i==0 and j>0:
dp[i][j]=dp[0][j-1]+arr[i][j]#The first line
s[i][j]=[i,j-1]
if i>0 and j==0:
dp[i][j]=dp[i-1][j]+arr[i][j]#The first column
s[i][j]=[i-1,j]
if i>0 and j>0:#Intermediate point
if dp[i-1][j]<dp[i][j-1]:
dp[i][j]=dp[i-1][j]+arr[i][j]
s[i][j]=[i-1,j]
else:
dp[i][j]=dp[i][j-1]+arr[i][j]
s[i][j]=[i,j-1]
find_path()
N=3
x=N-1
y=N-1
line=[]
#Push back from the end
while True:
point=[x,y]
line.insert(0,point)#Add from the beginning every time
prePoint=s[point[0]][point[1]]#Previous point coordinates
x=prePoint[0]
y=prePoint[1]
if x==0 and y==0:#Terminate at the origin
print([0,0])
break
print(f"The shortest path value is:{dp[N-1][N-1]}，The path is:{line}")
```
Keywords: Python Algorithm Dynamic Programming
Added by kol090 on Tue, 08 Feb 2022 21:32:00 +0200 | 2,810 | 9,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.912627 |
https://library.fiveable.me/introductory-probability-and-statistics-for-business/unit-4 | 1,726,382,673,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00775.warc.gz | 324,857,926 | 139,128 | # 📈Intro to Probability for Business Unit 4 – Discrete Probability Distributions
Discrete probability distributions are essential tools in business statistics, assigning probabilities to discrete random variables. They help model various scenarios, from quality control to inventory management, providing a framework for quantifying uncertainty and making informed decisions. Key distributions include Bernoulli, binomial, Poisson, geometric, hypergeometric, and negative binomial. Understanding their properties, probability mass functions, expected values, and variances enables businesses to analyze data, assess risks, and optimize operations effectively.
## Key Concepts and Definitions
• Discrete probability distributions assign probabilities to discrete random variables
• Random variables are variables whose values are determined by the outcomes of a random experiment
• Discrete random variables can only take on a finite or countably infinite number of distinct values
• Probability is a measure of the likelihood of an event occurring, expressed as a number between 0 and 1
• 0 indicates an impossible event, while 1 indicates a certain event
• The sum of probabilities for all possible outcomes in a discrete probability distribution equals 1
• Support of a discrete random variable is the set of all possible values it can take
• Probability distribution functions (PDF) describe the probability of a discrete random variable taking on a specific value
## Types of Discrete Probability Distributions
• Bernoulli distribution models a single trial with two possible outcomes (success or failure)
• Characterized by a single parameter $p$, representing the probability of success
• Binomial distribution models the number of successes in a fixed number of independent Bernoulli trials
• Characterized by parameters $n$ (number of trials) and $p$ (probability of success in each trial)
• Poisson distribution models the number of events occurring in a fixed interval of time or space
• Characterized by a single parameter $\lambda$, representing the average number of events per interval
• Geometric distribution models the number of trials until the first success in a series of independent Bernoulli trials
• Characterized by a single parameter $p$, representing the probability of success in each trial
• Hypergeometric distribution models the number of successes in a fixed number of draws from a finite population without replacement
• Characterized by parameters $N$ (population size), $K$ (number of successes in the population), and $n$ (number of draws)
• Negative binomial distribution models the number of failures before a specified number of successes in a series of independent Bernoulli trials
• Characterized by parameters $r$ (number of successes) and $p$ (probability of success in each trial)
## Probability Mass Functions (PMF)
• A probability mass function (PMF) is a function that gives the probability of a discrete random variable taking on a specific value
• For a discrete random variable $X$, the PMF is denoted as $P(X = x)$, where $x$ is a possible value of $X$
• The PMF satisfies two conditions:
1. $P(X = x) \geq 0$ for all $x$ in the support of $X$
2. $\sum_{x} P(X = x) = 1$, where the sum is taken over all possible values of $X$
• The PMF can be used to calculate probabilities of events involving the discrete random variable
• $P(a \leq X \leq b) = \sum_{x=a}^{b} P(X = x)$
• The cumulative distribution function (CDF) of a discrete random variable is the sum of its PMF up to a given point
• $F(x) = P(X \leq x) = \sum_{t \leq x} P(X = t)$
## Expected Value and Variance
• The expected value (or mean) of a discrete random variable $X$ is a weighted average of its possible values, weighted by their probabilities
• $E(X) = \sum_{x} x \cdot P(X = x)$
• The expected value represents the long-run average value of the random variable over many trials
• The variance of a discrete random variable $X$ measures the average squared deviation from its expected value
• $Var(X) = E((X - E(X))^2) = \sum_{x} (x - E(X))^2 \cdot P(X = x)$
• The standard deviation is the square root of the variance and measures the average deviation from the mean
• $SD(X) = \sqrt{Var(X)}$
• Linearity of expectation: For any two random variables $X$ and $Y$, $E(X + Y) = E(X) + E(Y)$
• Properties of variance:
• $Var(aX + b) = a^2 Var(X)$ for constants $a$ and $b$
• $Var(X + Y) = Var(X) + Var(Y)$ if $X$ and $Y$ are independent
## Common Discrete Distributions in Business
• Binomial distribution is used to model the number of successes in a fixed number of trials (e.g., number of defective items in a batch)
• Poisson distribution is used to model the number of rare events occurring in a fixed interval (e.g., number of customer arrivals per hour)
• Geometric distribution is used to model the number of trials until the first success (e.g., number of sales calls until a sale is made)
• Hypergeometric distribution is used to model the number of successes in a fixed number of draws from a finite population without replacement (e.g., number of defective items in a sample drawn from a batch)
• Negative binomial distribution is used to model the number of failures before a specified number of successes (e.g., number of unsuccessful product launches before a successful one)
• Inventory management: Poisson distribution can model the demand for a product during a fixed time period
• Quality control: Binomial and hypergeometric distributions can model the number of defective items in a batch or sample
• Marketing: Geometric distribution can model the number of sales calls required to make a sale
• Project management: Negative binomial distribution can model the number of unsuccessful projects before a successful one
• Risk assessment: Discrete probability distributions can help quantify the likelihood and impact of various risks
• Decision-making: Expected values and variances can guide decisions by comparing the long-run average outcomes and variability of different options
## Calculation Techniques and Examples
• To calculate probabilities using a PMF, substitute the given values into the formula and simplify
• Example: For a binomial distribution with $n=5$ and $p=0.3$, find $P(X=2)$
• $P(X=2) = \binom{5}{2} (0.3)^2 (0.7)^3 \approx 0.309$
• To calculate expected values, multiply each possible value by its probability and sum the results
• Example: For a discrete random variable $X$ with PMF $P(X=1)=0.4$, $P(X=2)=0.3$, and $P(X=3)=0.3$, find $E(X)$
• $E(X) = 1 \cdot 0.4 + 2 \cdot 0.3 + 3 \cdot 0.3 = 1.9$
• To calculate variances, subtract the expected value from each possible value, square the differences, multiply by the probabilities, and sum the results
• Example: For the same random variable $X$ as above, find $Var(X)$
• $Var(X) = (1-1.9)^2 \cdot 0.4 + (2-1.9)^2 \cdot 0.3 + (3-1.9)^2 \cdot 0.3 \approx 0.69$
## Key Takeaways and Review
• Discrete probability distributions assign probabilities to discrete random variables, which can only take on a finite or countably infinite number of distinct values
• The sum of probabilities for all possible outcomes in a discrete probability distribution equals 1
• Common discrete probability distributions include Bernoulli, binomial, Poisson, geometric, hypergeometric, and negative binomial distributions
• Probability mass functions (PMF) give the probability of a discrete random variable taking on a specific value and satisfy two conditions: non-negativity and summing to 1
• The expected value is a weighted average of the possible values, while variance measures the average squared deviation from the expected value
• Discrete probability distributions have various applications in business, such as inventory management, quality control, marketing, project management, risk assessment, and decision-making
• To calculate probabilities, expected values, and variances, substitute given values into the appropriate formulas and simplify
• Understanding discrete probability distributions is crucial for making informed decisions and quantifying uncertainty in business contexts | 1,845 | 8,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 48, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.832797 |
http://lambdapage.org/27762/letter-oo-worksheets/letter-o-tracing-worksheets-allfreeprintable-regarding-letter-oo-worksheets/ | 1,618,756,454,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00048.warc.gz | 56,422,990 | 7,197 | # letter o tracing worksheets allfreeprintable regarding letter oo worksheets
##### letter o tracing worksheets allfreeprintable regarding letter oo worksheets.
Learning to write the letter is the first skill that should be done for kids. This worksheet has a simple instruction for tracing the letter. You can make a demonstration to the kid before instruct them to do it by themselves. Remember, this is need a good patience.
The worksheet contains of several letter with the word. This is a little bit challenging because you also need to teach about spelling the word. However, this activity can be fun and give a satisfying result if you make a proper learning process.
If they done with their drawing, you also can ask them to color their drawing. Remember, you can’t expect the color or drawing in a really good picture. Usually, kids in preschool age draw some random objects.
You also can remind them to spell the word while tracing the letter. By doing this activity, kids will get additional skill of spelling the words. This can be a fun activity.
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### Education Com Worksheets
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The next worksheet for playgroup that can be used is the shape tracing worksheets. It’s also has a nice function for stimulating the soft motoric skill in children.
It helps to understand that the number they will write has a sequence. If your kid start recognizing the number, then you can start to write the number. The simplest number is 1.
Similar with the alphabet tracing, the alphabet coloring also can be fun activity to do. However, the main purpose of this exercise is to introducing the alphabet.
Introducing the writing method for early kids is actually recommended. It can be done by using the worksheet for kindergarten that specific for the writing progress.
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### Precursive Handwriting Worksheets
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This will help them to memorize about the shape of the letter. Improve their skill by continuing the exercise. As the final evaluation, you can make all the alphabet sets and ask them to color the letter that you want.
If you think that your kids coloring process is not really tidy, then you can repeat the same exercise using the different pictures. It will help them to improve their coloring skill.
This can be very fun and make your kid enjoy to get more exercise. This learning progress, of course is recommended for upgrading kids’ soft motoric skills.
Start by teaching your kid about the word that they want to practice. You also can teach them to spell the word. After that, make an example on how to trace the word.
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March 9, 2021 | 684 | 2,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.947851 |
https://forum.objectivismonline.com/index.php?/topic/2878-philosophy-riddle/&tab=comments | 1,619,038,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00214.warc.gz | 371,510,229 | 39,846 | Jump to content
Objectivism Online Forum
# Philosophy Riddle
Rate this topic
## Recommended Posts
Was in my Philosophy of Science class (good class on Logical Positism and Hume) and my prof had a good riddle to open class tonight. I'll try and post as accuratly as possible (didn't write it down).
There's a dollar making machine that can make unlimited amount of dollars at an ever faster rate.
At t=0, the man comes up to you and gives you two options:
a/ I'll give you 10 dollar bills with serial numbers 1-9, but you must burn #1
b/ I'll give you only 1 dollar bill
At t=1/2 he gives you the same option
Again at 3/4, 7/8, 15/16, 31/32, 63/64.. and so on. The machine only works faster and faster to churn out the bills at a higher rate of speed.
He didn't give an answer to the riddle, but gave a couple arguments. I was wondering how Objectivists would answer the question.
I'll put my feet in and say that since there is no infinity, you'd choose A. Peikoff writes in OPAR that infinity means "larger than any specific quantity" and since A=A, everything is finite (page 31). Is it an unanswerable riddle to an Objectivist then? BTW, he has said that although he doesn't personally consider himself a logical positivist/empiricist he does lean that way (if that helps).
It's not a quiz or anything, just something to chew over in our heads over the weekend.
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I don't really understand the riddle. From what you've written, I don't understand what role the money making machine plays.
All I see is that a man offers to give you two different pay outs each time period. One payout equals \$8 and one equals \$1. Since it's written as a perpetuity (an annuity that gives pay outs forever), you'd take the present value of each perpetuity ( = PMT/(going interest rate). But, the one that pays \$8 will obviously have a greater present value than the \$1.
But again, I think I'm misunderstanding the problem. What is the deal with 't' - what is the defined period? What is the deal with the serial numbers?
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I would collect gold and then refuse to accept dollar bills in payment for any of my goods or services.
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All I see is that a man offers to give you two different pay outs each time period. One payout equals \$8 and one equals \$1. Since it's written as a perpetuity (an annuity that gives pay outs forever), you'd take the present value of each perpetuity ( = PMT/(going interest rate). But, the one that pays \$8 will obviously have a greater present value than the \$1.
That's correct in some sense. Remember, you have to "burn" the lowest value dollar every time with option #1.
But again, I think I'm misunderstanding the problem. What is the deal with 't' - what is the defined period? What is the deal with the serial numbers?
"T' merely defines time as a variable.
The serial numbers identify the bill so the lowest number can be burnt.
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That's correct in some sense. Remember, you have to "burn" the lowest value dollar every time with option #1.
"T' merely defines time as a variable.
The serial numbers identify the bill so the lowest number can be burnt.
I still don't understand the riddle. Perhaps you could give an example of the first few iterations?
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I think the idea is that if you pick option 1) then every dollar bill gets burnt? Imagine the first 10 bills were numbered 1 through 10, the next lot numbered 11 through 20 and so on. Youd initially have to burn the dollar bill marked 1, then after you got given the next lot you'd have to burn 2, then 3, and so on. If you imagined that the 'giving/burning' thing was carried out infinite times t, then every dollar bill would eventually be burned and you'd have nothing left (the nth bill would be burnt after n*10 iterations).
I took a maths class once on analysis since it seemed interesting, and we done something quite similar to this - its basically a question about summing infinite series. In maths, an infinite series is defined to be a sum of infinitely many numbers, eg 1+1+1+1+1+..., or whatever. Even though you are adding infinitely many terms, some series eventually converge to a finite value (eg 0.9 + 0.09 + 0.009 + ... converges to 1, and 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 (ie 1/n_squared) ... converges to pi squared divided by 6). However a series such as 1+1+1+1+1... will never converge to a finite value since it keeps on getting bigger, nor will the series (1+1/2+1/3+1/4+...).
Now, some series can have negative terms as well as positive terms, such as the series 1 + (-2) + 3 + (-4) + ... If you replace all the negative terms with positive terms (in this case it would be 1+2+3+4+...) and the series converges, then it is said to be absolutely convergent. The important part is that if a series is absolutely convergent, then it doesnt matter in what order you sum the terms - it will always converge to the same value. However if the series isnt absolutely convergent then you cant do this - you will get different answers depending upon how you carry out the infinite sums - in fact you can prove that you can make the series sum up to any value whatsoever. Therefore a non-absolutely convergent series with infinitely many negative terms has no limiting value - you can prove that you can get any value
Your teacher is basically asking you to sum the series 10 - 1 + 10 - 1 + 10 - 1 + ... Since this series isnt absolutely convergent, this is impossible, and you can show that the sum can be equal to anything, depending on how you add the terms. Your teacher is basically arranging the terms in a way that the sum is equal to zero in the limit. It could just as well come out to be 27 or 2343, if they were arranged in a different way.
For a clearer example of what I mean, consider doing the following sum:
1 - 1 + 1 - 1 + 1 - 1 + ...
This series isnt absolutely convergent, so you can get different answers depending on how you do the sum. For instance:
(1 - 1) + (1 - 1) + (1 - 1) + ... is going to be equal to zero, whereas 1 - (1 - 1) - (1 - 1) - (1 - 1) - ..., which is the exact same sum, is going to be equal to 1. This kind of thing caused difficulties for mathematicians before the theory of series was properly formalized, and led to people putting forward 'proofs' that 0 = 1 and so on.
Edited by Hal
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(1 - 1) + (1 - 1) + (1 - 1) + ... is going to be equal to zero, whereas 1 - (1 - 1) - (1 - 1) - (1 - 1) - ..., which is the exact same sum, is going to be equal to 1.
But in the first case you have 6 numbers and in the second you have 7. You've stopped the two series at different times--that's why your claimed sum differs in each case. In other words, all you've said is that:
1 - 1 + 1 - 1 = 0
whereas
1 - 1 + 1 = 1
This isn't at all unusual. Obviously, if you keep going, you will have a sum of 0 if there are an even number of terms and a sum of 1 if there are an odd number, no matter how long a series one takes.
----
I still don't get the original riddle. It seems like in (A) one gets 9 additional dollars bills with each iteration, as opposed to 1 additional dollar bill with each iteration in (. Not a tough choice, if that's all that's going on.
Edited: to change a smiley face into the letter B.
Edited by danielshrugged
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But in the first case you have 6 numbers and in the second you have 7. You've stopped the two series at different times--that's why your claimed sum differs in each case. In other words, all you've said is that:
1 - 1 + 1 - 1 = 0
whereas
1 - 1 + 1 = 1
This isn't at all unusual. Obviously, if you keep going, you will have a sum of 0 if there are an even number of terms and a sum of 1 if there are an odd number, no matter how long a series one takes.
What I think your excluding here is that the series is infinate. There is no such thing as odd/versus even amount of numbers in infinity. There is no stopping of the series.
As far as the original riddle, I have no idea, I've never had a formal education in series and I haven't read about them in a while. Though, you may want to read up on Zeno's paradox to understand how a series works (and how they can be used fallaciously), as I recall, that helped me.
~Amanda
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What I think your excluding here is that the series is infinate. There is no such thing as odd/versus even amount of numbers in infinity. There is no stopping of the series.
~Amanda
I didn't claim there was. I merely said that however long the string of numbers, you can determine the sum by whether there is an even or odd amount. As the amount of numbers approaches infinity, therefore, it approaches NEITHER 0 nor 1. Neither of those is the sum of the series at infinity.
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I didn't claim there was. I merely said that however long the string of numbers, you can determine the sum by whether there is an even or odd amount. As the amount of numbers approaches infinity, therefore, it approaches NEITHER 0 nor 1. Neither of those is the sum of the series at infinity.
But it does approach either 0 or 1, depending upon how you are adding the terms. Let's say we take the terms in the above series and rearrange them to give: 1 + 1 + 1 + 1 + 1 + (1-1) + (1-1) + ... which appears to sum to 3 (the fact that some 1's are being 'borrowed' from the 'end' of the series doesnt matter - you have infinite 1's infinite -1's, so you arent going to run out of either "infinity minus infinity equals infinity" (scare quotes intended)). Since the series doesnt converge absolutely, the order in which you add the terms will determine the sum in the limit.
More formally:
Let S be the sum of the series (1 - 1) + (1 - 1) + ..., ie
S = (1 - 1) + (1 - 1) + ... = 0
but then S = 1 - { (1 - 1) + (1 - 1) + (1 - 1) + ...} = 1 - S = 1 - 0 = 0
Hence 1 = 0.
here is a slightly more indepth explanation.
Edited by Hal
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[url=http://courses.ncssm.edu/math/TCMConf/TCM2002/talks2002/Alt%20Series.pdf)</div><div class='quotemain'><!--QuoteEBegin-->hereis a slightly more indepth explanation.
I'll say this much: whatever concept of limit being used here, it does not seem to be Newton's, and I find it hard to see how any of that math applies to the real world. That said, I have not considered whether there might be some mathematical use for this.
Edited to fix quote.
Edited by danielshrugged
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If I understand it correctly, the riddle is asking the question, "What is more: 8 x Infinity or 1 x Infinity?"
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If I understand it correctly, the riddle is asking the question, "What is more: 8 x Infinity or 1 x Infinity?"
Well, what about the 1 bill that you burn. Aren't you burning infinate bills, even if you burn one. So, wouldn't it be 8 infinity - 1 infinity?
I'll know on Monday.
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Theres no such thing as "8infinity" or "1infinity" and infinity - infinity is undefined.
From the way you set the problem up it looked like T was approaching a limit of 1.
From what I can tell or how you set the problem up it just looks like our real choice is with which choice is approaching infinity quicker.
This would have to be choice b since it's growing 9 times quicker than choice a.
So if the question is: which choice will get me rich quicker it would have to be choice b.
If the question is as each choice approaches infinity which one is "bigger" (larger in number) it would always be the second choice because it grows quicker. But when your dealing with each choices infinities qua infinity that one set is not larger than the other. Infinity is Infinity.
Maybe realizing each infinity is the same "size" is this assignments philosphical purpose?
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I just wanted to point out to a few people that infinity is not a number. It is a concept. It is used to represent the fact that the number series is non-terminating. You cannot count to infinity. If you say to yourself I am going to count to infinity, and then begin: 1,2,3,4,5,6,7--you have stopped at seven. In essence infinity has become seven. Not to say that infinity=seven, but that is where you stopped. "Infinity" is the potential number to which you could have gotten and "seven" is the actual. The concept 'infinity' bears no relation to things that exist, because everything that exists is finite.
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Well, what about the 1 bill that you burn. Aren't you burning infinate bills, even if you burn one. So, wouldn't it be 8 infinity - 1 infinity?
I'll know on Monday.
I'm not the only one who has trouble understanding the problem. Well, actually, the answer is so obvious, that I assume I have misunderstood it. Working out the first few iterations from the two options:
Option 1:
Time 0: (10 - 1) = \$9
Time 1/2: 9 + (10 - 1) = \$18
Time 3/4: 18 + (10 - 1) = \$27
Option 2:
Time 0: 1 = \$1
Time 1/2: 1 + 1 = \$2
Time 3/4: 2 + 1 = \$3
You see why I guess that I do not understand the problem? The way I've understood it, even the village idiot can tell which one is better. (Can't vouch for for college philosophy professors, though . )
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You see why I guess that I do not understand the problem? The way I've understood it, even the village idiot can tell which one is better.
Try Time 1.
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Try Time 1.
How is that different?
Either, Time 1 does not happen, in which case why worry about it?
Or, more likely, I approach it and am still better off.
(I say likely, because I do not know enough of the math to tell you what a mathematician would say. All I really care about is reality.)
Edited by softwareNerd
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• 1 month later...
Either, Time 1 does not happen, in which case why worry about it?
Time 1 happens when the machine runs for a second.
At this point both options have given you an amount of money that are sums of diverging series: infinite. Infinity is not defined well enough, so it doesn't matter which do you choose, you get a undefined amount of money anyway.
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Time 1 happens when the machine runs for a second.
And therein lies the error in the problem itself. Once the machine has run for one second, it cannot have made an infinite number of bills. Machines operating in this universe have a finite capacity for production, especially over a short time period.
The whole problem is arbitrary and of no concern to anyone with a head for reality and for solving problems which actually exist in this universe. This one doesn't.
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... Infinity is not defined well enough, so it doesn't matter which do you choose, you get a undefined amount of money anyway.
Since you seem to understand the question, let me ask this: at any particular point that is less than 1 second, which option is better? Also, at any point after one second, which option is better?
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And therein lies the error in the problem itself. Once the machine has run for one second, it cannot have made an infinite number of bills. Machines operating in this universe have a finite capacity for production, especially over a short time period.
That is true. There are (at least) two impossible paths the events can go: we have infinite amount of money, or time stops at 1.
The whole problem is arbitrary and of no concern to anyone with a head for reality and for solving problems which actually exist in this universe. This one doesn't.
It has a concern to anyone with a head for reality who chooses so, and the solution these look for is that the problem is arbitrary.
Since you seem to understand the question, let me ask this: at any particular point that is less than 1 second, which option is better? Also, at any point after one second, which option is better?
I really don't know the answer to either. If a series of events has an impossible event (when t=1), do the other events really happen either?
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Since you seem to understand the question, let me ask this: at any particular point that is less than 1 second, which option is better?
At t=0, option a is better because it leaves you with \$8, while option b would yield you \$1. At t=1/2, option a would increase your earnings by \$8, while option b by only \$1; the same is true for any t=(n-1)/n.
So,
• for n=1 (that is, t=0), option a is better;
• if option a is better for any n, it is better for n+1 as well;
• therefore, option a is better for any n >= 1.
A nice, if pointless, example of induction.
Also, at any point after one second, which option is better?
That is a theology question.
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• Create New... | 4,397 | 17,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.967265 |
http://www.mathworks.com/help/finance/ecmnfish.html?requestedDomain=www.mathworks.com&nocookie=true | 1,519,361,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814393.4/warc/CC-MAIN-20180223035527-20180223055527-00469.warc.gz | 500,292,190 | 14,125 | # Documentation
### This is machine translation
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# ecmnfish
Fisher information matrix
## Syntax
```Fisher = ecmnfish(Data,Covariance,InvCovariance,MatrixFormat)
```
## Arguments
`Data` `NUMSAMPLES`-by-`NUMSERIES` matrix of observed multivariate normal data `Covariance` `NUMSERIES`-by-`NUMSERIES` matrix with covariance estimate of `Data` `InvCovariance` (Optional) Inverse of covariance matrix: `inv(Covariance)` `MatrixFormat` (Optional) Character vector that identifies parameters included in the Fisher information matrix. If `MatrixFormat` = `[]` or `''`, the default method `full` is used. The parameter choices are `full` — (Default) Compute full Fisher information matrix.`meanonly` — Compute only components of the Fisher information matrix associated with the mean.
## Description
`Fisher = ecmnfish(Data,Covariance,InvCovariance,MatrixFormat)` computes a `NUMPARAMS`-by-`NUMPARAMS` Fisher information matrix based on current parameter estimates, where
```NUMPARAMS = NUMSERIES*(NUMSERIES + 3)/2 ```
if `MatrixFormat = 'full'` and
```NUMPARAMS = NUMSERIES ```
if `MatrixFormat = 'meanonly'`.
The data matrix has `NaNs` for missing observations. The multivariate normal model has
```NUMPARAMS = NUMSERIES + NUMSERIES*(NUMSERIES + 1)/2 ```
distinct parameters. Therefore, the full Fisher information matrix is of size `NUMPARAMS`-by-`NUMPARAMS`. The first `NUMSERIES` parameters are estimates for the mean of the data in `Mean` and the remaining `NUMSERIES*(NUMSERIES + 1)/2 `parameters are estimates for the lower-triangular portion of the covariance of the data in `Covariance`, in row-major order.
If `MatrixFormat = 'meanonly'`, the number of parameters is reduced to `NUMPARAMS = NUMSERIES`, where the Fisher information matrix is computed for the mean parameters only. In this format, the routine executes fastest.
This routine expects the inverse of the covariance matrix as an input. If you do not pass in the inverse, the routine computes it. You can obtain an approximation for the lower-bound standard errors of estimation of the parameters from
```Stderr = (1.0/sqrt(NumSamples)) .* sqrt(diag(inv(Fisher))); ```
Because of missing information, these standard errors can be smaller than the estimated standard errors derived from the expected Hessian matrix. To see the difference, compare to standard errors calculated with `ecmnhess`. | 614 | 2,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-09 | latest | en | 0.626888 |
http://dukespace.lib.duke.edu/dspace/handle/10161/3370?show=full | 1,371,699,943,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368710196013/warc/CC-MAIN-20130516131636-00032-ip-10-60-113-184.ec2.internal.warc.gz | 79,801,581 | 6,619 | DukeSpace
# A mathematical theory of stochastic microlensing. II. Random images, shear, and the Kac-Rice formula
## DukeSpace
dc.contributor.author Petters, Arlie en_US dc.contributor.author Teguia, A. M. en_US dc.date.accessioned 2011-04-15T16:46:39Z dc.date.available 2011-04-15T16:46:39Z dc.date.issued 2009 en_US dc.identifier.citation Petters,A. O.;Rider,B.;Teguia,A. M.. 2009. A mathematical theory of stochastic microlensing. II. Random images, shear, and the Kac-Rice formula. Journal of Mathematical Physics 50(12): 122501-122501. en_US dc.identifier.issn 0022-2488 en_US dc.identifier.uri http://hdl.handle.net/10161/3370 dc.description.abstract Continuing our development of a mathematical theory of stochastic microlensing, we study the random shear and expected number of random lensed images of different types. In particular, we characterize the first three leading terms in the asymptotic expression of the joint probability density function (pdf) of the random shear tensor due to point masses in the limit of an infinite number of stars. Up to this order, the pdf depends on the magnitude of the shear tensor, the optical depth, and the mean number of stars through a combination of radial position and the star's mass. As a consequence, the pdf's of the shear components are seen to converge, in the limit of an infinite number of stars, to shifted Cauchy distributions, which shows that the shear components have heavy tails in that limit. The asymptotic pdf of the shear magnitude in the limit of an infinite number of stars is also presented. All the results on the random microlensing shear are given for a general point in the lens plane. Extending to the general random distributions (not necessarily uniform) of the lenses, we employ the Kac-Rice formula and Morse theory to deduce general formulas for the expected total number of images and the expected number of saddle images. We further generalize these results by considering random sources defined on a countable compact covering of the light source plane. This is done to introduce the notion of global expected number of positive parity images due to a general lensing map. Applying the result to microlensing, we calculate the asymptotic global expected number of minimum images in the limit of an infinite number of stars, where the stars are uniformly distributed. This global expectation is bounded, while the global expected number of images and the global expected number of saddle images diverge as the order of the number of stars. en_US dc.language.iso en_US en_US dc.publisher AMER INST PHYSICS en_US dc.relation.isversionof doi:10.1063/1.3267859 en_US dc.subject functional analysis en_US dc.subject gravitational lenses en_US dc.subject morse potential en_US dc.subject probability en_US dc.subject random processes en_US dc.subject star clusters en_US dc.subject stochastic processes en_US dc.subject random complex zeros en_US dc.subject random polynomials en_US dc.subject number en_US dc.subject caustics en_US dc.subject physics, mathematical en_US dc.title A mathematical theory of stochastic microlensing. II. Random images, shear, and the Kac-Rice formula en_US dc.type Article en_US dc.description.version Version of Record en_US duke.date.pubdate 2009-12-0 en_US duke.description.endpage 122501 en_US duke.description.issue 12 en_US duke.description.startpage 122501 en_US duke.description.volume 50 en_US dc.relation.journal Journal of Mathematical Physics en_US | 770 | 3,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.777764 |
https://www.jiskha.com/questions/78982/a-sphere-of-radius-a-carries-a-volume-charge-density-p-p-r-a-2-find-the-electric-field | 1,623,619,310,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610841.7/warc/CC-MAIN-20210613192529-20210613222529-00306.warc.gz | 783,740,271 | 6,073 | # physics
A sphere of radius a carries a volume charge density p=p (r/a)^2. Find the electric field inside and outside the sphere.
I don't know what equation to use. Do I use electirc field due to point charge?
1. 👍
2. 👎
3. 👁
1. Inside the sphere, any imaginary volume you think of will have no charge inside it. Therefore no Electric Field is coming out of it. Thus there is no electric field within the sphere.
The instant you get outside the sphere, the charge looks like it is all at the center, and it is the usual 9*10^9 q/r^2 kind of field.
1. 👍
2. 👎
2. so the electic field outside the sphere is 9*10^9 p (r/a)^2/a^2?
1. 👍
2. 👎
3. Whoops, sorry, was thinking charge on surface of sphere.
If uniformly distributed throughout the volume, then my answer is the same outside the sphere
However inside the sphere you will surround chagres the will look as if they are at the center
The amount you surround will be the volume at radius r times the charge density.
V = (4/3) pi r^3
q =V (charge density)= (4/3) pi r^3 (p) (r^2/a^2)
which is
(4/2) p *pi ^5/a^2
that times is q
E = 9*10^9 q/r^2
= 9 * 10^9 (4/3)p* pi r^3/a^2
when you reach r = a
you get
E = 9*10^9 (4/3) p* pi r^3 /a^2
which is of course
9*10^9 ( q total)/r^2 from then on
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2. 👎
4. could i get some help damon
1. 👍
2. 👎
5. (4/3) p * pi r^5 /a^2
1. 👍
2. 👎
6. The idea is that inside the sphere, the only charge that counts is what is inside the radius you are at. That looks like it is at the center and is the charge density times the volume inside the radius you are at.
Outside the sphere, the entire charge, the density times the entire volume of the sphere (4/3) pr a^3 * density is inside your radius and no longer increases/
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2. 👎
7. The answers Dan H gave you looked reasonable to me Rory.
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2. 👎
8. yep i noticed that =P
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2. 👎
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### A look at the second Feynman cipher - fractionated morse [part 4]
The Feynman ciphers are a set of 3 ciphers given to Richard Feynman, the first of which has been solved, but the second 2 remain unsolved. This is part 4 of a discussion about the second cipher.
I have talked about this cipher three times previously in Part 1, Part 2 and Part 3. In part 1 I was playing around with vigenere type ciphers, in part 2 I decided to step back a little bit a figure out what the ciphertext is not i.e. identify the cipher algorithms it can't be. Part 3 looked at the 3x3 Hill cipher and some of its variants.
So far I have been working my way through the cipher types I think it may be. In part 2 I identified fractionated morse as a candidate cipher. It turns out breaking fractionated morse is not too difficult, the procedure is similar to breaking standard substitution ciphers. I will start with a short description of Fractionated morse, then we will look at how to break it.
# Description of Fractionated Morse
There are many descriptions of the algorithm around, the ACA has a good description, I will reproduce a part of G. worley's description. It assumes you know morse code.
If Alice wanted to send Bob the
message "I love you." she would tap over the telegraph:
Pt: I love you.
Et: ..xx.-..x---x...-x.xx-.--x---x..-xx.-.-.-
After encoding the plaintext in Morse Code, the encoded text is broken into
n-graphs, usually trigraphs because there are 26 Morse Code trigraphs (you never
have three dividers in a row, so you don't count that one). Here's a sample
trigraph mapping (using a keyword):
Keyword: LOOKINGGLASS
L O K I N G A S B C D E F H J M P Q R T U V W X Y Z
. . . . . . . . . - - - - - - - - - x x x x x x x x
. . . - - - x x x . . . - - - x x x . . . - - - x x
. - x . - x . - x . - x . - x . - x . - x . - x . -
So, breaking our message into trigraphs we get:
. x . - x . . - - - . x - .
. . . - . - x . x - . x . -
x - x - . x x - - x - . - x
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ct: K T K H R G B D P J O Y D G
Cracking this is much harder because the dividers are obfuscated. The only way
to attack this is to do a statistical analysis for common Morse Code trigraphs.
The cipher assumes 'x' is placed between each letter and 'xx' is placed between each word.
# Cryptanalysis of Fractionated Morse
Lets look at the following Fractionated Morse table:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
. . . . . . . . . - - - - - - - - - x x x x x x x x
. . . - - - x x x . . . - - - x x x . . . - - - x x
. - x . - x . - x . - x . - x . - x . - x . - x . -
If you look at the table above, you can see that certain ciphertext letter combinations are not possible e.g. RS, RT, ... RZ cannot occur because we can't have xxx in the plaintext Morse. Also IS, IT, ... IZ can't happen for the same reason. Other impossible combinations: CY, CZ, FY, FZ, OY, OZ etc. We also can't have e.g. ABD because there are no Morse letters that go '.....-.-.'. You should be able to come up with many other impossible combinations. We can also use frequency information, since THE occurs very often in English, we expect 'xx-x....x.xx'='ZSCI' to be quite common in the ciphertext (with this particular key, if you look at the counts provided down below, we see ZSCI is actually the most common quadgram). This means certain keys will be very unlikely if they lead to impossible transitions, others more likely if they lead to likely transitions.
The trick to breaking Fractionated Morse is that finding the key can be done completely independently of the English->Morse translation. Knowing the statistics of a particular Fractionated Morse key means we can just find the key that translates to it, without ever checking to see if the putative plain text even decodes properly. The Idea is to start with a random key, then continuously swap characters in the key trying to make the ciphertext look like it has been enciphered with the key "ABCDEFGHIJKLMNOPQRSTUVWXYZ". Once this is done, we can just decrypt it. To do this we need quadgram statistics for a lot of English text that has been enciphered with the key "ABCDEFGHIJKLMNOPQRSTUVWXYZ". I did this for 40 million sentences, or around 1 billion characters. The results are linked below.
Code for breaking Fractionated Morse ciphers is here, with the required statistics here. I have run the code on many possible routes through the second Feynman cipher, and nothing resulted in any good looking plaintexts. I would expect it to work, since breaking length 100 Fractionated Morse ciphertexts is not too difficult using the code above, and F2 is 261 characters. This leads me to believe that F2 is not a Fractionated Morse cipher.
From here I'll try to write an efficient running key solver for vigenere, beaufort, variant and porta rules and see how that goes. | 1,276 | 4,874 | {"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} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.907557 |
https://www.simplyfreshers.com/temnos-paper-pattern/ | 1,701,822,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00004.warc.gz | 1,095,845,484 | 26,189 | TEMNOS Placement Paper
Company: temenos
TEMNOS Paper Pattern
No Negative Marks So you can attend all the questions
1) Given that black->blue
blue – green
green – yellow
yellow – red
red – violet
violet – black
what is the color of blood?
Ans: Red
2) Five books A,B,C,D,E are arranged not in order, D is below E, B is above C, A is below B. which book is not top?
Ans: E
3)A man has to travel 5Km. he takes 5min break every hour. His speed is 10Km/hr. what is the time to reach the destinaion ?
Ans: 50 min
4) A frog climbs a 20ft wall, every hour it clim,s 3ft and decliens 2ft. how much time it takes ?
Ans: 18(it reached) not 20.
5) Nepo is three time as old as his son,s age. his father is 40 more than twice is age. what is nepo,s age ?
Ans: 360 (not 120=son,s age)
6) A man travel 3Km north from home, 2Km west,1km north, 5km east. what is the distance from his home now ?
Ans: 3(use pythogorus theorm, draw some extensions, diagonal is 3)
7) A Deer and hare had a race, hare was 16m ahead during start. Distance covered by hare in 6 jumps is equal to 4 jumps of Deer. distance covered in one jump by hare is 1m, and that of deer is 2.5m.find the distance covered by deer to catch hare.
Ans: don,t know
Cool A contractor has to complete, construction of a road in 15 days of length 12km, initialy done by 100 people After 9 days only 5km of work is completed. Now the contractor has to complete the remaning job in same time period. now how many more people he has to take.
Ans: i think 10(don,t forget to subtract 100)
9) There are 6 people A,B,C,D,E,F. A and D are husband and wife, F is the sister of D, c is the grandfather of D, D,s son is police. The two ladies are unemployed now say how is the pilot?
Ans: I don,t know( but E is police, because only E is not mentioned) i think D.
10) if ANTICIPATION is said as ICITNAOINTAP. then what is PRODUCTAVITY.
ANS: i don,t know.
11) A man consumes 20% more, when the product rate decress by 20%. what is his profit?
Ans: -4%( that is he has 4% loss) gain or loss% = (x/10)^2.
12) A train is moving at a speed of 84km/hr, and a man is moving at a speed of 6km/hr in opp. direction. They cross each other a 5sec. What is the length of train ?
TEMNOS Placement Paper
Over 3,000 firms across the globe, including 41 of the world’s top 50 banks, rely on Temenos to process the client interactions and daily transactions of more than 1.2 billion banking customers.
Temenos offers cloud-native, cloud-agnostic, API-first digital banking, core banking, payments, fund management, and wealth management software products, enabling banks to deliver consistent, frictionless customer journeys and achieve market-leading cost/income performance.
We have a relentless focus on innovation and consistently reinvest 20% of our revenues into R&D, the highest in banking software.
We have the largest and most dynamic global community of banks, FinTechs, developers and partners in the industry and together we make banking better.visit offical website of TEMNOS for more details. | 833 | 3,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-50 | longest | en | 0.962073 |
www.tsm-resources.com | 1,579,781,053,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610004.56/warc/CC-MAIN-20200123101110-20200123130110-00301.warc.gz | 1,096,844,085 | 3,039 | TSM LONGER TUTORIALS Autograph Home Page TSM Autograph Resources - TSM Images- TSM Data TSM Short Videos
AUTOGRAPH TUTORIALS by Douglas Butler
Tutorial 1. Geometry and Measure (14 min)
A basic introduction
to how Autograph works.
The Whiteboard Mode
and multiple selection
dependent objects
Using the scribble tool
Reflection and rotation
Circles (coordinate geometry)
Circles (Euclidian geometry)
Tutorial 2. Algebra and Graphing (20 min)
The Straight Line
Using the constant controller
The parabola y = x²; slow plot
Transformations of y = x²:
y = ax², y = (ax)²,
y = (x – a)² and y = x² + a
Fitting a parabola to an image
(human canon-ball)
Other topics to cover:
Linear Programming
The Autograph 'Extras' pages
Tutorial 3. Advanced Topics (20 min)
Introduction to differentiation
Introducing trigonometry
Introducting integration
3D Autograph: conic sections
Volume of revolution of an area
Other topics to cover:
Vectors (2D and 3D)
Differential Eqns (1st/2nd order)
Parametric and Polar equations
(2D and 3D)
Tutorial 4. Statistics and Handling Data (16 min)
Introduction to bivariate data
The principle of least squares
Pasting 2 columns of data
The Statistics page
Pasting a single column of data
The TSM Data resource
The dot plot
Grouping data for a histogram
Cumulative frequency diagram
Exporting a table of statistics
and a screen image
Tutorial 5. Probability Distributions (13 min)
The binomial distribution
Fitting a dependent normal
Calculating binomial probabilities
The normal distribution
and calculations
Type 1 and Type 2 errors
The Central Limit Theorem
Tutorial 6. Online Resources (7 min)
www.autograph-maths.com
The TSM Resources site
www.tsm-resources.com
including
TSM Data
and
TSM Images
Tutorial 7. Onscreen Keyboard (9 min)
The onscreen keyboard allows users to control objects in Autograph, and to enter one-line mathematical expressions using a wide variety of symbols that are correctly interpreted, eg:
sin²2θ, 2x − 3y ≤ 6,
−b ± √(b²−4ac), y = |x|
It is also invaluable when using a whiteboard or a walk-about tablet. This tutorial also looks at changing languages
=> TSM Short Videos
Oundle, UK - March 2012 | 562 | 2,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-05 | longest | en | 0.706788 |
https://www.studypool.com/services/20901/statistics-250 | 1,542,408,195,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743216.58/warc/CC-MAIN-20181116214920-20181117000920-00233.warc.gz | 999,657,644 | 21,669 | # Statistics 250
Apr 30th, 2015
Studypool Tutor
University of Michigan - Ann Arbor
Course: General Education
Price: \$10 USD
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Statistics: numbers measured for some purpose; collection of procedures and principles for gathering data and analyzing data to help people make decisions when faced w/ uncertainty
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Section 1.1Statistics: numbers measured for some purpose; collection of procedures and principles for gathering data and analyzing data to help people make decisions when faced w/ uncertaintySection 1.2dotplot: plot in which each dot represents response of individualfive-number summary: lowest/highest values; cutoff points for , , and of datamedian: value in middle when numbers are put in orderlower quartile: median of lower halve of dataupper quartile: median of upper halve of datarisk: estimation that bad outcome will occur in future based on pastbase rate (baseline risk): rate or risk at beginning time period or under specific conditionspopulation: collection of all individuals about which information is desiredrandom sample: subset derived from population so that each individual has specified probability of being part of samplesample survey: investigative gathering of opinion or information from each sample individualmargin of error: number added to/subtracted from sample information to produce 95% certain intervalself-selected sample (volunteer sample) => ask anyone who wishes to respond to do soex. magazines, television stations, Internet websites, etc.observational study: study in which participants are observed and measuredvariable: characteristic that differs from one individual to nextconfounding variable: variable that is not main concern of study, but may be partially responsiblerandomized experiment: study in which treatments
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December 19, 2023
Last Updated:
May 18, 2023
# APR vs APY
by
André Ringdorfer
SHARE
Key Takeaways:
• APR stands for annual percentage rate and does not include compound interest
• APY on the other hand, stands for annual percentage yield
• APY reflects interest paid on interest, which means action is needed to compound interest
• APY always show higher than APR
This article is going to shed some light on the differences between APR and APY. Although both terms look very similar and both of them have to do with interest rates, it is very important to know the difference between these two. Understanding the difference between these terms will help everyone make better decisions regarding money management.
In essence, APR stands for annual percentage rate and does not include compound interest. APY on the other hand, stands for annual percentage yield, and takes compound interest into account.
APR will tell you how much interest you will earn on your initial investment over a given period of time. To provide a more practical example: if your APR is 20% and you invest 100 USD for one year, at the end of this year your investment will be worth 120 USD.
APY will tell you how much interest you will earn on your initial investment and also the interest on your interest over a given period of time. To showcase this even better, if your investment bears a 20% APR and gets compounded on a daily basis (meaning that every day your interest also starts earning interest) your investment of 100 USD will yield 122.13 USD. This would signify an APY of 22.13%.
Looking at this example from an alternative perspective: If an investment opportunity offers you a 20% APR rate over one year you would end up with 120 USD on a 100 USD investment. If the same opportunity offers you 20% APY you would also end up with 120 USD but, your APR would have only been 18.24% (when applying daily compounding).
Why is the displayed APY always higher than APR for the same pool?
As mentioned, the difference between APR and APY is that one form of investment (APY) gets compounded every time period, whereas the other one shows you your potential gains without compounding effect (APR).
APY reflects interest paid on interest, which mean action is needed to compound interest. This action can be done either manually (i.e. users withdraw interest and stake it), or automatically, meaning the action will be implemented by the protocol. Due to this particular reason, APY is always higher than APR.
What is compound interest?
Compound interest is the interest rate one gets on their initial investment and also the interest that the earned interest earns. This means that the accumulated interest from previous time periods also earns interest.
## HELI Fast, HELI Green, HELI Affordable
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https://www.coursehero.com/file/5808819/SOL20/ | 1,496,017,283,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612003.36/warc/CC-MAIN-20170528235437-20170529015437-00398.warc.gz | 1,059,842,162 | 52,901 | SOL20 - Thevalingam Donald Homework 20 Due Mar 4 2007...
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Thevalingam, Donald – Homework 20 – Due: Mar 4 2007, midnight – Inst: Eslami 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Hey guys, instead oF giving me ten defni- tions For a little grade improvement in the last minute, you need to solve fve problems, but not For me, For yourselves. I do not need your solving problems; you need it! I have solved mine years ago when you were still crawling :)) 001 (part 1 oF 1) 10 points A bobsled slides down an ice track starting (at zero initial speed) From the top oF a(n) 194 m high hill. The acceleration oF gravity is 9 . 8 m / s 2 . Neglect Friction and air resistance and de- termine the bobsled’s speed at the bottom oF the hill. Correct answer: 61 . 6636 m / s. Explanation: By conservation oF energy: mgh = mv 2 2 hence the velocity at the bottom oF the hill is: v = p 2 gh = q 2(9 . 8 m / s 2 )(194 m) = 61 . 6636 m / s keywords: 002 (part 1 oF 1) 10 points A(n) 662 kg elevator starts From rest. It moves upward For 4 . 3 s with a constant acceleration until it reaches its cruising speed oF 1 . 91 m / s. The acceleration oF gravity is 9 . 8 m / s 2 . ±ind the average power delivered by the elevator motor during the period oF this accel- eration. Correct answer: 6
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Ask a homework question - tutors are online | 560 | 2,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.857343 |
https://la.mathworks.com/matlabcentral/cody/problems/58-tic-tac-toe-ftw/solutions/1564425 | 1,606,652,792,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00614.warc.gz | 359,904,274 | 17,089 | Cody
# Problem 58. Tic Tac Toe FTW
Solution 1564425
Submitted on 19 Jun 2018 by Ente
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [ 1 0 0 0 -1 0 -1 0 1]; b = [0]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
2 Pass
a = [ 1 0 0 0 1 -1 1 -1 -1]; b = [2 7]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
3 Pass
a = [ 1 0 0 -1 1 -1 1 -1 0]; b = [7 9]; out = ticTacToe(a); assert(isequal(out(:), b(:)))
### Community Treasure Hunt
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https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-1-equations-and-inequalities-1-4-rewrite-formuilas-and-equations-1-4-exercises-skill-practice-page-31/30 | 1,726,638,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00171.warc.gz | 732,331,648 | 15,257 | ## Algebra 2 (1st Edition)
Published by McDougal Littell
# Chapter 1, Equations and Inequalities - 1.4 Rewrite Formuilas and Equations - 1.4 Exercises - Skill Practice - Page 31: 30
#### Answer
$y=\frac{x}{x-1}$
#### Work Step by Step
After multiplying both sides by $xy$: $y+x=xy\\xy-y=x\\y(x-1)=x\\y=\frac{x}{x-1}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 147 | 483 | {"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} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.789662 |
https://www.hackmath.net/en/math-problem/18713?tag_id=37 | 1,591,337,030,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00157.warc.gz | 728,534,399 | 11,522 | # An example
An example is playfully for grade 6 from Math and I don't know how to explain it to my daughter when I don't want to use the calculator to calculate the cube root. Thus:
A cuboid was made from a block of 16x18x48 mm of modeline. What will be the edge of the cube? If I calculate the cuboid volume, would I have to cube root or not? Thank you for your answer.
Result
a2 = 24 mm
#### Solution:
$a=16 \ \text{mm} \ \\ b=18 \ \text{mm} \ \\ c=48 \ \text{mm} \ \\ \ \\ V=a \cdot \ b \cdot \ c=16 \cdot \ 18 \cdot \ 48=13824 \ \text{mm}^3 \ \\ \ \\ V=a_{2}^3 \ \\ x_{0}=\dfrac{ a+b+c }{ 3 }=\dfrac{ 16+18+48 }{ 3 } \doteq \dfrac{ 82 }{ 3 } \doteq 27.3333 \ \text{mm} \ \\ V_{0}=27^3=19683 \ \text{mm}^3 \ \\ V_{1}=28^3=21952 \ \text{mm}^3 \ \\ V_{3}=26^3=17576 \ \text{mm}^3 \ \\ V_{4}=25^3=15625 \ \text{mm}^3 \ \\ V_{5}=24^3=13824 \ \text{mm}^3 \ \\ V_{5}=V \ \\ a_{2}=24 \ \text{mm} \ \\ \ \\ \text{ Correctness test: } \ \\ 13824=2^9 \cdot \ 3^3=( 2^3 \cdot \ 3)^3=(8 \cdot \ 3)^3=24^3 \ \\ a_{2}=24 \ \\ \ \\ a_{2}=\sqrt[3]{ V}=\sqrt[3]{ 13824 }=24 \ \text{mm}$
Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
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The shape of the observatory dome is close to the hemisphere. Its outer diameter is 11 m. How many kilograms of paint and how many liters of paint is used for its double coat if you know that 1 kg of paint diluted with 1 deciliter of paint will paint an a
17. Alien ship
The alien ship has the shape of a sphere with a radius of r = 3000m, and its crew needs the ship to carry the collected research material in a cuboid box with a square base. Determine the length of the base and (and height h) so that the box has the large | 1,391 | 4,756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-24 | latest | en | 0.803167 |
https://math.stackexchange.com/questions/4295312/solve-the-angle-angledcb-in-triangle-triangleabc-with-anglea-84 | 1,714,059,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00027.warc.gz | 342,369,881 | 38,362 | # Solve the angle $\angle{DCB}$ in triangle $\triangle{ABC}$ with $\angle{A}=84^{\circ}$
Where $$\angle{A}=84^{\circ}, \angle{ACD}=42^{\circ}, BD=AC$$, find $$\angle{BCD}$$.
Wonder if there is solution without using trigonometric functions.
I tried with getting circumcenter of triangle ABC, but seems hard to prove it forms an equilateral triangle with side AC.
Also if trying from equilateral and form an Isosceles triangle with two angles of $$24^{\circ}$$, and then form another isosceles triangle with top angle to be $$\angle{B}=24^{\circ}$$, it is not easy to prove that $$\angle{ADC}=54^{\circ}$$.
• Please tell what you have tried and where you are stuck
– user987987
Nov 3, 2021 at 8:14
• I tried with getting circumcenter of triangle ABC, but seems hard to prove it forms an equilateral triangle with side AC.
– r ne
Nov 3, 2021 at 8:50
• There is only one relationship: $\angle{B}+\angle{BCD}=54^{\circ}$. This problem is not solvable by angle chasing, at least not this way.
– r ne
Nov 3, 2021 at 9:11
– r ne
Nov 3, 2021 at 10:31
Draw isosceles triangle $$ACE$$ such that $$CA=CE$$.
Let the circumcentre of $$\triangle DEC$$ be $$F$$ and draw the equilateral triangle $$DEF$$.
$$\angle CEF=\angle ECF=36^\circ$$.
Now construct another equilateral triangle with side $$CE$$ as in the figure.
$$\angle DEG=36^\circ$$
$$\therefore\triangle DEG\cong\triangle FEC\text{ (S-A-S)}\\ \implies DE=DG.$$
Also $$\angle BDG=72^\circ.$$
Then mark point $$H$$ on $$CE$$ such that $$CF=CH$$.
$$\therefore \angle CHF=\angle CFH=72^\circ$$.
Also $$\triangle EHF$$ is isosceles.
Now we can see $$\triangle EDG\sim\triangle EHF$$. From there we can prove that $$\triangle BGE\sim \triangle CFE$$.
From similar isosceles triangles above, $$\frac{ED}{EG}=\frac{EH}{EF}.$$
Adding $$1$$ to both sides, $$\frac{EB}{EG}=\frac{EC}{EF}\implies\small \triangle BGE\sim \triangle CFE.$$
It shows that $$BG=GE$$ and then we can easily show $$\angle EBC=\angle ABC=30^\circ.$$
• (+1) Very nice! How did you come up with the idea of adding equilateral triangles and the first isosceles triangle, if I may ask? Nov 16, 2021 at 5:55
• “Let the circumcentre of △DEC be F and draw the equilateral triangle DEF.” But how do we know that △DEF is equilateral, not merely isosceles like △ECF and △DCF? Dec 4, 2021 at 14:10
• The solution is great. Thanks and sorry for my late response!
– r ne
Dec 6, 2021 at 22:52
• Got a simpler solution, starting with the same idea but ending with shorter approach: i.imgur.com/SmmC6B2.png
– r ne
Dec 7, 2021 at 0:41
• @TimPederick , $\angle DFE=2\angle DCE$ (why?)
– ACB
Dec 12, 2021 at 11:56
Working "the other way around" sometimes helps. Consider triangle $$\triangle ACD$$, and construct the regular pentagon $$PQRDC$$ as shown above, noting that $$AD$$ bisects $$\angle CDR$$. Finally produce $$AD$$ to $$B^\prime$$ in such a way that $$\triangle CRB^\prime$$ is equilateral.
We will show that $$B \equiv B^\prime$$, by proving that $$B^\prime D \cong AC$$, as in the triangle in OP's hypotheses.
1. Angle chasing yields $$\measuredangle ACP = \measuredangle CAP = 76^\circ$$. Thus $$\measuredangle CPA = 48^\circ$$.
2. Working similarly on $$\triangle ARQ$$ allows to claim that $$\triangle PAQ$$ is equilateral.
3. Angle chasing and $$CQ \cong RB^\prime$$ imply that $$\triangle B^\prime DR \cong \triangle ACQ$$ (SAS), and, in particular $$B^\prime D \cong AC$$.
So, as stated, $$B^\prime\equiv B$$, and, in particular $$\boxed{\measuredangle DCB = 24^\circ}.$$ | 1,117 | 3,503 | {"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": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | latest | en | 0.770308 |
https://codereview.stackexchange.com/questions/60323/inefficient-slow-loop-with-calculations-in-worksheet/60331 | 1,696,331,100,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00861.warc.gz | 203,863,352 | 44,172 | Inefficient (slow) loop with calculations in worksheet
I am looking for best practice help. Slow loops seem to be a recurring problem for me and I'd like to learn a better way. The code itself works as it should, except it is far too slow.
The problem is the worksheet needs to calculate after each B & i value is dropped into "N13" so that "U12", "V12", and "W12" update before being deposited into wsRepository. If I turn Calculation on Manual then my values are no good because they are contingent upon the other worksheet formulas updating (calculating). I think I can copy my worksheets "off-screen", calculate, and then paste my values back "on-screen", but I am not sure how to do this. I've used variants in the paste to do similar things, but I not comfortable with them. There may even be more efficient ways of achieving my desired result that I am unaware of.
Application.ScreenUpdating = False
Dim wsRepository As Worksheet
Dim wsInput As Worksheet
Dim i As Integer
Set wsRepository = ThisWorkbook.Sheets("Repository")
Set wsInput = ThisWorkbook.Sheets("Input")
For i = 4 To 2004
wsInput.Range("N13").Value = wsRepository.Range("B" & i).Value
'copy back amounts
wsRepository.Range("E" & i).Value = wsInput.Range("U12").Value
wsRepository.Range("C" & i).Value = wsInput.Range("V12").Value
wsRepository.Range("D" & i).Value = wsInput.Range("W12").Value
Next i
wsInput.Activate
• How complicated are the formulas in U12, etc.? Can you share them? It would be easier to help if you can. Aug 17, 2014 at 23:02
Application.ScreenUpdating = False
I find it scary that the corresponding = True is nowhere in your code, for reasons already mentioned. Whenever I turn off screen updating, I find it's good UX to also specify a status bar message, and change the mouse cursor to a hourglass. Something along these lines:
Public Sub ToggleWaitMode(Optional ByVal waitMode As Boolean = False)
Application.ScreenUpdating = waitMode
Application.Calculation = IIf(waitMode, xlCalculationManual, xlCalculationAutomatic)
Application.StatusBar = IIf(waitMode, "Please wait...", vbNullString)
Application.Cursor = IIf(waitMode, xlWait, xlDefault)
End Sub
Which makes your procedure stub look like this:
Option Explicit
Public Sub DoSomething()
On Error GoTo ErrHandler
ToggleWaitMode True
'do that thing
CleanExit:
If Not Application.ScreenUpdating Then ToggleWaitMode
Exit Sub
ErrHandler:
ToggleWaitMode
' handle errors here
Resume CleanExit
End Sub
If I turn Calculation on Manual then my values are no good because...
If I understand properly, you need to update $N$13 some 2,000 times with a value that's in "$B$" & i, and then I guess $U$12, $V$12 and $W$12 need to be recalculated accordingly.
You haven't shown us what these cells contain and what cells their formula is referring to, but if they're the only cells that need to be recalculated when $N$13 changes, then you can force calculation like this:
wsInput.Range("$U$12").Calculate
wsInput.Range("$V$12").Calculate
wsInput.Range("$W$12").Calculate
But that might not speed up anything. You're pretty much stuck, since you need to recalculate these three cells before you can do anything, and you need to do that 2,000 times.
I think you're somewhat misusing VBA here, it looks like you could use 3 hidden columns (say, $AA$4:$AC$2004) and use Excel formulas to automatically calculate the would-be "U", "V" and "W" values for each row; the VBA macro could then just copy values from Input!$AA$4:$AC$2004 to Repository!$C$4:$E$2004.. if a macro is even needed for that.
I would suggest naming the ranges/cells in row 12 - anytime you have a specific cell with a specific meaning, it's always better for the VBA code to refer to the meaning rather than the cells' addresses.
I have no clue what these cells mean, but picture this:
Dim interestRate As Double
interestRate = wsInput.Range("InterestRate").Value
This extra abstraction level somewhat decouples the VBA code from the worksheet structure, which allows you to modify [at least parts of] the worksheet without having to modify the VBA code - for example you could insert another row and now InterestRate is read in row 13 instead of 12, and the VBA code couldn't care less.
You can define names in the [Formulas] Ribbon tab, under the [Defined Names] section. Or you can just select the cell and type its name in the address/names dropdown, just left of the formula bar.
This also has the advantage of making your Excel formulas more readable: instead of =$X$12*$N42 a formula can now look like =InterestRate*$N42
One last thing, I know it's common to call a worksheet variable like wsInput, but I find it sounds backwards and looks Hungarian. I'd call it inputSheet instead; wsRepository would be repositorySheet. Also it wouldn't hurt to rename i for row.
Two short things that even may be unneeded if yo u excluded them.
Use Option Explicit
For one it helps immensely when searching for errors with spelling. Been there done that. It's gruesome.
Secondly it helps you with writing code that's more similar to "real" programming languages. Nothing against vba, but I much prefer languages where you need to declare your variables.
Use an error handler.
Everytime you turn off screen updating you get into the dangerous zone of "breaking" the application in case somethin goes wrong.
Instead do something along the lines of:
Option Explicit
On Error Goto ErrorHandler
' a whole lot of code
:ErrorHandler
Application.ScreenUpdating = true
MsgBox "An error has occurred"
'Whatever else you need to do and probably something like
Exit Sub
I'm not sure this will work for you, but consider replacing or duplicating the formulas in your "12" cells with an Evaluate call. It's a little tricky to avoid runtime errors, so I suggest reading this. It might look something like this.
'wsRepository.Range("E" & i).Value = wsInput.Range("U12").Value
wsRepository.Range("E" & i).Value = Evaluate("SUM(A1:A10)")
Of course, using the formula in U12. Your mileage may vary, but this should let you set calculation to manual (I think). I would prefer the method Mat's Mug described though. This is just another option to try.
Some other notes
• i is typically used as a loop counter, but row would be more meaningful.
• 4 and 2004 are mysterious hardcoded numbers. What a lot of programmers refer to as magic numbers. It would good for readability/maintainability to replace them with startRow and lastRow constants.
• Great comments in my opinion. They're short and clear. Not too much, not too little.
• I'm not sure why you activate wsInput at the end, but you do a great job of avoiding it elsewhere, make sure you're not needlessly activating the sheet there.
• I wouldn't expect Evaluate to speed things up though. Aug 18, 2014 at 1:09
• No. It wouldn't, but it may allow OP to use manual calculation @Mat'sMug and that would speed it up. (Assuming there are many other cells that wouldn't need to be recalculated). It's a stretch, but may be worth a shot. Aug 18, 2014 at 1:13 | 1,680 | 7,034 | {"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} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.829807 |
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What is the time and space complexity of fuzzy c-means? Of SOM? How do these complexities compare to those of K-means?
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<br/><br/> <br/> -What is the time and space complexity of fuzzy c-means?
Of SOM? How do these complexities compare to those of K-means?
-Explain the difference between likelihood and probability.
-Give an example of a set of clusters in which merging based on the closeness of clusters leads to a more natural set of clusters than merging based on the strength of connection (interconnectedness) of clusters.
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http://mathhelpforum.com/pre-calculus/52872-graphical-method-print.html | 1,529,722,592,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864919.43/warc/CC-MAIN-20180623015758-20180623035758-00609.warc.gz | 200,295,981 | 2,588 | # graphical method?
• Oct 9th 2008, 01:26 PM
alex_2008
graphical method?
hey i cant think of a simple answer to this. im looking for an obvious anwswer
t=switching time
V= applied voltage
A=constant
B=constant
the formula:
t=A/(V^2-B)
how can I extract the constants A and B by a simple graphical method from the data
• Oct 9th 2008, 02:06 PM
skeeter
I can't give you an "obvious" answer ... but I do have an "answer".
solving for $\displaystyle V^2$ ...
$\displaystyle V^2 = A\left(\frac{1}{t}\right) + B$
if you graph $\displaystyle \frac{1}{t}$ vs. $\displaystyle V^2$ as x vs. y, you should get a linear graph.
A will be the slope of the graph
B will be the y-intercept. | 221 | 684 | {"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} | 3.0625 | 3 | CC-MAIN-2018-26 | latest | en | 0.856126 |
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# 253-hwk-30a - HPHY 253 Homework 30a Magnetic Fields Dr A E...
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HPHY 253 Homework 30a Magnetic Fields Dr. A. E. Bak Name 1. Serway & Jewett Problem 30 : 2 . Calculate the magnitude of the magnetic °eld at a point 100 cm perpendicularly from an in°nitely long, thin, straight wire carrying a current of 1 : 00 A . 2. Serway & Jewett Problem 30 : 5 . Determine the magnetic °eld at a point P located a distance x from the corner of an in°nitely long, thin wire bent at a right angle as shown in the Figure P 30 : 5 . The wire carries a steady current I . 3. Serway & Jewett Problem 30 : 8 . A long, thin, straight wire carries current I . A right-angle bend is made in the middle of the wire. The bend forms a circular arc of radius r as shown in Figure P 30 : 8 . Determine the magnetic °eld at the center of the arc. 4. Serway & Jewett Problem 30 : 18 . Two long, thin, parallel conductors, separated by 10 : 0 cm , carry currents in the same direction. The °rst wire carries current I 1 = 5 : 00 A , and the second wire carries current I 2 = 8 : 00 A .
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http://library.kiwix.org/stats.stackexchange.com_eng_all_2018-08/A/tag/variance/1.html | 1,550,527,875,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247488490.40/warc/CC-MAIN-20190218220415-20190219002415-00608.warc.gz | 163,015,378 | 9,109 | ## Tag: variance
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10 Variance inflation factor for generalized additive models 2014-04-08T15:40:19.657
10 Determinant of Fisher information 2014-08-30T09:12:13.683 | 2,552 | 7,754 | {"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} | 3.015625 | 3 | CC-MAIN-2019-09 | latest | en | 0.72697 |
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# UNIT G481 Module 1 1.1.
1 Physical Quantities & Units When quoting the measurement of a physical quantity it is essential 1
state the unit as well as the numerical value.
Candidates should be able to :
Explain that some physical quantities consist of a numerical The scientific system of units is called the Systeme Internationale
magnitude and a unit. DUnites (S.I.System). The seven base quantities and their units
are listed in the table below :
Use correctly the named units listed in this specification as
appropriate.
BASE QUANTITY SYMBOL BASE UNIT SYMBOL
Use correctly the following prefixes and their symbols to
mass m kilogram kg
indicate decimal sub-multiples or multiples of units :
length l metre m
time t second s
pico (p), nano (n), micro (), milli (m), centi (c), kilo (k),
mega (M), giga (g), tera (T). electric I ampere A
current
temperature T, kelvin K
Make suitable estimates of physical quantities included
within this specification. amount of n mole mol
substance
PHYSICAL QUANTITIES luminous candela cd
intensity
A PHYSICAL QUANTITY (e.g. mass, density..) is a measurable
property whose meaning is precisely defined so that everyone can
have the same understanding of the term.
All other quantities and units can be derived from the seven base
The meaning of a physical quantity can be represented by : quantities and units.
## Density = mass Examples
A DEFINING EQUATION -
Volume = l x w x b is measured in m3.
volume
Density = mass/volume is measured in kg m-3.
Acceleration = velocity change/time is measured in m s-2.
A WORD DEFINITION - The Density of a substance is Momentum = mass x velocity is measured in kg m s-1.
Charge = current x time is measured in A s-1 (called coulomb, C).
the mass per unit volume of
2008 FXA
UNIT G481 Module 1 1.1.1 Physical Quantities & Units
## STANDARD PREFIXES FOR S.I. UNITS Mass of a person
Height of a person
In Physics we are often faced with very large and very small numbers.
To cope with this, numbers are written using powers of 10. This is Walking speed
called scientific notation. Standard prefixes ,such as those shown in Speed limit on motorways
the table below, are used as an abbreviation for some of the powers
Volume of a can of coke
of 10.
Density of water
PREFIX SYMBOL VALUE
Weight of an apple
pico p 10-12
Weight of a saloon car
nano n 10-9
Diameter of the Earth
micro 10-6
Mass of the Earth
milli m 10-3
Current in a domestic
centi c 10-2 appliance
kilo k 103 e.m.f. of a car battery
mega M 106 Voltage of the mains
giga G 109 supply
## tera T 1012 Diameter of a sewing
needle
Maximum speed of a
modern fighter plane
ESTIMATION
## In problem solving or calculations carried out in experiments you
should always look at your answer to see if it seems reasonable. The
only way you can know if an answer is absurd is if you have some
awareness of some benchmarks. So lets try to make some estimates :
2008 FXA | 741 | 2,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-45 | latest | en | 0.88451 |
https://www.scribd.com/doc/47178838/Examples-for-Hamming-code | 1,508,464,356,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00667.warc.gz | 981,691,038 | 20,954 | # Examples for Hamming code
:
The message you want to send is 4-bits string:
0000
0100
1000
1100
0001
0101
1001
1101
0010
0110
1010
1110
0011
0111
1011
1111
There is 16 different messages.
The 4-bits messages are mapped to the following Sixteen Valid Codewords
0
1
2
3
4
5
6
7
0000000
0000111
0011001
0011110
0101010
0101101
0110011
0110100
8
9
A
B
C
D
E
F
1001011
1001100
1010010
1010101
1100001
1100110
1111000
1111111
The Hamming Code essentially defines 16 valid codewords. The sixteen words are
arranged such that the minimum distance between any two words is 3.
Check the hamming equation:
M=4, R=3, N=7
Left side: (M+R+1)*(2^M)=8*16=128
Right side: 2^N=128
Perfect match!
Exercise 1:
Calculate the Hamming distance between any two codewords in the
above table.
The send will only send one of these 16 valid codewords. For example, the send will
never send 0000001, which is not a valid codeword.
Due to the transmission error, the receiver might receive invalid codewords. Since the
code transmitted is 7-bit long, total amount of possible codes is 128.
When received a code, the receiver will look for the closest valid codeword as a guess for
what might be actually transmitted.
The receiver will calculate the Hamming distance between r and all valid codewords. the table of D(0000001.) 3 4 4 3 2 5 5 6 Thus the receiver conclude that the actual transmitted code is 0000000.1000111 0011001. and the last bit is inverted due to transmission error.0001111.0000101 0000011. 0111001. Exercise 2: Check that when 0000000 is sent.0001000. 0101011.0010111 0100111.0011011 0011101. 0011111. 0111110.0101000 After decoding Received codes After decoding 0000000 1001011 0000111 1001100 0011001 1010010 0011110 1010101 0101010 1100001 . x) is Code word 0000000 0000111 0011001 0011110 0101010 0101101 0110011 0110100 D(r.0010000 0100000.1000000 0000111. 1011110. 0010001. 0011010.Decoding at the Receiver Side For example: if the sender send m=0000000.0000110. In fact. 0011100. The codeword with the smallest Hamming distance will be the one. 1011001 0011110. any 1 bit error can be corrected by above procedure.0001001. which is the correct.) 1 2 2 5 4 3 3 4 Code word 1001011 1001100 1010010 1010101 1100001 1100110 1111000 1111111 D(r. The receiver actually might be implemented with a working sheet: Received codes 0000000. 0001110. 0010110. 0101010. the receiver received r=0000001.0000001.0011000.0000010 0000100.
1101101 0110011. 0111101 0001101. 0111010 0001010. 0110110. 0111011. 110011 0110100. 0101111 0101001. 0111100. 0100011 000011. 110100 0101101 1100110 0110011 1111000 0110100 1111111 The right half of the table will be left as an exercise. 0101100. 000100. 0100010. 1101010 0101101. 0110010.110101. 0110001 0110111. . 0100101. 0100100. 0110000.0101110. | 1,011 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-43 | latest | en | 0.714139 |
https://www.physicsforums.com/threads/reversible-adiabatic-expansion-for-an-ideal-gas.473911/ | 1,719,274,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00140.warc.gz | 819,160,639 | 15,770 | # Reversible Adiabatic Expansion for an Ideal Gas
In summary, for a monoatomic gas undergoing reversible expansion from 30 L and 400K to 60 L with a molar heat capacity of (3/2)R, independent of temperature, the final pressure is 0.55 atm and the final temperature remains at 400 K as the process is adiabatic. The equation Q=0 is used to solve for the final pressure and the equation P1V1= P2V2 is not applicable for an adiabatic process.
## Homework Statement
1 mole of a monoatomic gas undergoes reversible expansion from 30 L and 400K to 60 L. The molar heat capacity in this situation is (3/2)R, independent of temperature. Calculate the final pressure and temperature of this process if it is adiabatic.
## Homework Equations
Q= 0 if it is adiabatic
Q= nCv(dT)
PV=nRT
P1V1= P2V2
monoatomic Cv= (3/2)R <-- heat capacity at const. vol.
## The Attempt at a Solution
Using Q=0, 0= nCv(dT)
0= (1 mol)(3/2)(R)(dT)
dT= 0
P1V1=nRT or (P1)(30L)=(1)(8.314 J/mol K)(400 K)
P1= 1.1 atm (using conversion factor)
Substituting into P1V1=P2V2, I get 0.55 atm.
Temperature is still 400 K at the end
I wanted to make sure if this was correct (the fact that the change in temp is zero while it was adiabatic confused me).
Last edited:
Welcome to Physics Forums.
P1V1= P2V2
That's for an isothermal process, but this one is adiabatic. There is a different, but similar-looking, equation to use instead.
Last edited:
## What is reversible adiabatic expansion for an ideal gas?
Reversible adiabatic expansion for an ideal gas is a process in which an ideal gas expands without any heat transfer and without any energy losses. This process is also known as isentropic expansion.
## What is the equation for reversible adiabatic expansion?
The equation for reversible adiabatic expansion is given by: PV^γ = constant, where P is the pressure, V is the volume, and γ is the adiabatic index (also known as the ratio of specific heats).
## What is the difference between reversible adiabatic expansion and irreversible adiabatic expansion?
The main difference between reversible adiabatic expansion and irreversible adiabatic expansion is that reversible expansion occurs at a slow and controlled rate, while irreversible expansion occurs quickly and is not controlled. Additionally, reversible expansion results in no energy losses, while irreversible expansion results in energy losses due to friction and other factors.
## What is the significance of reversible adiabatic expansion in thermodynamics?
Reversible adiabatic expansion is significant in thermodynamics because it is a theoretical process that can be used to measure the maximum amount of work that can be extracted from a system. It is also a key concept in understanding the principles of heat engines and refrigeration cycles.
## What are some real-life examples of reversible adiabatic expansion?
Some real-life examples of reversible adiabatic expansion include the compression and expansion of air in a bicycle pump, the expansion of gases in a jet engine, and the compression and expansion of gases in a refrigerator or air conditioner.
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351 | 914 | 3,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.914345 |
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# Which type of heat transfer would appear in a hot pot of coffee?
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Heat transfers. The hotter loses heat energy to the cooler.
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A cup of coffee is an example of convection because the heat from the water warms up everything in the cup and all the atoms are bouncing off the cup[ in the coffee] of coffee because the coffee is hot.
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Heat from the coffee goes to the surroundings. The coffee gets colder, the surroundings get warmer.
### How is heat transferred when a cup of hot coffee is placed in a freezer?
Heat can beradiated,conducted or travel byconvection or a combination of these three.The base of the cup would conduct the heat to the surface it is sitting on and the body of the cup would radiate the heat and the air in the freezer would transfer the heat by convection currents.
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Conduction
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Convection heat transfer is the transfer of heat by the movement of a fluid.
### The heat from a campfire is transferred by what energy?
Radiant heat transfer is the primary way you feel heat from a campfire. There would also be convective and conductive heat transfer to the air. | 574 | 2,662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.958032 |
https://math-master.org/calculator/basic | 1,725,868,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00391.warc.gz | 351,249,010 | 28,327 | # The Basic Calculator
### How to use a Basic Calculator in math-master.org
The online basic calculator works just like a small handheld calculator. It performs basic calculator functions including addition, subtraction, multiplication, and division. However, the ease of use and readiness make this online calculator a useful tool for everyday use.
Using the online calculator free tool is quite easy. All the numbers and functions have been strategically placed to help in the accurate and quick processing of math problems. When using a desktop, control the calculator using a mount, number pad, or keyboard. When on the phone and other touch devices, simply tap to process functions.
## Functions of the Online Basic Calculator
AC All clear + Addition - Subtraction % Percentage ÷ Division × Multiplication +/- Plus/minus toggle = Calculate
## How to Use Online Basic Calculator Operations
The basic calculator simplifies functions, allowing you to process operations quickly and easily. Browse the example calculations below as a guide to find the calculator’s functions and processes. Follow the specific steps used to input numbers and symbols, prompting the calculator with operator buttons. The examples also show the various types of operations you can perform with this standard calculator.
Calculator Operation Steps4 4 + 2 = 6 Tap/click 4, followed by the [+] sign and 2. Then click [=] to generate the results 8 – 5 = 3 Tap/click 8, followed by the [-] sign and 5. Then click [=] to generate the results 2 + $-5$ = -3 Tap/click 2, followed by the [+/-] sign and 5. Then click [=] to generate the results 1.4 + 3.5 – 2.1 = Tap/click 1.4, followed by the [+] sign and 3.5. Then [-] followed by 2.1 and click [=] to generate the results
The basic online calculator stands out with its simple and easy-to-use interface. As a standard calculator for everyday use, you will find it simple and intuitive. You will also find calculator functions useful anytime and anywhere. In addition, the online calculator's free access makes it ideal for users of all ages and experience levels. You can perform simple arithmetic functions without the need for additional hardware or software.
## Additional Functions – Copy and Delete
The copy function allows you to replicate an operation to another place in your device or build upon it on the basic calculator. To copy:
• Highlight the results in the display and then copy it to your clipboard. Use the Ctrl + C command or right-click to select the copy option
• When using your phone or table, hold and tap to select the character. Then choose copy from the pop-up menu.
Use the backspace or delete key to delete one character at a time to the left:
• When using a desktop or PC, the backspace button serves as the delete button
• When using a smartphone or tablet, tap into the display. On the virtual keyboard that appears, use the delete button.
## Conclusion
The basic online calculator is a simple-to-use tool to complete simple mathematical operations. The simplified version of the online standard calculator comes with addition, subtraction, division, and multiplication functions. This makes it perfect for anyone looking to perform basic arithmetic calculations easily and quickly.
### FAQs
Which is the best simple online calculator? The basic online calculator is one of the best and easiest-to-use tools for arithmetic functions
Is the basic online calculator free? The basic online calculator is completely free to use, allowing you to perform calculations anytime and anywhere.
Can I use a basic calculator? Yes, the basic online calculator allows you to complete basic arithmetic functions quickly and easily. | 757 | 3,695 | {"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} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.867246 |
https://essayhint.com/category/mathematics/ | 1,657,046,273,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00258.warc.gz | 296,356,417 | 9,405 | +1443 776-2705 panelessays@gmail.com
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## homework 5 1. Compute the antiderivative of the following funcitons: (a) x^(2/3) + x^(-1/3) (b) x^2
homework 5 1. Compute the antiderivative of the following funcitons: (a) x^(2/3) + x^(-1/3) (b) x^2 exp (x^3) (c) tan (2x) (d) x^2/(1+x^3) 2. Compute the definite integrals: (a) integrate x^2 + x over [0, 1] (b) integrate sin(x/2) over [0, pi] (c) integrate x/(1+x^2)...
## Write a summary of the Article: The impact of covid 19 on the economy Utilize articles written in th
Write a summary of the Article: The impact of covid 19 on the economy Utilize articles written in the Wall Street Journal: which can be found in ProQuest, also use articles found on the internet like google news. Link the Article to macroeconomic data; link the...
## Emmanuel kai Math Scheduling exercises s 1. Create a digraph for the following set of tasks:
Emmanuel kai Math Scheduling exercises s 1. Create a digraph for the following set of tasks: Task Time required Tasks that must be Completed first A 3 B 4 C 7 D 6 A, B E 5 B F 5 D, E G 4 E 3. Using the priority list T4, T1, T7, T3, T6, T2, T5, schedule the project...
## See Attached Math Reply DF 3 Choose a classmate without a reply. No credit will be given for a repl
See Attached Math Reply DF 3 Choose a classmate without a reply. No credit will be given for a reply if the classmate’s post already has a reply with the correct answers. Calculate the distance around the figure (use the base of the figure for three-dimensional...
## homework 4 Differentiate the following functions 1. y = cos(x^2+12x) 2. y = (2x+1)^2/(x^2+x+1) 3. y
homework 4 Differentiate the following functions 1. y = cos(x^2+12x) 2. y = (2x+1)^2/(x^2+x+1) 3. y = e^(sin(x^2)) 4. y = cos^2 (x^2) 5. y = ln (cos x)
## Of the 5 glm() models you ran, which appears to be the best model? Support your answer by providing
Of the 5 glm() models you ran, which appears to be the best model? Support your answer by providing the relevant information from your R output. You should also provide ΔAIC values for all 5 models in your answer. Of your 5 models, what was the model weight (Akaike... | 672 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.819436 |
https://physics.stackexchange.com/questions/367185/information-contained-in-a-quantum-simulation/367242#367242 | 1,660,524,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572089.53/warc/CC-MAIN-20220814234405-20220815024405-00219.warc.gz | 424,367,345 | 68,077 | # Information contained in a quantum simulation
The system in the picture has a box filled with $N$ classical particles which can either be on the left or the right side. Let this system have a Hamiltonian $$H= \frac{\mathbf p^2}{\sum 2m_i} + V(x, t)$$ where V becomes infinite at $\pm \frac{L}{2}$. Let it be represented by a point in $2N$-dimensional phase space $$(x_1, p_1, ... , x_N, p_N)$$ If we say that a positive x coordinate corresponds to 1, and negative to 0, then this system will encode a string of bits $N$ digits long.
Suppose I want a quantum computer to track the information in the string of bits above. The computer will contain $M$ qubits where $M = \log N$. Assuming this is a perfect quantum computer with no errors, the general state of the computer will be a density matrix in $2^M$-dimensional Hilbert space. $$\rho = \begin{bmatrix} c_{1,1} & c_{1,2} & \dots & c_{1,N} \\ c_{2,1} & c_{2,2} \\ \vdots & & \ddots \\ c_{N,1} & & & c_{N,N} \end{bmatrix}$$
Now we will quantify the amount of information in both systems.
In order to quantify how much information the box of particles contains, we can use the Shannon Entropy. For a random variable $P_i$ representing the probability of finding a one of a zero in the string at position $i$, the information entropy is$$S = - \sum_i P_i \log P_i$$ If we assume the particles are sufficiently mixed up in the box, then every $P_i$ will have a probability $\frac{1}{2}$, and then total entropy will be $\frac{N}{2} \log 2$
Next we turn the quantum computer, where the measure of the amount of information is Von Neumann Entropy. Given a density matrix $\rho$ for the quantum computer, the amount of information is given by $$S = - Tr(\rho \log \rho)$$
Question 1: Does there exist a density matrix that contains the same amount of information as the classical system?
Question 2: Does there exist a Hamiltonian to control the evolution of the quantum computer so that the information contained in both systems are always equal?
• Actually, the classical phase space grows exponentially with N as well (assuming constant amount of free energy per particle.) Can you rephrase your question more precisely?
– TLDR
Nov 5, 2017 at 18:23
• Really? I thought there was a point in 2N-dimensional phase space for N particles. Why does it grow exponentially due to free energy? Nov 5, 2017 at 18:26
• Because as the dimension increases linearly, the 'number' (or volume) of configurations grows exponentially.
– TLDR
Nov 5, 2017 at 23:22
• That's right. But couldn't you say the same thing about the vector in Hilbert space, that it grows super exponentially in M, because the dimensions grow exponentially in M? Nov 6, 2017 at 1:46
• Not exactly. There are way more configurations of the wave function than there are states of the system from a specific choice of measurement basis. In a sense this is because the wave function describes the state of the system according to an arbitrary choice of measurement basis. There is no classical analogue to the wave function, and there is no simple classical analogue of the discrete configuration space of q-bits. The dimension of the quantum system configuration space increases linearly with $M$ in the sense that the dimension of $(\mathbb Z_2)^M$ increases linearly with $M$.
– TLDR
Nov 6, 2017 at 2:06
What do you mean by "information"?
There is classical information, which lives in binary strings, and there is quantum information, which lives in Hilbert spaces. They aren't the same thing.
Holevo's theorem says that the most classical information you can extract from a system of $t$ qubits is $t$ classical bits. However, if you want to approximately describe an arbitrary quantum state on $t$ qubits, you need $c^t$ classical bits to do so ($c$ is a constant that depends on how good you want the approximation to be).
So for your question, the amount of information you need to prepare the system is the same: $N$ bits, and the amount of information you can extract from the quantum system is only $\log N$ bits. Which of these is the real, bona fide information? That depends on your definition of information.
The standard definition of information in quantum information theory is von Neumann information, which for a system with density matrix $\rho$ is $-\mathrm{Tr} \,\rho \log \rho$. When you extend the theorems of classical information theory to quantum mechanics, this is generally the correct definition. By this measure, the quantum system has $\log N$ bits of information. So by the standard definition, they're different.
But as for how much "information" there really is in the system — this seems like a philosophy question to me.
• Hi Peter. By simulate I mean that the dynamics of the quantum system correspond to the dynamics of the classical system, so that, assuming that the phase space and Hilbert same have the same dimension, you could infer the position of one from the other. Nov 6, 2017 at 2:52
• Also, what if I'm not worried about extracting the information. Is the fact that it's there, even if it's not recoverable, physically meaningful? Nov 6, 2017 at 2:53
• Most $n$-bit classical systems need $n$ qubits to simulate them in any meaningful sense, because if you want to do it with fewer, the dynamics of the classical system are very constrained. If someone can identify a system where you can do it with less, then your question gets very interesting. Nov 6, 2017 at 2:56
• I will update my question to show the specific correspondence between the two spaces. Nov 6, 2017 at 2:58
• I've edited my question again. It seems to me that both systems have information proportional no N log N. Nov 12, 2017 at 6:58
In general, the entropy of a system is written $$S_\text{tot}=S_{EE}+S_\text{thermal}$$ where EE denotes the entanglement entropy. In a thermal classical system, you have no ignorance associated with the possibility of states being entangled. However, a quantum computer simulates the classical system using some quantum circuit that has non-zero entanglement entropy so clearly the information content of the two systems are different.
Ignoring the quantum computer itself, the information pertaining to the classical simulation the quantum computer performs contains the same amount of information as the classical system being simulated.
• For the purposes of my question, let's assume that we a have a perfect, error-free quantum computer. In that case, even though there will exist entanglement between qubits within the computer, the wavefunction of the entire computer will be a pure state, and hence the entanglement entropy will be zero. For reference, quantiki.org/wiki/entropy-entanglement-0 Nov 6, 2017 at 2:27 | 1,617 | 6,702 | {"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} | 3.390625 | 3 | CC-MAIN-2022-33 | latest | en | 0.829645 |
https://competitive-exam.in/post/square-square-root-and-cube-cube-root | 1,713,113,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00825.warc.gz | 160,433,465 | 14,838 | # Square - Square Root and Cube - Cube Root
Friday 1st of January 2016
### Sharing is caring
Square, when any number is multiplied by itself, then the produce is known as square of the number.
Square Root, Square may be define the number whose square is equal to given number. In other words, square root of a given number is the number when multiplied by itself, gives the product equal to the given number.
Cube, cube of a number is the triple product obtained on multiplying the number by itself.
Cube Root, cube root of a number is the number whose cube is the given number. | 128 | 584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.940665 |
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# Number of subsequences of an array with given function value
Given an array A[] of length N and integer F, the task is to find the number of subsequences where the average of the sum of the square of elements (of that particular subsequence) is equal to the value F.
Examples:
Input: A[] = {1, 2, 1, 2}, F = 2
Output: 2
Explanation: Two subsets with value F = 2 are {2} and {2}. Where 2^2/1 = 2
Input: A[] = {1, 1, 1, 1}, F = 1
Output: 15
Explanation: All the subsets will return the function value of 1 except the empty subset. Hence the total number of subsequences will be 2 ^ 4 – 1 = 15.
Approach: This can be solved using the following idea:
Using Dynamic programming, we can reduce the overlapping of subsets that are already computed.
Follow the steps below to solve the problem:
• Initialize a 3D array, say dp[][][], where dp[i][k][f] indicates the number of ways to select k integers from first i values such that their sum of squares or any other function according to F is stored in dp[i][k][f]
• Traverse the array, we still have 2 options for each element i.e. to include or to exclude. The transition will be as shown at the end:
• If included then we will have k + 1 elements with functional sum value equal to s+ sq where sq is the square value i.e. arr[i]^2.
• If it is excluded we simply traverse to the next index storing the previous state in it as it is.
• Finally, the answer will be the sum of dp[N][j][F*j] as the functional value is the squared sum average.
Transitions for DP:
• Include the ith element: dp[i + 1][k + 1][s + sq] += dp[i][k][s]
• Exclude the ith element: dp[i + 1][k][s] += dp[i][k][s]
Below is the implementation of the code:
## C++
`// C++ program for the above approach` `#include ` `using` `namespace` `std;` `// Storing the DP states` `// dp[i][k][f] indicates that the number` `// of ways to select k integers from first` `// i values such that their sum of squares` `// or any other function according to F` `// is stored in dp[i][k][F]` `int` `dp[101][101][1001];` `// Calculate the number of subsequences` `// with given function value` `int` `countValue(``int` `n, ``int` `F, vector<``int``> v)` `{` ` ``// Base condition` ` ``dp[0][0][0] = 1;` ` ``// Three loops for three states` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `k = 0; k < n; k++) {` ` ``for` `(``int` `s = 0; s <= 1000; s++) {` ` ``long` `long` `sq` ` ``= (``long` `long``)v[i] * (``long` `long``)v[i];` ` ``// Recurrence relation` ` ``// Include the ith element` ` ``dp[i + 1][k + 1][s + sq] += dp[i][k][s];` ` ``// Exclude the ith element` ` ``dp[i + 1][k][s] += dp[i][k][s];` ` ``}` ` ``}` ` ``}` ` ``int` `cnt = 0;` ` ``// Iterate over the range [1, N]` ` ``for` `(``int` `j = 1; j <= n; j++) {` ` ``cnt += dp[n][j][F * j];` ` ``}` ` ``// Return the final count` ` ``return` `cnt;` `}` `// Driver Code` `int` `main()` `{` ` ``vector<``int``> v = { 1, 2, 1, 2 };` ` ``int` `F = 2, N = v.size();` ` ``// Function call` ` ``cout << countValue(N, F, v);` ` ``return` `0;` `}`
## Java
`import` `java.util.*;` `public` `class` `Main {` ` ``// Storing the DP states` ` ``// dp[i][k][f] indicates that the number` ` ``// of ways to select k integers from first` ` ``// i values such that their sum of squares` ` ``// or any other function according to F` ` ``// is stored in dp[i][k][F]` ` ``static` `long``[][][] dp = ``new` `long``[``101``][``101``][``1001``];` ` ``// Calculate the number of subsequences` ` ``// with given function value` ` ``static` `long` `countValue(``int` `n, ``int` `F, List v) {` ` ``// Base condition` ` ``dp[``0``][``0``][``0``] = ``1``;` ` ``// Three loops for three states` ` ``for` `(``int` `i = ``0``; i < n; i++) {` ` ``for` `(``int` `k = ``0``; k < n; k++) {` ` ``for` `(``int` `s = ``0``; s <= ``1000``; s++) {` ` ``long` `sq = (``long``) v.get(i) * (``long``) v.get(i);` ` ``// Recurrence relation` ` ``// Include the ith element` ` ``if` `(i + ``1` `<= n && k + ``1` `<= n && s + sq <= ``1000``) {` ` ``dp[i + ``1``][k + ``1``][s + (``int``) sq] += dp[i][k][s];` ` ``}` ` ``// Exclude the ith element` ` ``if` `(i + ``1` `<= n && k <= n && s <= ``1000``) {` ` ``dp[i + ``1``][k][s] += dp[i][k][s];` ` ``}` ` ``}` ` ``}` ` ``}` ` ``long` `cnt = ``0``;` ` ``// Iterate over the range [1, N]` ` ``for` `(``int` `j = ``1``; j <= n; j++) {` ` ``if` `(n <= ``100` `&& j <= ``100` `&& F * j <= ``1000``) {` ` ``cnt += dp[n][j][F * j];` ` ``}` ` ``}` ` ``// Return the final count` ` ``return` `cnt;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args) {` ` ``List v = Arrays.asList(``1``, ``2``, ``1``, ``2``);` ` ``int` `F = ``2``, N = v.size();` ` ``// Function call` ` ``System.out.println(countValue(N, F, v));` ` ``}` `}`
## Python3
`# Python program for the above approach` `# Storing the DP states` `# dp[i][k][f] indicates that the number` `# of ways to select k integers from first` `# i values such that their sum of squares` `# or any other function according to F` `# is stored in dp[i][k][F]` `dp ``=` `[[[``0` `for` `i ``in` `range``(``1005``)] ``for` `j ``in` `range``(``101``)] ``for` `k ``in` `range``(``101``)]` `# Calculate the number of subsequences` `# with given function value` `def` `countValue(n, F, v):` ` ``# Base condition` ` ``dp[``0``][``0``][``0``] ``=` `1` ` ``# Three loops for three states` ` ``for` `i ``in` `range``(n):` ` ``for` `k ``in` `range``(n):` ` ``for` `s ``in` `range``(``1001``):` ` ``sq ``=` `v[i] ``*` `v[i]` ` ``# Recurrence relation` ` ``# Include the ith element` ` ``dp[i ``+` `1``][k ``+` `1``][s ``+` `sq] ``+``=` `dp[i][k][s]` ` ``# Exclude the ith element` ` ``dp[i ``+` `1``][k][s] ``+``=` `dp[i][k][s]` ` ``cnt ``=` `0` ` ``# Iterate over the range [1, N]` ` ``for` `j ``in` `range``(``1``, n ``+` `1``):` ` ``cnt ``+``=` `dp[n][j][F ``*` `j]` ` ``# Return the final count` ` ``return` `cnt` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ``v ``=` `[``1``, ``2``, ``1``, ``2``]` ` ``F, N ``=` `2``, ``len``(v)` ` ``# Function call` ` ``print``(countValue(N, F, v))` `# This code is contributed by Susobhan Akhuil`
## C#
`// C# program for the above approach` `using` `System;` `public` `class` `GFG {` ` ``// Storing the DP states` ` ``// dp[i][k][f] indicates that the number` ` ``// of ways to select k integers from first` ` ``// i values such that their sum of squares` ` ``// or any other function according to F` ` ``// is stored in dp[i][k][F]` ` ``static` `int``[,,] dp = ``new` `int``[101,101,1005];` ` ``// Calculate the number of subsequences` ` ``// with given function value` ` ``static` `int` `countValue(``int` `n, ``int` `F, ``int``[] v) {` ` ``// Base condition` ` ``dp[0,0,0] = 1;` ` ``// Three loops for three states` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `k = 0; k < n; k++) {` ` ``for` `(``int` `s = 0; s <= 1000; s++) {` ` ``long` `sq = (``long``)v[i] * (``long``)v[i];` ` ``// Recurrence relation` ` ``// Include the ith element` ` ``dp[i + 1,k + 1,s + sq] += dp[i,k,s];` ` ``// Exclude the ith element` ` ``dp[i + 1,k,s] += dp[i,k,s];` ` ``}` ` ``}` ` ``}` ` ``int` `cnt = 0;` ` ``// Iterate over the range [1, N]` ` ``for` `(``int` `j = 1; j <= n; j++) {` ` ``cnt += dp[n,j,F * j];` ` ``}` ` ``// Return the final count` ` ``return` `cnt;` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `Main() {` ` ``int``[] v = { 1, 2, 1, 2 };` ` ``int` `F = 2, N = v.Length;` ` ``// Function call` ` ``Console.WriteLine(countValue(N, F, v));` ` ``}` `}`
## Javascript
`// Storing the DP states` `// dp[i][k][f] indicates that the number` `// of ways to select k integers from first` `// i values such that their sum of squares` `// or any other function according to F` `// is stored in dp[i][k][F]` `const dp = Array.from(Array(101), () => Array.from(Array(101), () => ``new` `Array(1005).fill(0)));` `// Calculate the number of subsequences` `// with given function value` `function` `countValue(n, F, v) {` ` ``// Base condition` ` ``dp[0][0][0] = 1;` ` ``// Three loops for three states` ` ``for` `(let i = 0; i < n; i++) {` ` ``for` `(let k = 0; k < n; k++) {` ` ``for` `(let s = 0; s <= 1000; s++) {` ` ``const sq = v[i] * v[i];` ` ``// Recurrence relation` ` ``// Include the ith element` ` ``dp[i + 1][k + 1][s + sq] += dp[i][k][s];` ` ``// Exclude the ith element` ` ``dp[i + 1][k][s] += dp[i][k][s];` ` ``}` ` ``}` ` ``}` ` ``let cnt = 0;` ` ``// Iterate over the range [1, N]` ` ``for` `(let j = 1; j <= n; j++) {` ` ``cnt += dp[n][j][F * j];` ` ``}` ` ``// Return the final count` ` ``return` `cnt;` `}` `// Driver Code` `const v = [1, 2, 1, 2];` `const F = 2;` `const N = v.length;` `// Function call` `console.log(countValue(N, F, v));`
Output
`2`
Time Complexity: O(F*N2)
Auxiliary Space: O(F*N2)
Efficient approach : Space optimization
In previous approach the current value dp[i][j][s] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 2D vector to store the computations.
Implementation:
## C++
`#include ` `using` `namespace` `std;` `// Calculate the number of subsequences` `// with given function value` `int` `countValue(``int` `n, ``int` `F, vector<``int``> v)` `{` ` ``// Initialize the DP table` ` ``vector > dp(n + 1, vector<``int``>(1001, 0));` ` ``dp[0][0] = 1;` ` ``// Compute the DP table` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `k = i + 1; k >= 1; k--) {` ` ``for` `(``int` `s = 0; s <= 1000; s++) {` ` ``int` `sq = v[i] * v[i];` ` ``// Recurrence relation` ` ``if` `(s + sq <= 1000) {` ` ``dp[k][s + sq] += dp[k - 1][s];` ` ``}` ` ``}` ` ``}` ` ``}` ` ``// Compute the final count` ` ``int` `cnt = 0;` ` ``for` `(``int` `k = 1; k <= n; k++) {` ` ``cnt += dp[k][k * F];` ` ``}` ` ``return` `cnt;` `}` `// Driver Code` `int` `main()` `{` ` ``vector<``int``> v = { 1, 2, 1, 2 };` ` ``int` `F = 2, N = v.size();` ` ``// Function call` ` ``cout << countValue(N, F, v);` ` ``return` `0;` `}`
Output
`2`
Time Complexity: O(F*N^2)
Auxiliary Space: O(N^2)
My Personal Notes arrow_drop_up | 4,202 | 11,649 | {"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} | 3.96875 | 4 | CC-MAIN-2023-23 | latest | en | 0.843715 |
https://www.greaterwrong.com/posts/jzf4Rcienrm6btRyt/priors-as-mathematical-objects | 1,591,247,611,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439019.86/warc/CC-MAIN-20200604032435-20200604062435-00224.warc.gz | 740,132,834 | 20,433 | # Priors as Mathematical Objects
Followup to: “Inductive Bias”
What exactly is a “prior”, as a mathematical object? Suppose you’re looking at an urn filled with red and white balls. When you draw the very first ball, you haven’t yet had a chance to gather much evidence, so you start out with a rather vague and fuzzy expectation of what might happen—you might say “fifty/fifty, even odds” for the chance of getting a red or white ball. But you’re ready to revise that estimate for future balls as soon as you’ve drawn a few samples. So then this initial probability estimate, 0.5, is not repeat not a “prior”.
An introduction to Bayes’s Rule for confused students might refer to the population frequency of breast cancer as the “prior probability of breast cancer”, and the revised probability after a mammography as the “posterior probability”. But in the scriptures of Deep Bayesianism, such as Probability Theory: The Logic of Science, one finds a quite different concept—that of prior information, which includes e.g. our beliefs about the sensitivity and specificity of mammography exams. Our belief about the population frequency of breast cancer is only one small element of our prior information.
In my earlier post on inductive bias, I discussed three possible beliefs we might have about an urn of red and white balls, which will be sampled without replacement:
• Case 1: The urn contains 5 red balls and 5 white balls;
• Case 2: A random number was generated between 0 and 1, and each ball was selected to be red (or white) at this probability;
• Case 3: A monkey threw balls into the urn, each with a 50% chance of being red or white.
In each case, if you ask me—before I draw any balls—to estimate my marginal probability that the fourth ball drawn will be red, I will respond “50%”. And yet, once I begin observing balls drawn from the urn, I reason from the evidence in three different ways:
• Case 1: Each red ball drawn makes it less likely that future balls will be red, because I believe there are fewer red balls left in the urn.
• Case 2: Each red ball drawn makes it more plausible that future balls will be red, because I will reason that the random number was probably higher, and that the urn is hence more likely to contain mostly red balls.
• Case 3: Observing a red or white ball has no effect on my future estimates, because each ball was independently selected to be red or white at a fixed, known probability.
Suppose I write a Python program to reproduce my reasoning in each of these scenarios. The program will take in a record of balls observed so far, and output an estimate of the probability that the next ball drawn will be red. It turns out that the only necessary information is the count of red balls seen and white balls seen, which we will respectively call R and W. So each program accepts inputs R and W, and outputs the probability that the next ball drawn is red:
• Case 1: return (5 - R)/(10 - R—W) # Number of red balls remaining / total balls remaining
• Case 2: return (R + 1)/(R + W + 2) # Laplace’s Law of Succession
• Case 3: return 0.5
These programs are correct so far as they go. But unfortunately, probability theory does not operate on Python programs. Probability theory is an algebra of uncertainty, a calculus of credibility, and Python programs are not allowed in the formulas. It is like trying to add 3 to a toaster oven.
To use these programs in the probability calculus, we must figure out how to convert a Python program into a more convenient mathematical object—say, a probability distribution.
Suppose I want to know the combined probability that the sequence observed will be RWWRR, according to program 2 above. Program 2 does not have a direct faculty for returning the joint or combined probability of a sequence, but it is easy to extract anyway. First, I ask what probability program 2 assigns to observing R, given that no balls have been observed. Program 2 replies “1/2”. Then I ask the probability that the next ball is R, given that one red ball has been observed; program 2 replies “2/3″. The second ball is actually white, so the joint probability so far is 12 * 13 = 16. Next I ask for the probability that the third ball is red, given that the previous observation is RW; this is summarized as “one red and one white ball”, and the answer is 12. The third ball is white, so the joint probability for RWW is 112. For the fourth ball, given the previous observation RWW, the probability of redness is 25, and the joint probability goes to 130. We can write this as p(RWWR|RWW) = 25, which means that if the sequence so far is RWW, the probability assigned by program 2 to the sequence continuing with R and forming RWWR equals 25. And then p(RWWRR|RWWR) = 12, and the combined probability is 160.
We can do this with every possible sequence of ten balls, and end up with a table of 1024 entries. This table of 1024 entries constitutes a probability distribution over sequences of observations of length 10, and it says everything the Python program had to say (about 10 or fewer observations, anyway). Suppose I have only this probability table, and I want to know the probability that the third ball is red, given that the first two balls drawn were white. I need only sum over the probability of all entries beginning with WWR, and divide by the probability of all entries beginning with WW.
We have thus transformed a program that computes the probability of future events given past experiences, into a probability distribution over sequences of observations.
You wouldn’t want to do this in real life, because the Python program is ever so much more compact than a table with 1024 entries. The point is not that we can turn an efficient and compact computer program into a bigger and less efficient giant lookup table; the point is that we can view an inductive learner as a mathematical object, a distribution over sequences, which readily fits into standard probability calculus. We can take a computer program that reasons from experience and think about it using probability theory.
Why might this be convenient? Say that I’m not sure which of these three scenarios best describes the urn—I think it’s about equally likely that each of the three cases holds true. How should I reason from my actual observations of the urn? If you think about the problem from the perspective of constructing a computer program that imitates my inferences, it looks complicated—we have to juggle the relative probabilities of each hypothesis, and also the probabilities within each hypothesis. If you think about it from the perspective of probability theory, the obvious thing to do is to add up all three distributions with weightings of 13 apiece, yielding a new distribution (which is in fact correct). Then the task is just to turn this new distribution into a computer program, which turns out not to be difficult.
So that is what a prior really is—a mathematical object that represents all of your starting information plus the way you learn from experience.
• I’m confused when you say that the prior represents all your starting information plus the way you learn from experience. Isn’t the way you learn from experience fixed, in this framework? Given that you are using Bayesian methods, so that the idea of a prior is well defined, then doesn’t that already tell how you will learn from experience?
• Hal, with a poor prior, “Bayesian updating” can lead to learning in the wrong direction or to no learning at all. Bayesian updating guarantees a certain kind of consistency, but not correctness. (If you have five city maps that agree with each other, they might still disagree with the city.) You might think of Bayesian updating as a kind of lower level of organization—like a computer chip that runs programs, or the laws of physics that run the computer chip—underneath the activity of learning. If you start with a maxentropy prior that assigns equal probability to every sequence of observations, and carry out strict Bayesian updating, you’ll still never learn anything; your marginal probabilities will never change as a result of the Bayesian updates. Conversely, if you somehow had a good prior but no Bayesian engine to update it, you would stay frozen in time and no learning would take place. To learn you need a good prior and an updating engine. Taking a picture requires a camera, light—and also time.
This probably deserves its own post.
• Another thing I don’t fully understand is the process of “updating” a prior. I’ve seen different flavors of Bayesian reasoning described. In some, we start with a prior, get some information and update the probabilities. This new probability distribution now serves as our prior for interpreting the next incoming piece of information, which then causes us to further update the prior. In other interpretations, the priors never change; they are always considered the initial probability distribution. We then use those prior probabilities plus our sequence of observations since then to make new interpretations and predictions. I gather that these can be considered mathematically identical, but do you think one or the other is a more useful or helpful way to think of it?
In this example, you start off with uncertainty about which process put in the balls, so we give 13 probability to each. But then as we observe balls coming out, we can update this prior. Once we see 6 red balls for example, we can completely eliminate Case 1 which put in 5 red and 5 white. We can think of our prior as our information about the ball-filling process plus the current state of the urn, and this can be updated after each ball is drawn.
• Hal,
You are being a bad boy. In his earlier discussion Eliezer made it clear that he did not approve of this terminology of “updating priors.” One has posterior probability distributions. The prior is what one starts with. However, Eliezer has also been a bit confusing with his occasional use of such language as a “prior learning.” I repeat, agents learn, not priors, although in his view of the post-human computerized future, maybe it will be computerized priors that do the learning.
The only way one is going to get “wrong learning” at least somewhat asymptotically is if the dimensionality is high and the support is disconnected. Eliezer is right that if one starts off with a prior that is far enough off, one might well have “wrong learning,” at least for awhile. But, unless the conditions I just listed hold, eventually the learning will move in the right direction and head towards the correct answer, or probability distribution, at least that is what Bayes’ Theorem asserts.
OTOH, the reference to “deep Bayesianism” raises another issue, that of fundamental subjectivism. There is this deep divide among Bayesians between the ones that are ultimately classical frequentists but who argue that Bayesian methods are a superior way of getting to the true objective distribution, and the deep subjectivist Bayesians. For the latter, there are no ultimately “true” probability distributions. We are always estimating something derived out of our subjective priors as updated by more recent information, wherever those priors came from.
Also, saying a prior should the known probability distribution, say of cancer victims, assumes that this probability is somehow known. The prior is always subject to how much information the assumer of a prior has when they being their process of estimation.
• Eliezer may not approve of it, but almost all of the literature uses the phrase “updating a prior” to mean exactly the type of sequential learning from evidence that Eliezer discusses. I prefer to think of it as ‘updating a prior’. Bayes’ theorem tells you that data is an operator on the space of probability distributions, converting prior information into posterior information. I think it’s helpful to think of that process as ‘updating’ so that my prior actually changes to something new before the next piece of information comes my way.
• Eliezer ,
Just to be clear . . . going back to your first paragraph, that 0.5 is a prior probability for the outcome of one draw from the urn (that is, for the random variable that equals 1 if the ball is red and 0 if the ball is white). But, as you point out, 0.5 is not a prior probability for the series of ten draws. What you’re calling a “prior” would typically be called a “model” by statisticians. Bayesians traditionally divide a model into likelihood, prior, and hyperprior, but as you implicitly point out, the dividing line between these is not clear: ultimately, they’re all part of the big model.
• Barkley, I think you may be regarding likelihood distributions as fixed properties held in common by all agents, whereas I am regarding them as variables folded into the prior—if you have a probability distribution over sequences of observables, it implicitly includes beliefs about parameters and likelihoods. Where agents disagree about prior likelihood functions, not just prior parameter probabilities, their beliefs may trivially fail to converge.
Andrew’s point may be particularly relevant here—it may indeed be that statisticians call what I am talking about a “model”. (Although in some cases, like the Laplace’s Law of Succession inductor, I think they might call it a “model class”?) Jaynes, however, would have called it our “prior information” and he would have written “the probability of A, given that we observe B” as p(A|B,I) where I stands for all our prior beliefs including parameter distributions and likelihood distributions. While we may often want to discriminate between different models and model classes, it makes no sense to talk about discriminating between “prior informations”—your prior information is everything you start out with.
• Eliezer, I am very interested in the Bayesian approach to reasoning you’ve outlined on this site, it’s one of the more elegant ideas I’ve ever run into.
I am a bit confused, though, about to what extent you are using math directly when assessing truth claims. If I asked you for example “what probability do you assign to the proposition ‘global warming is anthropogenic’ ?” (say), would you tell me a number?
Or is this mostly about conceptually understanding that P(effect|~cause) needs to be taken into account?
If it’s a number, what’s your heuristic for getting there (i.e., deciding on a prior probability & all the other probabilities)?
If there’s a post that goes into that much detail, I haven’t seen it yet, though your explanations of Bayes theorem generally are brilliant.
• My reason for writing this is not to correct Eliezer. Rather, I want to expand on his distinction between prior information and prior probability. Pages 87-89 of Probability Theory: the Logic of Science by E. T. Jaynes (2004 reprint with corrections, ISBN 0 521 59271 2) is dense with important definitions and principles. The quotes below are from there, unless otherwise indicated.
Jaynes writes the fundamental law of inference as
`````` P(H|DX) = P(H|X) P(D|HX) / P(D|X) (4.3)
``````
Which the reader may be more used to seeing as
`````` P(H|D) = P(H) P(D|H) / P(D)
``````
Where
`````` H = some hypothesis to be tested
D = the data under immediate consideration
X = all other information known
``````
X is the misleadingly-named ‘prior information’, which represents all the information available other than the specific data D that we are considering at the moment. “This includes, at the very least, all it’s past experiences, from the time it left the factory to the time it received its current problem.”—Jaynes p.87, referring to a hypothetical problem-solving robot. It seems to me that in practice, X ends up being a representation of a subset of all prior experience, attempting to discard only what is irrelevant to the problem. In real human practice, that representation may be wrong and may need to be corrected.
“ … to our robot, there is no such thing as an ‘absolute’ probability; all probabilities are necessarily conditional on X at the least.” “Any probability P(A|X) which is conditional on X alone is called a prior probability. But we caution that ‘prior’ … does not necessarily mean ‘earlier in time’ … the distinction is purely a logical one; any information beyond the immediate data D of the current problem is by definition ‘prior information’.”
“Indeed, the separation of the totality of the evidence into two components called ‘data’ and ‘prior information’ is an arbitrary choice made by us, only for our convenience in organizing a chain of inferences.” Please note his use of the word ‘evidence’.
Sampling theory, which is the basis of many treatments of probability, “ … did not need to take any particular note of the prior information X, because all probabilities were conditional on H, and so we could suppose implicitly that the general verbal prior information defining the problem was included in H. This is the habit of notation that we have slipped into, which has obscured the unified nature of all inference.”
“From the start, it has seemed clear how one how one determines numerical values of of sampling probabilities¹ [e.g. P(D|H) ], but not what determines prior probabilities [AKA ‘priors’ e.g. P(H|X)]. In the present work we shall see that this s only an artifact of the unsymmetrical way of formulating problems, which left them ill-posed. One could see clearly how to assign sampling probabilities because the hypothesis H was stated very specifically; had the prior information X been specified equally well, it would have been equally clear how to assign prior probabilities.”
Jaynes never gives up on that X notation (though the letter may differ), he never drops it for convenience.
“When we look at these problems on a sufficiently fundamental level and realize how careful one must be to specify prior information before we have a well-posed problem, it becomes clear that … exactly the same principles are needed to assign either sampling probabilities or prior probabilities …” That is, P(H|X) should be calculated. Keep your copy of Kendall and Stuart handy.
I think priors should not be cheaply set from an opinion, whim, or wish. “ … it would be a big mistake to think of X as standing for some hidden major premise, or some universally valid proposition about Nature.”
The prior information has impact beyond setting prior probabilities (priors). It informs the formulation of the hypotheses, of the model, and of “alternative hypotheses” that come to mind when the data seem to be showing something really strange. For example, data that seems to strongly support psychokinesis may cause a skeptic to bring up a hypothesis of fraud, whereas a career psychic researcher may not do so. (see Jaynes pp.122-125)
I say, be alert for misinformation, biases, and wishful thinking in your X. Discard everything that is not evidence.
I’m pretty sure the free version Probability Theory: The Logic of Science is off line. You can preview the book here: http://books.google.com/books?id=tTN4HuUNXjgC&printsec=frontcover&dq=Probability+Theory:+The+Logic+of+Science&cd=1#v=onepage&q&f=false .
FOOTNOTES
1. There are massive compendiums of methods for sampling distributions, such as
• Feller (An Introduction to Probability Theory and its Applications, Vol1, J. Wiley & Sons, New York, 3rd edn 1968 and Vol 2. J. Wiley & Sons, New York, 2nd edn 1971) and Kendall and
• Stuart (The Advanced Theory of Statistics: Volume 1, Distribution Theory, McMillan, New York 1977).
** Be familiar with what is in them.
Edited 05/05/2010 to put in the actual references.
• Then the task is just to turn this new distribution into a computer program, which turns out not to be difficult.
Can someone please provide a hint how?
• Here’s some Python code to calculate a prior distribution from a rule for assigning probability to the next observation.
A “rule” is represented as a function that takes as a first argument the next observation (like “R”) and as a second argument all previous observations (a string like “RRWR”). I included some example rules at the end.
EDIT: oh man, what happened to my line spacing? my indents? jeez.
EDIT2: here’s a dropbox link: https://www.dropbox.com/s/16n01acrauf8h7g/prior_producer.py
``````from functools import reduce
def prod(sequence):
‴Product equivalent of python’s “sum”‴
return reduce(lambda a, b: a*b, sequence)
def sequence_prob(rule, sequence):
‴Probability of a sequence like “RRWR” using the given rule for
computing the probability of the next observation.
To put it another way: computes the joint probability mass function.‴
return prod([rule(sequence[i], sequence[:i]) \
for i in range(len(sequence))])
def number2sequence(number, length):
‴Convert a number like 5 into a sequence like WWRWR.
The sequence corresponds to the binary digit representation of the
number: 5 --> 00101 --> WWRWR
This is convenient for listing all sequences of a given length.‴
binary_representation = bin(number)[2:]
seq_end = binary_representation.replace(‘1’, ‘R’).replace(‘0’, ‘W’)
if len(seq_end) > length:
raise ValueError(‘no sequence of length {} with number {}‘\
.format(length, number))
# Now add W’s to the beginning to make it the right length -
# like adding 0’s to the beginning of a binary number
return ″.join(‘W’ for i in range(length—len(seq_end))) + seq_end
def prior(rule, n):
‴Generate a joint probability distribution from the given rule over
all sequences of length n. Doesn’t feed the rule any background
knowledge, so it’s a prior distribution.‴
sequences = [number2sequence(i, n) for i in range(2**n)]
return [(seq, sequence_prob(rule, seq)) for seq in sequences]
``````
And here’s some examples of functions that can be used as the “rule” arguments.
``````def laplaces_rule(next, past):
R = past.count(‘R’)
W = past.count(‘W’)
if R + W != len(past):
raise ValueError(‘knowledge is not just of red and white balls’)
red_prob = (R + 1)/(R + W + 2)
if next == ‘R’:
return red_prob
elif next == ‘W’:
return 1 - red_prob
else:
raise ValueError(‘can only predict whether next will be red or white’)
def antilaplaces_rule(next, past):
return 1 - laplaces_rule(next, past)
``````
• So just to be clear. There are two things, the prior probability, which is the value P(H|I), and the back ground information which is ‘I’. So P(H|D,I_1) is different from P(H|D,I_2) because they are updates using the same data and the same hypothesis, but with different partial background information, they are both however posterior probabilities. And the priors P(H_I_1) may be equal to P(H|I_2) even if I_1 and I_2 are radically different and produce updates in opposite directions given the same data. P(H|I) is still called the prior probability, but it is smething very differnet from the background information which is essentially just I.
Is this right? Let me be more specific.
Let’s say my prior information is case1, then P( second ball is R| first ball is R & case1) = 49
If my prior information was case2, then P( second ball is R| first ball is R & case2) = 23 [by the rule of succession]
and P( first ball is R| case1) = 50% = P( first ball is R|case2)
This is why different prior information can make you learn in different directions, even if two prior informations produce the same prior probability?
Please let me know if i am making any sort of mistake. Or if I got it right, either way.
• You got it right. The three different cases correspond to different joint distributions over sequences of outcomes. Prior information that one of the cases obtains amounts to picking one of these distributions (of course, one can also have weighted combinations of these distributions if there is uncertainty about which case obtains). It turns out that in this example, if you add together the probabilities of all the sequences that have a red ball in the second position, you will get 0.5 for each of the three distributions. So equal prior probabilities. But even though the terms sum to 0.5 in all three cases, the individual terms will not be the same. For instance, prior information of case 1 would assign a different probability to RRRRR (0.004) than prior information of case 2 (0.031).
So the prior information is a joint distribution over sequences of outcomes, while the prior probability of the hypothesis is (in this example at least) a marginal distribution calculated from this joint distribution. Since multiple joint distributions can give you the same marginal distribution for some random variable, different prior information can correspond to the same prior probability.
When you restrict attention to those sequences that have a red ball in the first position, and now add together the (appropriately renormalized) joint probabilities of sequences with a red ball in the second position, you don’t get the same number with all three distributions. This corresponds to the fact that the three distributions are associated with different learning rules.
• No really, i really want help. Please help me understand if I am confused, and settle my anxiety if I am not confused.
• One can update one’s beliefs about one’s existing beliefs and the ways in which one learns from experience too – click.
• Under standard assumptions about the drawing process, you only need 10 numbers, not 1024: P(the urn initially contained ten white balls), P(the urn initially contained nine white balls and one red one), P(the urn initially contained eight white balls and two red ones), and so on through P(one white ball and nine red ones). (P(ten red balls) equals 1 minus everything else.) P(RWRWWRWRWW) is then P(4R, 6W) divided by the appropriate binomial coefficient.
• So then this initial probability estimate, 0.5, is not repeat not a “prior”.
This really confuses me. Considering the Universe in your example, which consists only of the urn with the balls, wouldn’t one of the prior hypotheses(e.g. case 2) be a prior and have all the necessary information to compute the lookup table?
In other words aren’t the three following equivalent in the urn-with-balls universe?
1. Hypothesis 2 + bayesian updating
2. Python program 2
3. The lookup table generated from program 2 + Procedure for calculating conditional probability(e.g. if you want to know the probability that the third ball is red, given that the first two balls drawn were white.)
• Unless I am misunderstanding you, yes, that’s precisely the point.
I don’t understand why you are confused, though. None of these are, after all, numbers in (0,1), which would not contain any information as to how you would go about doing your updates given more evidence. | 9,602 | 28,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-24 | latest | en | 0.901494 |
https://math.stackexchange.com/questions/199080/calculating-the-shapley-value-in-a-weighted-voting-game | 1,716,163,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00096.warc.gz | 335,944,548 | 35,742 | # Calculating the Shapley value in a weighted voting game.
Given a special case of WVG (Weighted Voting Game) of $a$ 1s and $b$ 2s and a quota q, $[q:1,1,1,1..1,2,2,..2]$. I need help with calculating the Shapley value of a player with a weight of $2$ and a player with a weight of $1$ as a function of $a$,$b$ and $q$.
I know how to calculate the Shapely value in general, but I would like to get a simple closed form for the result. I tried to think about what happens if there are only players with weights of $1$ (simple scenario $a=2k-1,b=0,q=k$ I think you will agree that $\phi_{1}(v)=\frac{1}{a}=\frac{1}{2k-1}$. the simple calculation is $\binom{2k-2}{k-1}(k-1)!(k-1)!=\frac{1}{2k-1}=\frac{1}{a}$ which is $\frac{1}{2k-1}$. The calculation for a player of weight 1 (when there are 1s and 2s) would be a sum of choosing i players of weight 1 and $\frac{q-i}{2}$ players of weight 2. similar calculation for a player of weight 2.
I tagged binomial coefficients because I thought they could be useful for counting the occurrences.
Thanks in advance, Mati
• Joriki, please refer to the edit in my question. Thanks. Sep 19, 2012 at 14:06
• Thanks for the quick response. I've removed my comments and downvote (and in fact upvoted). I doubt that there's a nice closed form, though. The result will depend on the quota if the quota is close to $0$ or $a+2b$, but for small $b$ there will be a plateau of intermediate quota where the result doesn't change with the quota. Sep 19, 2012 at 14:23
## 1 Answer
Note that the Shapley value for this game is the probability that a player is pivotal in a random ordering. Now the probability that a 1 player is pivotal is the probability that the sum is q-1 when she is reached. While for a 2 player it is the probability that it is q-1 or q-2. This should simplify the analysis, but I haven't worked out the details.
However a simple conclusion is that the Shapley value for 1 players is approximately 1/2 that for 2 players for reasonable size values and away from extremes, so the values are approximately 1/(a+2b) and 2/(a+2b) respectively. In particular this holds in the limit where a,b go to infinity and q=y(a+2b) for 05 and 2/a | 644 | 2,187 | {"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} | 3.84375 | 4 | CC-MAIN-2024-22 | latest | en | 0.951667 |
http://www.hindawi.com/journals/aaa/2013/405258/ | 1,394,405,421,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-10/segments/1394010450813/warc/CC-MAIN-20140305090730-00085-ip-10-183-142-35.ec2.internal.warc.gz | 378,259,140 | 121,685 | `Abstract and Applied AnalysisVolume 2013 (2013), Article ID 405258, 17 pageshttp://dx.doi.org/10.1155/2013/405258`
Research Article
## Mathematical Analysis of a Malaria Model with Partial Immunity to Reinfection
1Department of Mathematics, Xinyang Normal University, Xinyang 464000, China
2Centre for Advanced Mathematics and Physics, National University of Sciences and Technology, H-12, Islamabad 44000, Pakistan
3Department of Mathematics, Pusan National University, Busan 609-735, Republic of Korea
4Department of Mathematics, Vaal University of Technology, Andries Potgieter Boulevard, X021, Vanderbijlpark 1900, South Africa
5National Fisheries Research and Development Institute, Busan 619-705, Republic of Korea
Received 5 October 2012; Accepted 7 December 2012
Academic Editor: Sanyi Tang
Copyright © 2013 Li-Ming Cai et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
A deterministic model with variable human population for the transmission dynamics of malaria disease, which allows transmission by the recovered humans, is first developed and rigorously analyzed. The model reveals the presence of the phenomenon of backward bifurcation, where a stable disease-free equilibrium coexists with one or more stable endemic equilibria when the associated reproduction number is less than unity. This phenomenon may arise due to the reinfection of host individuals who recovered from the disease. The model in an asymptotical constant population is also investigated. This results in a model with mass action incidence. A complete global analysis of the model with mass action incidence is given, which reveals that the global dynamics of malaria disease with reinfection is completely determined by the associated reproduction number. Moreover, it is shown that the phenomenon of backward bifurcation can be removed by replacing the standard incidence function with a mass action incidence. Graphical representations are provided to study the effect of reinfection rate and to qualitatively support the analytical results on the transmission dynamics of malaria.
#### 1. Introduction
Malaria is a mosquito-borne disease caused by a parasite. It is endemic and widespread in tropical and subtropical regions, including much of sub-Saharan Africa, Asia, and the Americas. Malaria is still a public health problem today. Every year, there are more than 225 million cases of malaria, killing around 781,000 people according to the World Health Organization’s 2010 World Malaria Report [1].
In humans, malaria is caused due to infection by one of four Plasmodium species [2, 3]. Transmission from mosquito to human occurs during a bite by an infectious mosquito. A mosquito becomes infected when it takes a blood meal from an infected human. Once ingested, the parasite gametocytes taken up in the blood will further differentiate into gametes and then fuse in the mosquito’s gut. Gametocytes are responsible for transmission of the parasite from humans by mosquitoes bite. Fertilization of the parasite occurs in the mosquito gut, and after a short period of replication and development, the cycle of transmission may begin anew.
One of the most complex features of the epidemiology of malaria is the dynamic interaction between infection and immunity. A better understanding of this interaction is important for evaluating the impact of malaria control activities. An important phenomenon is noticed that the changes with age reflect the slow acquisition of an immunity that reduces illness but does not completely block infection [4, 5]. In endemic areas, children younger than five years have repeated and often serious attacks of malaria. The survivors develop and maintain partial immunity that reduces the severity of the disease but does not prevent subsequent infections. Thus, in these areas older children and adults often have become asymptomatic carriers of infection [6]. In areas of low malaria transmission, immunity develops slowly and may take years or decades and probably never results in sterile immunity [7]. Therefore, humans are susceptible to reinfections. Incomplete immunity to malaria complicates disease control strategies [8, 9] as the partially immune individuals suffer only mild infections, and might not seek medical attention but continue to transmit the parasite in the community.
The enormous public health burden inflicted by malaria disease necessitates the use of mathematical modeling and analysis to gain insights into its transmission dynamics, and to determine effective control strategies. The earliest malaria transmission models can be traced to the model formulated by Ross in 1911 [9]. He used a mathematical model and showed that bringing a mosquito population below a certain threshold was sufficient to eliminate malaria. This threshold naturally depended on biological factors such as the biting rate and vectorial capacity. To estimate infection and recovery rates, MacDonald extended the Ross model in 1957 [9]. Macdonald's model shows that reducing the number of mosquitoes is an inefficient control strategy. Moreover, this would have little effect on the epidemiology of malaria in areas of intense transmission. Since then, the emergence and reemergence of malaria diseases have promoted many author’s interest in mathematical modeling to describe and to predict the transmission dynamics of malaria in the literature (see, e.g., [1016] and the references therein). In paper [12], Dietz et al. applied the Garki model to show that the duration of acquired immunity in humans in malaria depends on repeated exposure. In paper [13], Niger and Gumel constructed a mathematical model that includes multiple infected and recovered classes, to assess the role of the partial immunity on the transmission dynamics of malaria in a human population. Their results reveals the presence of the phenomenon of backward bifurcation in the standard incidence model with the disease-induced death in the human population. Recently, a transmission model of human malaria in a partially immune population is formulated in Wan and Cui’s paper [14]. They established the basic reproduction number and explicit subthreshold conditions for the model, and showed that if the disease induced death rate is large enough, the model undergoes a backward bifurcation. Li [15] formulated a malaria transmission model with partial immunity in humans and showed that the established model having the same reproductive number but different numbers of progression stages can exhibit different transient dynamics. Thus, the above mentioned models always let the recovered individuals return into the susceptible class to explore the transmission dynamics of diseases. But this only takes states of complete immunity and full susceptibility in consideration. In addition, various vector-borne disease model concerning malaria transmission have been established and discussed [1720]. For example, in paper [17], Yang et al have investigated global stability of an epidemic model for vector-borne disease, however, they assumed that the immunity of the recovered population have never lose.
Motivated by the recent work of [13, 15], in this paper, we shall continue to construct a malaria transmission model with partial immunity to reinfection in the recovered human population. Our purpose is to explore the transmission dynamics of the malaria and to assess the role of partial immunity to reinfection on the transmission dynamics of malaria in a human population.
The organization of this paper is as follows: in the next section, the standard incidence malaria model, which incorporates the partial immunity to reinfection, is formulated. The existence and stability of the equilibria, and the phenomena of the backward bifurcation are, respectively, explored in Sections 2.2 and 2.3. Graphical representations are provided to study the effect of reinfection rate in Section 2.4. In Section 3, the associated mass action incidence model is formulated, and mathematical results such as existence and local stability of equilibria are provided in Section 3.2. Our main theorems for the global stability of equilibria for the mass action model and the proofs are given in Section 3.3. The paper ends with a conclusion in Section 4.
#### 2. Model Formulation
We formulate a model for the spread of malaria in the human and mosquito population, with the total population size at time given by and , respectively. The total human population is divided into three epidemiological classes: , and , which denote, respectively, the number of the susceptible, infective, and immune class at time . Thus, . The susceptible human population is generated by the recruitment of humans (assumed susceptible) into the community at a rate , , and are, respectively, the natural death rate and recovery rate in human hosts population. Also, some disease-induced death in human population contributes to an additional population decrease at the constant rate .
Due to its short life, a mosquito never recovers from the infection, and we may not consider the recovered class in this population. Thus, the total vector population is divided into the susceptible class, , and infective class, , so that . Susceptible mosquitoes vectors are generated at a rate by birth, is the per capita mortality rate of mosquitoes. Let be the transmission probability from vector to human, and be the transmission probability from human to vector. The parameter is the average number of bites per mosquito per day. This rate depends on a number of factors, in particular, climatic ones, but for simplicity in this paper we assume to be a constant. The parameter determines the degree of partial protection for the recovered individuals given by a primary infection: implies complete protection, and implies no protection. Taking into account the assumptions made above, the interaction between human hosts and the mosquito vector population with partial immunity to reinfection in host population is described by the following system of equations: The total humans host and mosquitoes vector populations and are governed, respectively, by It is easily seen that for the mosquitoes vector population the corresponding total population size is asymptotically constant: . This implies that in our model we assume without loss of generality that , for all , provided that . Let Since and . Using and , system (1) is reduced to the following four-dimensional nonlinear system of ODEs: where and .
##### 2.1. Basic Properties of the Model
Since the model (4) monitors humans host and mosquitoes vector populations, it is plausible to assume that all its state variables and parameters are nonnegative for all . Further, it can be shown that the region given by is positively invariant with respect to system (4). Thus, every solution of the model (4), with initial conditions in remains there for . Therefore, it is sufficient to consider the dynamics of the flow generated by (4) in . In this region, the model can be considered as been epidemiologically and mathematically well posed.
##### 2.2. Stability of Disease-Free Equilibria
The stability of the disease-free equilibrium state can be obtained from studying the eigenvalues of the Jacobian matrix evaluated at the equilibrium point. If all the eigenvalues have negative real parts, then the equilibrium point is stable. The disease-free equilibrium for the system (4) is . The Jacobian matrix at the disease-free equilibrium is The characteristic equation of the above matrix is where . There are four eigenvalues corresponding to (7). Two of the eigenvalues , have negative real parts. The other two eigenvalues can be obtained from the equation Applying the Routh-Hurwitz criteria for a quadratic polynomial. It is easy to see that both the coefficients of (8) are positive if an only if . Thus, all roots of (8) are with negative real parts if , and one of its roots is with positive real part if . Therefore, the disease-free equilibrium (DFE) is locally asymptotically stable if and unstable if . Thus, we have the following result.
Theorem 1. The uninfected equilibrium is locally asymptotically stable if and unstable if in .
From Theorem 1, the threshold quantity , is called the basic reproduction number of system (4). The basic reproduction number, measures the average number of new malaria infections generated by a single infected individual in a completely susceptible population [21]. Theorem 1 also implies that malaria can be eliminated from the community (when ) if the initial sizes of the subpopulations of the model are in the basin of attraction of the disease-free equilibrium (DFE) (). To ensure that disease elimination is independent of the initial sizes of the subpopulations, it is necessary to show that the DFE is globally asymptotically stable (GAS) if . This is explored for a special case in Section 3.2.
##### 2.3. The Existence of Endemic Equilibria and Backward Bifurcation
In this section, conditions for the existence of endemic equilibria and phenomenon of backward bifurcation in system (4) will be determined. In order to do this, we let represent an arbitrary endemic equilibrium of the model (4). Setting the right-hand sides of the equations in (4) to zero and solving them in terms of gives the following expressions for the state variables of the model where , and is determined from the following equation:
where and .
It follows from (13) that . Further, whenever . Thus, the number of possible positive real roots for (12) depends on the signs of , , and . This can be analyzed using the Descartes Rule of Signs on the quartic . The various possibilities for the roots of are tabulated in Table 1. We have the following result from the various possibilities enumerated in Table 1.
Table 1: Number of possible positive real roots of for and .
Theorem 2. The system (4) has a unique endemic equilibrium if and Cases 1–3 and are satisfied; it could have more than one endemic equilibrium if and Cases , , , and are satisfied; it could have or more endemic equilibria if and Cases 2–8 are satisfied.
The existence of multiple endemic equilibria when (is shown in Table 1). Table 1 suggests the possibility of backward bifurcation (see [2224]), where the stable DFE coexists with a stable endemic equilibrium, when the reproduction number is less than unity. Thus, the occurrence of a backward bifurcation has an important implications for epidemiological control measures, since an epidemic may persist at steady state even if . This is explored below by using Centre Manifold Theory (see, e.g., [25] and the references therein).
Now, we shall establish the conditions on parameter values that cause a backward bifurcation to occur in system (4), based on the use of Center Manifold theory, of the paper in Castillo-Chavez and Song [25].
Theorem 3. Let one consider the following general system of ordinary differential equations with a parameter : Without loss of generality, it is assumed that is an equilibrium for system (14) for all values of the parameter . Assume that is the linearized matrix of system (14) around the equilibrium with evaluated at . Zero is a simple eigenvalues of and all other eigenvalue of have negative real parts; Matrix has a nonnegative right eigenvector and a left eigenvector corresponding to the zero eigenvalue.
Let be the kth component of and The local dynamics of system (14) around are totally determined by and . (i)In the case where , , one has that when with close to zero, is unstable; when , is unstable and there exists a negative and locally asymptotically stable equilibrium; (ii)In the case where , , one has that when with close to zero, is locally asymptotically stable and there exists a positive unstable equilibrium; when , is locally asymptotically stable, and there exists a positive unstable equilibrium; (iii)In the case where , , one has that when with close to zero, is unstable and there exists a locally asymptotically stable negative equilibrium; when , is stable and a positive unstable equilibrium appears; (iv)In the case where , , one has that when changes from negative to positive, changes its stability from stable to unstable. Correspondingly negative unstable equilibrium becomes positive and locally asymptotically stable. Particularly, if and , then a backward bifurcation occurs at .
To apply the center manifold method, the following simplification and change of variables are made on the model (4). First of all, let , , , , and , so that and . Further, by using the vector notation , the system (4) can be written in the form as follows:
Choose as a bifurcation parameter and solving gives The Jacobian matrix evaluated at disease-free equilibrium with is It can be easily seen that the Jacobian of the linearized system has a simple zero eigenvalue and all other eigenvalues have negative real parts. Hence, the center manifold theory can be used to analyze the dynamics of the system (16). For the case when , it can be shown that the Jacobian matrix has a right eigenvector (corresponding to the zero eigenvalue) given by , where Similarly, the components of the left eigenvector of (corresponding to the zero eigenvalue), denoted by , are given by Computation of : for the transformed system (16), the associated non-zero partial derivatives of (evaluated at the DFE) which we need in the computation of are given by Direct calculations shows that
Computation of : Substituting the vectors and and the respective partial derivatives (evaluated at the DFE ) into the expression gives . Since the coefficient is automatically positive, it follows that the sign of the coefficient decides the local dynamics around the disease-free equilibrium for . Based on Theorem 3, system (4) will undergo backward bifurcation if the coefficient is positive. The coefficient is positive if and only if Thus, we have the following result.
Theorem 4. The system (4) exhibits backward bifurcation whenever the condition (24) holds.
The backward bifurcation phenomenon is illustrated by simulating the system (4) with the following set of parameter values , , , , , , , (so that, and . Figure 1 depicts the associated backward bifurcation diagram.
Figure 1: Simulations of the model (4) illustrating the phenomenon of backward bifurcation.
##### 2.4. The Effect of the Reinfection
We further investigate the effect of the reinfection parameter and the transmission probability from an infectious human to a susceptible vector on the associated backward bifurcation region, as a function of the average life span of mosquitoes . The backward bifurcation region is illustrated (Figures 24) by simulating the model (4) with the following set of parameter values (note that the parameters are chosen in order to illustrate the backward bifurcation region, and may not all be realistic epidemiologically), . Also to be noted is, the parameter values are chosen such that and (so that backward bifurcation occurs).
Figure 2: Backward bifurcation regions for the model (4) in the - parameter space corresponding to and various ranges of . Parameter values used are: , . In (a) (backward bifurcation region for is , (b) (backward bifurcation region for is , and (c) (backward bifurcation region for is . With the above set of parameter values, , and .
Figure 3: Backward bifurcation regions for the model (4) in the - parameter space corresponding to and various ranges of . Parameter values used are: ,. In (a) (backward bifurcation region for is , (b) (backward bifurcation region for is , and (c) (backward bifurcation region for is . With the above set of parameter values, , and .
Figure 4: Backward bifurcation regions for the model (4) in the - parameter space corresponding to and various ranges of . Parameter values used are: ,. In (a) (backward bifurcation region for is , (b) (backward bifurcation region for is , and (c) (backward bifurcation region for is . With the above set of parameter values, , and .
Solving for in terms of and (i.e., fixing all parameters in the expression for except and ) we obtained the backward bifurcation region for . Figure 2 depicted the results obtained for , it shows that the region for backward bifurcation (for ) increases as the average life span of vectors decreases. For instance, when the average life span of vectors is 20 days , the backward bifurcation region for is , as shown in Figure 2(a). When the average life span of vectors is decreased to 10 days , the backward bifurcation region for increases to (Figure 2(b)). Furthermore, when the average life span of vectors is decreased to 5 days , the backward bifurcation region for increases to (Figure 2(c)). Similar results are obtained for the cases (Figures 3(a), 3(b), and 3(c)) and (Figures 4(a), 4(b), and 4(c)), from which it is evident that the backward bifurcation regions for increase with increasing values of the reinfection rate . These results are tabulated in Table 2 ( represents average life span of vectors.)
Table 2: Backward Bifurcation Ranges for for Various Values of and .
#### 3. The Mass Action Model
In this section, we shall investigate the dynamics of system (1) if mass action incidence is used instead of the standard incidence function. Thus the resulting (mass action) model is where the prime stands for the derivative with respect to time and initial conditions , and .
##### 3.1. Basic Properties: Positivity and Invariant Regions
The dynamics of the total human population, obtained by adding first three equations in the model (25), is given by Thus, we have Thus, for a low level of disease induced death rate () total human population could eventually assume a steady-state value. Motivated by this, we consider a human population which assumes a steady-state value stationary. Similarly, the dynamics of the total mosquito population, obtained by adding last two equations in the model (25), is given by, , so that, as .
Let and . Based on the above discussion, we define a region It is easy to verify that is positively invariant with respect to the system (25). In this part, it is sufficient to consider the dynamics of the flow generated by (25) in .
##### 3.2. Equilibrium and Local Stability
In this section, we investigate the existence and local stability of equilibria of system (25). Obviously, the system (25) always has a disease-free equilibrium . Let represents any arbitrary endemic equilibrium of the model (25). Solving the equations in (25) at steady state gives where is the positive root of the following quadratic equation, with The dynamics of the model (25) are analyzed by given by The threshold quantity is the basic reproduction number of the system (25). It can be derived from the Jacobian matrix of the system (25) at the disease-free equilibrium together with the assumption of local asymptotical stability of [21]. From (31), we see that if and only if, . Since , (30) has a unique positive root in feasible region . If , then . Also , is equivalent to . Hence, . Thus, by considering the shape of the graph (and noting that ), we have that there will be zero endemic equilibria in this case. Therefore, we can conclude that if , (30) has no positive root in the feasible region . If, , (30) has a unique positive root in the feasible region . This result is summarized below.
Theorem 5. System (25) always has the infection-free equilibrium . If , system (25) has a unique endemic equilibrium defined by (29) and (30).
Linearizing the system (25) around the disease-free equilibrium yields the following characteristic equation: Two of the roots of the characteristic equation (33) , have negative real parts. The other two roots can be determine from the quadratic term in (33) and have negative real parts if and only if . Therefore, the disease-free equilibrium is locally asymptotically stable for . When , becomes an unstable equilibrium point, and the endemic equilibrium emerges in . This result is summarized below.
Theorem 6. The disease-free equilibrium of system (25) is locally asymptotically stable if and unstable if .
In order to discuss the stability of the endemic equilibrium and to simplify our calculations, we assume both humans and mosquitoes populations are at steady state. Thus, using , and , system (25) in the invariant space can be written as the following equivalent three dimensional nonlinear system of ODEs: Now, linearization of system (34) about an endemic equilibrium gives the following characteristic equation: where, . Expanding (35) gives where From the second equation of system (34) at steady state , we have Using (37), direct calculations show that Obviously, the term in first bracket times the term in the second bracket is . Multiplying the terms under straight line and using (30), we have . Hence, . Thus, by Routh Hurwitz criteria, the following result is established.
Theorem 7. The endemic equilibrium of the reduced model (34) is locally asymptotically stable if .
##### 3.3. Global Stability of the Equilibria
In this section, the global stability of the equilibria of system (34) will be explored. First, we claim the following theorem.
Theorem 8. If , then the infection-free-equilibrium of system (34) is globally asymptotically stable in .
Proof. To establish the global stability of the disease-free equilibrium , we construct the following Lyapunov function Calculating the derivative of (where a dot represents differentiation with respect to ) along the solutions of (34) we obtain
Thus, if with if and only if . Thus, from the second and the first equation of system (34), we have , and . Therefore, the largest compact invariant set in is the singleton in . Using the LaSalle’s invariant principle [26], the infection-free equilibrium is globally asymptotically stable for in . The epidemiological implication of the above result is that malaria will be eliminated from the population if can be brought to (and maintained at) a value less than unity. Thus, the substitution of standard incidence with mass action incidence in the model (1) removes the phenomenon of backward bifurcation. The result of Theorem 8 is illustrated numerically by simulating the model (30), for the case , using various initial conditions. The solution profiles obtained shows convergence to the DFE, as depicted in Figure 5.
Now, we investigate the global stability of the endemic equilibrium . We notice that when the incomplete immunity term , system (30) is no longer competitive. To investigate the global stability of , we adopted a general approach of Li and Muldowney [27, 28], which is developed for higher dimensional systems irrespective if they are competitive. While the approach of Li and Muldowney has been successfully applied to many classes of epidemic models, we demonstrated in the present paper, for the first time, that this approach is also applicable to vector-host model which is non-competitive.
Figure 5: Simulations of the model (5) showing (a) the number of susceptible humans, (b) the number of infected humans, (c) the number of infected mosquitoes, as a function of time using the parameters: , , , , , , , , and (so that ).
We briefly state the approach developed recently in Li and Muldowney as follows:
Let be an open set and be a function for . Consider the following differential equation: Let denote the solution of (42) satisfying . We make the following two assumptions. There exists a compact absorbing set . Eqution (42) has a unique equilibrium in .
Let be an matrix-valued function, that is, it is and exists for . Let be a Lozinski measure on , where . Define a quantity as where , the matrix is obtained by replacing each entry of by its derivative in the direction of , , and is the second additive compound matrix of the Jacobian matrix of system (42). The following results have been established in Li and Muldowney [27, 28].
Theorem 9. For system (42), assume that is a simple connected and that the assumptions () and () hold. Then, the unique equilibrium is globally asymptotically stable in if there exist a function and a Lozinski measure such that .
From the above discussion, we know that is simply connect and is the unique positive equilibrium for in .
To apply the result of Theorem 9 to investigate the global stability of the infective equilibrium , we first state and prove the following result.
Theorem 10. If , then system (34) is uniformly persistent, that is, there exists (independent of initial conditions), such that , and .
Proof. Similar to the proof of Theorem 3.4 in [29], we choose . It is easy to obtain that , and is an isolated compact invariant set in . Furthermore, let , thus, is an acyclic isolated covering of .
Now, we only need to show that is a weak repeller for . Suppose that there exists a positive orbit of (34) such that Since , there exists a small enough , such that From (34), we choose large enough such that when , we have Consider the following matrix defined by Since admits positive off-diagonal element, the Perron-Frobenius Theorem [26] implies that there is a positive eigenvector for the maximum eigenvalue of . From (45), we see that the maximum eigenvalue is positive. Let us consider the following system: Let be a solution of (48) through at , where satisfies . Since the semiflow of (48) is monotone and , it follows that are strictly increasing and as , contradicting the eventual boundedness of positive solutions of system (48). Thus, is weak repeller for . The proof is completed.
From Theorem 10 and the boundedness of solutions, it follows that a compact set exists in system (34). Therefore, in Theorem 9, both assumptions () and () are satisfied for .
Now, we apply Theorem 9 to investigate the global stability of the endemic equilibrium in the feasible region . For the global stability of the endemic equilibrium , we have the following theorem.
Theorem 11. If , then the infective equilibrium of system (48) is globally asymptotically stable in .
Proof. The Jacobian matrix evaluated at a general solution of system (34) is and its corresponding second compound matrix takes the form Set the function . Then . Moreover, where , , , and Let be the vectors in . We choose a norm in as , and let be the corresponding Lozinski measure. From the paper [28], we have the following estimate: where Here, are matrix norm with respect to the vector norm, and is the Lozinski measure with respect to norm. Thus, we have According to paper [28], can be evaluated as follows: Thus, From system (34), we have Using (58) in (57) gives From Theorem 9, we know that for the uniform persistence constant , there exists a time independent of , the compact absorbing set, such that for . Thus, from (59) and (60), we get Therefore, from (61), we have for , where Then along each solution such that for , give the following: which implies that . This proves that the unique infective equilibrium is globally asymptotically stable whenever it exists.
This completes the proof.
Remark 12. The epidemiological implication of Theorem 11 is that malaria would persist in the population if . Theorem 11 is illustrated numerically by simulating the model (25), for the case using various initial conditions. The convergence of the solutions to for the case , is depicted in Figure 6.
Figure 6: Simulations of the model (5) showing (a) the number of susceptible humans, (b) the number of infected humans, (c) the number of infected mosquitoes, as a function of time using the parameters: , , , , , , , , , and | 6,612 | 31,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-10 | longest | en | 0.867854 |
https://stackoverflow.com/questions/33474206/add-custom-legend-without-any-relation-to-the-graph | 1,585,553,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00375.warc.gz | 711,632,194 | 32,834 | # Add custom legend without any relation to the graph
I wish to insert a legend that is not related to the graph whatsoever:
``````figure;
hold on;
plot(0,0,'or');
plot(0,0,'ob');
plot(0,0,'ok');
leg = legend('red','blue','black');
``````
Now I wish to add it to another figure:
``````figure;
t=linspace(0,10,100);
plot(t,sin(t));
%% ADD THE LEGEND OF PLOT ABOVE
``````
• this is a simple example of a bigger thing I have so what you suggest is not relevant, there must be a way to do it – jarhead Nov 2 '15 at 9:22
• So it is not clear what you want to achieve. Can you provide more information? Why do you need legend if there is no related data in the plot. – NKN Nov 2 '15 at 9:24
• what I want to achieve is exactly what I wrote, use the box with the legend of the first figure, and have it displayed on the second – jarhead Nov 2 '15 at 9:28
• 2 hacky options: (1) add empty `lineserie` (or any empty graphic object) in the `axes` so that there are enough to populate the legend. (2, better in my view) make your own `legend` object (it's only a special `axes` object after all) which you can then move/copy/modify freely within your figures. Example on how to do that can be found here: plot-legend-title. – Hoki Nov 2 '15 at 9:40
• You can even use `NaN` instead of `0` for your plot data. This way you don't even have to touch the `visible` property, nothing will be displayed anyway (the legend will still accept the data serie as a valid entry). – Hoki Nov 2 '15 at 9:50
This is how I have solved this problem in the past:
``````figure
t=linspace(0,10,100);
plot(t,sin(t));
hold on;
h = zeros(3, 1);
h(1) = plot(NaN,NaN,'or');
h(2) = plot(NaN,NaN,'ob');
h(3) = plot(NaN,NaN,'ok');
legend(h, 'red','blue','black');
``````
This will plot the additional points, but because the coordinates are at `NaN` they will not be visible on the plot itself:
EDIT 26/10/2016: My original answer results in greyed out legend entries in 2016b. The updated code above works, but the answer below is only relevant pre-2016b:
``````figure
t=linspace(0,10,100);
plot(t,sin(t));
hold on;
h = zeros(3, 1);
h(1) = plot(0,0,'or', 'visible', 'off');
h(2) = plot(0,0,'ob', 'visible', 'off');
h(3) = plot(0,0,'ok', 'visible', 'off');
legend(h, 'red','blue','black');
``````
This will plot the additional points, but they will not be visible on the plot itself.
You can also use `copyobj` to copy graphics elements from one figure to another if you have a lot of elements, then use `set(x, 'visible', 'off')` to hide them before showing the legend, but it depends on what your final application is.
• as commented above u can use NaN, and then u do not need to use the visible handle – jarhead Nov 2 '15 at 11:17
• @vindarmagnus, ouch, yes, you're right! I have updated my answer to use `NaN` coordinates instead of invisible plots, as suggested by jarhead and Hoki in the comments. – zelanix Oct 26 '16 at 10:59
Your question is a little unclear. However, the first thing I thought of when reading it was the `text` function in Matlab.
You can use the `text` function to add text to a Matlab figure. It's use is
``````>> text(x, y, str);
``````
where `x` and `y` are the coordinates in the figure where you want to add the text `str`. You can use the `Color` option of `text` for colours and TeX to draw lines or even `_`. I've gotten very creative with plots using text.
Here's a quick and dirty example of emulating a `legend` with `text`
``````x = 0:pi/20:2*pi;
y = sin(x);
plot(x,y)
axis tight
legend('sin(x)');
text(5.7, 0.75, 'sin(x)');
text(5.1, 0.78, '_____', 'Color', 'blue');
``````
which produces
For this specific case you could use the specific command (noted by @Hoki in the comments).
``````ht = text(5, 0.5, {'{\color{red} o } Red', '{\color{blue} o } Blue', '{\color{black} o } Black'}, 'EdgeColor', 'k');
``````
to produce
by retrieving the handle to the `text` object it becomes trivial to copy it to a new figure, `copyobj(ht, newfig)`. [1]
• However `text` does not provide the color and format that legend provides. – NKN Nov 2 '15 at 9:35
• You can use the `Color` option of `text` for colours and TeX to draw lines or even `_`. I've gotten very creative with plots using `text`. – IKavanagh Nov 2 '15 at 9:40
• Sounds interesting, could you give a small example of a `TeX` and `Color` syntax in `text` objects ? – Hoki Nov 2 '15 at 9:43
• Nice. You can also modify colour partially in one text zone (with TeX color markup), and have multiple lines of text in one text object. So you could define the legend in one go !. Try this : `ht = text(5,0,{'{\color{red} o } Red (serie 1)','{\color{blue} o } Blue (serie 2)','{\color{black} o } Black (serie 3)'},'EdgeColor','k') ;`. This way you don't have to define multiple `text` objects. – Hoki Nov 2 '15 at 10:10
• I definitely don't mind. I was hoping you'd reproduce the example ;-). It's not different enough from your answer to deserve a separate answer, but it gives extra info/options to build a nice looking legend so it is worth mentioning. Me too I miss a way to draw nice lines with text. Also, in case your line symbols are different, you may consider using a fixed size font in order to keep a nice alignment (it can get quite ugly). Last advice, I would retrieve the handle of the text object created, this way it becomes trivial to use the same object in another figure: `copyobj(ht,newfig)`. – Hoki Nov 2 '15 at 10:41 | 1,591 | 5,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.858552 |
http://cboard.cprogramming.com/cplusplus-programming/114617-struct-program-find-point-rectangle.html | 1,467,413,252,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403826.29/warc/CC-MAIN-20160624155003-00007-ip-10-164-35-72.ec2.internal.warc.gz | 50,098,409 | 11,954 | # Thread: Struct Program to find Point in Rectangle
1. ## Struct Program to find Point in Rectangle
I am working on an assignment where I have to create a program for users to input coordinates for a rectangle.
This program is intended to be a struct within a struct.
If invalid, I have to output an error message and let user try again, indefinitely, until the user gets it right.
The program should repeatedly ask users for coordinates of a point, and the program will quit when user enters 0 and 0 for x and y respectively.
The program has to say whether the point is inside of outside the rectangle. I also need to figure out where to put the main function, and what to put in it. Please let me know how exactly to complete this program, and I need to know ASAP. The program is due TONIGHT.
Here is my code:
Code:
```#define _CRT_SECURE_NO_DEPRECATE
#include <stdio.h>
typedef struct
{
int x;
int y;
} point_t;
typedef struct
{
point_t upper_left;
point_t lower_right;
} rectangle_t;
int is_inside (point_t* pPoint, rectangle_t* pRect)
{
return ((pPoint->x >= pRect->upper_left.x ) &&
(pPoint->x <= pRect->lower_right.x) &&
(pPoint->y >= pRect->upper_left.y ) &&
(pPoint->y <= pRect->lower_right.y));
}
point_t get_point(char* prompt)
{
point_t pt;
printf("Given a rectangle with a side parallel to the x axis and a series of points on the xy plane this program will say where each point lies in relation to the rectangle. It considers a point on the boundary of the rectangle to be inside the rectangle\n");
printf ("Enter coordinates for the upper left corner\n");
printf ("X: ");
scanf ("%d", &pt.x);
printf ("Y: ");
scanf ("%d", &pt.y);
return pt;
}
rectangle_t get_rect(char* prompt)
{
rectangle_t rect;
printf (prompt);
rect.upper_left = get_point("Upper left corner: \n");
rect.lower_right = get_point("Lower right corner: \n");
return rect;
}```
2. Originally Posted by DMJKobam
I also need to figure out where to put the main function...
After the function definitions...
Originally Posted by DMJKobam
...and what to put in it.
Function calls...
It looks like most of the work is done, and you just have to call the functions. I suspect these functions were already provided for you, and you just had to create main. If you have problems writing main, post what you've done and someone could probably help. However, it seems you didn't leave yourself enough time for this.
3. ## Functions where not provided
The functions where not provided. I am trying to figure out to how to call them. I have to submit this in an hour. | 609 | 2,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-26 | longest | en | 0.811435 |
https://help.scilab.org/docs/6.0.1/fr_FR/arma.html | 1,686,428,442,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646350.59/warc/CC-MAIN-20230610200654-20230610230654-00196.warc.gz | 329,595,544 | 5,234 | Change language to:
English - 日本語 - Português - Русский
See the recommended documentation of this function
# arma
Scilab arma library
### Description
Armax processes can be coded with Scilab tlist of type `'ar'`. `armac` is used to build `Armax` scilab object. An `'ar'` tlist contains the fields `['a','b','d','ny','nu','sig']`.
armac
this function creates a Scilab tlist which code an Armax process `A(z^-1)y= B(z^-1)u + D(z^-1)sig*e(t)`
```-->ar=armac([1,2],[3,4],1,1,1,sig);
-->ar('a')
ans =
! 1. 2. !
-->ar('sig')
ans =
1.```
armap(ar [,out])
Display the armax equation associated with `ar`
armap_p(ar [,out])
Display the armax equation associated with `ar` using polynomial matrix display.
[A,B,D]=armap2p(ar)
extract polynomial matrices from ar representation
armax
is used to identify the coefficients of a n-dimensional ARX process `A(z^-1)y= B(z^-1)u + sig*e(t)`
armax1
armax1 is used to identify the coefficients of a 1-dimensional ARX process `A(z^-1)y= B(z^-1)u + D(z^-1)sig*e(t)`
arsimul
armax trajectory simulation.
narsimul
armax simulation ( using rtitr)
odedi
Simple tests of ode and arsimul. Tests the option 'discret' of ode
prbs_a
pseudo random binary sequences generation
reglin
Linear regression
### Example
```// Example extracted from the demo arma3.dem.sce in the cacsd module
// Spectral power estimation
// ( form Sawaragi et all)
m = 18;
a = [1,-1.3136,1.4401,-1.0919,+0.83527];
b = [0.0,0.13137,0.023543,0.10775,0.03516];
u = rand(1,1000,'n');
z = arsimul(a,b,[0],0,u);
//----Using macro mese
[sm,fr]=mese(z,m);
//----The theoretical result
function gx=gxx(z, a, b)
w = exp(-%i*2*%pi*z*(0:4))'
gx = abs(b*w)^2/(abs(a*w)^2);
endfunction
res=[];
for x=fr
res=[ res, gxx(x,a,b)];
end
//----using armax estimation of order (4,4)
// it's a bit tricky because we are not supposed to know the order
[arc,la,lb,sig,resid]=armax(4,4,z,u);
res1=[];
for x=fr
res1=[ res1, gxx(x,la(1),lb(1))];
end
//-- visualization of the results
plot2d([fr;fr;fr]',[20*log10(sm/sm(1));20*log10(res/res(1));20*log10(res1/res1(1))]',[2,1,-1])
legend(["Using macro mese";"Theoretical value";"Arma identification"])
xtitle("Spectral power","frequency","spectral estimate")```
Report an issue << abcd Représentation de Systèmes Linéaires arma2p >> | 771 | 2,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-23 | latest | en | 0.635843 |
https://www.conservapedia.com/index.php?title=Fluid_statics&diff=next&oldid=1295056&printable=yes | 1,566,632,210,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00166.warc.gz | 763,311,245 | 8,602 | Difference between revisions of "Fluid statics"
Fluid statics is the branch of fluid mechanics which studies fluids at rest.
Equations
Pressure variation in a static Fluid
Applying Newton's laws of motions to a fluid at rest, it can be determined that the sum of forces must equal zero throughout the fluid, which means any arbitrary element of the fluid is subjected to forces that sums to zero. Since the fluid is not deforming while it is at rest, the only forces acting on the fluid are those due to gravity and pressure.
For a given fluid element, the pressure is given by the differential equation[1]
where is the density of the fluid, is the gravitational acceleration (vector), and is the pressure at the fluid element.
For small changes in altitude, can be assumed to be constant (A more accurate description for large change in altitude can be found here). Simplifications can also be made for constant density, and reducing the analysis to only the y dimension:
Examples
For system filled with an isothermal (constant temperature throughout) ideal gas, a relation between pressure altitude can be established using the following method: For ideal gases, the density is given by
Where is the molar mass, is the ideal gas constant and is the temperature (Absolute temperature, in Kelvin or Rankine) of the gas. we can then set up the equation as follows:
Note that we are using the scalar value of gravity (), so the minus sign is included due to gravity is downwards, in the negative direction of the y-axis. using the method of separation of variables, we can rearrange the equation so
Integrating both sides gives
Where is the reference pressure at point (often taken at the point which is the atmospheric pressure).
Rearranging gives
References
1. James R. Welty, Charles E. Wicks, Robert E. Wilson, and Gregory L. Rorrer, Fundamentals of Momentum, Heat, and Mass Transfer, 4th Ed. Toronto: John Wiley & Sons Inc, 2001. | 424 | 1,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 18, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-35 | latest | en | 0.918416 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/336/7/bh/h/ | 1,642,457,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00186.warc.gz | 936,604,597 | 60,898 | # Properties
Label 336.7.bh.h Level $336$ Weight $7$ Character orbit 336.bh Analytic conductor $77.298$ Analytic rank $0$ Dimension $24$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$336 = 2^{4} \cdot 3 \cdot 7$$ Weight: $$k$$ $$=$$ $$7$$ Character orbit: $$[\chi]$$ $$=$$ 336.bh (of order $$6$$, degree $$2$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$77.2981720963$$ Analytic rank: $$0$$ Dimension: $$24$$ Relative dimension: $$12$$ over $$\Q(\zeta_{6})$$ Twist minimal: no (minimal twist has level 168) Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$24 q + 324 q^{3} - 126 q^{5} + 12 q^{7} + 2916 q^{9} + O(q^{10})$$ $$\operatorname{Tr}(f)(q) =$$ $$24 q + 324 q^{3} - 126 q^{5} + 12 q^{7} + 2916 q^{9} + 1190 q^{11} - 2268 q^{15} - 1500 q^{17} + 13446 q^{19} - 2106 q^{21} - 21504 q^{23} + 22542 q^{25} - 85484 q^{29} - 6264 q^{31} + 32130 q^{33} - 32268 q^{35} - 46938 q^{37} + 17010 q^{39} + 19548 q^{43} - 30618 q^{45} + 167004 q^{47} + 250644 q^{49} - 13500 q^{51} - 258982 q^{53} + 242028 q^{57} - 744834 q^{59} - 390096 q^{61} - 59778 q^{63} - 19388 q^{65} - 62742 q^{67} + 1102984 q^{71} - 663534 q^{73} + 608634 q^{75} + 404298 q^{77} + 271032 q^{79} - 708588 q^{81} + 2540040 q^{85} - 1154034 q^{87} - 433740 q^{89} + 2142270 q^{91} - 56376 q^{93} - 2205360 q^{95} + 578340 q^{99} + O(q^{100})$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
145.1 0 13.5000 7.79423i 0 −160.491 92.6595i 0 338.827 + 53.3419i 0 121.500 210.444i 0
145.2 0 13.5000 7.79423i 0 −151.084 87.2284i 0 −180.229 + 291.833i 0 121.500 210.444i 0
145.3 0 13.5000 7.79423i 0 −126.269 72.9015i 0 −114.286 323.400i 0 121.500 210.444i 0
145.4 0 13.5000 7.79423i 0 −118.587 68.4663i 0 302.168 162.306i 0 121.500 210.444i 0
145.5 0 13.5000 7.79423i 0 −56.9387 32.8735i 0 −298.061 + 169.732i 0 121.500 210.444i 0
145.6 0 13.5000 7.79423i 0 −51.6596 29.8257i 0 −161.216 302.751i 0 121.500 210.444i 0
145.7 0 13.5000 7.79423i 0 44.9687 + 25.9627i 0 120.755 + 321.041i 0 121.500 210.444i 0
145.8 0 13.5000 7.79423i 0 59.9565 + 34.6159i 0 −9.46147 342.869i 0 121.500 210.444i 0
145.9 0 13.5000 7.79423i 0 76.6020 + 44.2262i 0 298.004 + 169.831i 0 121.500 210.444i 0
145.10 0 13.5000 7.79423i 0 112.001 + 64.6636i 0 −289.805 + 183.473i 0 121.500 210.444i 0
145.11 0 13.5000 7.79423i 0 127.654 + 73.7013i 0 −327.920 100.586i 0 121.500 210.444i 0
145.12 0 13.5000 7.79423i 0 180.847 + 104.412i 0 327.223 102.830i 0 121.500 210.444i 0
241.1 0 13.5000 + 7.79423i 0 −160.491 + 92.6595i 0 338.827 53.3419i 0 121.500 + 210.444i 0
241.2 0 13.5000 + 7.79423i 0 −151.084 + 87.2284i 0 −180.229 291.833i 0 121.500 + 210.444i 0
241.3 0 13.5000 + 7.79423i 0 −126.269 + 72.9015i 0 −114.286 + 323.400i 0 121.500 + 210.444i 0
241.4 0 13.5000 + 7.79423i 0 −118.587 + 68.4663i 0 302.168 + 162.306i 0 121.500 + 210.444i 0
241.5 0 13.5000 + 7.79423i 0 −56.9387 + 32.8735i 0 −298.061 169.732i 0 121.500 + 210.444i 0
241.6 0 13.5000 + 7.79423i 0 −51.6596 + 29.8257i 0 −161.216 + 302.751i 0 121.500 + 210.444i 0
241.7 0 13.5000 + 7.79423i 0 44.9687 25.9627i 0 120.755 321.041i 0 121.500 + 210.444i 0
241.8 0 13.5000 + 7.79423i 0 59.9565 34.6159i 0 −9.46147 + 342.869i 0 121.500 + 210.444i 0
See all 24 embeddings
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 241.12 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.d odd 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 336.7.bh.h 24
4.b odd 2 1 168.7.z.a 24
7.d odd 6 1 inner 336.7.bh.h 24
28.f even 6 1 168.7.z.a 24
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
168.7.z.a 24 4.b odd 2 1
168.7.z.a 24 28.f even 6 1
336.7.bh.h 24 1.a even 1 1 trivial
336.7.bh.h 24 7.d odd 6 1 inner
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$22\!\cdots\!03$$$$T_{5}^{18} -$$$$32\!\cdots\!94$$$$T_{5}^{17} +$$$$62\!\cdots\!86$$$$T_{5}^{16} +$$$$87\!\cdots\!90$$$$T_{5}^{15} -$$$$12\!\cdots\!75$$$$T_{5}^{14} -$$$$16\!\cdots\!50$$$$T_{5}^{13} +$$$$18\!\cdots\!25$$$$T_{5}^{12} +$$$$20\!\cdots\!00$$$$T_{5}^{11} -$$$$19\!\cdots\!00$$$$T_{5}^{10} -$$$$18\!\cdots\!00$$$$T_{5}^{9} +$$$$16\!\cdots\!00$$$$T_{5}^{8} +$$$$10\!\cdots\!00$$$$T_{5}^{7} -$$$$85\!\cdots\!00$$$$T_{5}^{6} -$$$$39\!\cdots\!00$$$$T_{5}^{5} +$$$$32\!\cdots\!00$$$$T_{5}^{4} +$$$$91\!\cdots\!00$$$$T_{5}^{3} -$$$$69\!\cdots\!00$$$$T_{5}^{2} -$$$$12\!\cdots\!00$$$$T_{5} +$$$$10\!\cdots\!00$$">$$T_{5}^{24} + \cdots$$ acting on $$S_{7}^{\mathrm{new}}(336, [\chi])$$. | 2,515 | 5,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.363271 |
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Mathematics (1,874)
MATH 136 (145)
Lecture 10
# Lecture 10.pdf
5 Pages
97 Views
School
University of Waterloo
Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter
Description
Monday, January 27 − Lecture 10 : Gauss-Jordan elimination algorithm for solving systems of linear equations. Concepts: 1. Gauss-Jordan elimination algorithm for solving systems of linear equations. 2. consistent and inconsistent system. 3. ERO’s We see that solving a system of linear equations which is in RREF is a relatively easy task. However most systems we encounter are not in RREF. We now look at a technique where we transform such systems in RREF without changing the solution. 10.1 Definition − The Gauss-Jordan elimination is an algorithm which transforms a system of linear equations onto an RREF system in such a way that the resulting system has the same solution as the original system of equations. These changes are done by performing 3 "Elementary row operations" (ERO’s) on the augmented matrix associated to the given system : - ERO of type I : Interchange two rows of the system. We denote this row operation by P ijmeaning we "permute" or exchange the i and j rows. th - ERO of type II : Multiply a row of the system by a non-zero real number. We denote this by cR, mianing we multiply the row i with the scalar c. - ERO of type III : Replace the i row R with the sum of R and the scalar multiple of i i row R j. We denote this by R + cR. i j Note: We could actually apply the above operations to the system of equations and referring to these as the 3 “equation-operations”: - “ P ij, interchange the order of two equations, - “cE”,imultiply both sides of equation E by a non-zeio scalar c, - “E +icE”, rejlace equation E, by the sui of the equation E and the equation i obtained by multiply both sides of the equation E jby c. Anyone of these 3 operations does not change the solution of the system. The elementary row operations essentially describe the operations performed above in solving the system of linear equations. 10.1.1 Note − Performing the elementary operations on a system always results in a system which has the same solution as the system we started off with. 10.1.2 Remark − Any ERO can be "undone" by an ERO: - To "undo" P apijy P ij - To "undo" cR appiy (1/c)R. i - To "undo" R + ci applyjR + (−c)Ri j. Verify this with a simple matrix. 10.1.3 Remark − For some there may be some confusion about the interpretation given to an ERO such as, for example, 2R + R . 3 5 - The ERO “2R + R 3 is t5 be interpreted in the same ways as “R + 2R ”. It 5 3 means “Replace row 5 with 2 times row 3” - Just to avoid confusion some may prefer to state the ERO as: R → R + 2R 5 5 3 which more clearly states “replace row 5 with 2 times row 3”. - But do not write “R → 23 + R ”,3which5will surely confuse the reader. 10.2 General strategy for solving linear systems by using the Gauss-Jordan elimination. System of linear equations → Obtain the augmented coefficient matrix [A|b] → Apply ERO’s to transform this matrix matrix [A|b] in RREF, [A RREF | d] → Transform back into the system of linear equations of [A RREF| d] → Isolate the basic variables. → Write out the solution set in scalar form. → Write out the solution in “vector equation form”. 10.2.1 Note − If the operations are done accurately the solution to the system of linear equation [A RREF | d] is identical
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So we can recommend you notes for your school. | 969 | 3,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.907921 |
https://electronics.stackexchange.com/questions/75470/why-arent-sram-modules-laid-out-in-a-matrix | 1,560,963,676,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999003.64/warc/CC-MAIN-20190619163847-20190619185847-00047.warc.gz | 433,831,812 | 35,927 | Why aren't SRAM modules laid out in a matrix?
I'm currently reading about RAM modules. Bigger DRAM module are laid out in a matrix. When retrieving data you first retrieve the row and then the column. One of the benefits of the matrix lay-out is that the row and column addresses are multiplexed on the same pins, so you only need 2^(n/2) pins for the addresses. By using a matrix the pin count can be reduced quite easily, so why isn't this done with SRAM modules?
First, don't confuse the interface (multiplexing the address lines) with the internal arrangement (laying things out in a matrix).
All RAM chips use a matrix a the the lowest (single bit cell) level, most use a less regular layout at the higher level (for instance two separate banks).
Multiplexing the address lines reduces the number of pins, but it has a cost: supplying the two halves of the address with the associated clock signals takes time. When this time can be matched with things that must be done inside the RAM chip anyway this is OK, but when it would slow the chip down this is a big performance problem.
SRAM can be 'directly addressed': the N bit address is decoded into 2^N select signals, that each activate a word of bit cells. (AFAIK this is not how modern SRAM really works, but it was true some time ago).
DRAM is addressed by row/column, because it eases the refreshing that is required: when you address one row (or was it a column?) all cells in that row are read and written back. Next you can (but don't have to) select one column and you get the values of the selected bit cell(s). So the row/column distinction maps nicely to the refresh hardware and process that must be present anyway.
• If I recall, on a DRAM chip, asserting /RAS causes the row to be read, and releasing it causes that row to be written back. One may if desired read or write one or more bits within the row before that happens, though on many chips there is a limit to how long /RAS may be asserted. – supercat Jul 10 '13 at 18:43
The architecture is different. Dynamic devices store a bit per device, so you see 8 or 16 chips in a line (or on a SIMM) to store a byte or word. (Plus one or two more, if parity bits are needed.) Static devices store a full byte, so you usually see one or two of them on a board. It's only if the data bus is wide (16 or 32 bits) that you are forced to have more than one SRAM chip.
The Apple ][ was a classic design. It had both types of devices, and played on their strengths. Dynamic memory was expensive, so you could choose how many rows you wanted to populate in the 3x8 array of 16Kbit devices. The DRAM refresh cycle was piggybacked onto the video display update -- each time the video frame was read out, it also refreshed all the DRAM devices. Then there were six 2Kbyte ROM/EPROM sockets that stored the resident part of the bootstrap and OS at the top of the memory map. EPROMs and SRAMs have very similar pinouts. Embedded designs can take advantage of this to let you design/program with a SRAM, and ship an EPROM in the same socket for the finished product. | 726 | 3,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | longest | en | 0.937324 |
http://mathhelpforum.com/advanced-statistics/214603-discrete-random-variables-pmf-print.html | 1,526,966,226,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864624.7/warc/CC-MAIN-20180522034402-20180522054402-00079.warc.gz | 190,547,857 | 3,129 | # Discrete Random Variables - pmf
• Mar 11th 2013, 01:13 PM
Matt1993
Discrete Random Variables - pmf
The discrete random variable R takes the values in S = { -3, -1, 1, 3 } with probabilities respectively,
( 1 - theta )/4, ( 1 - 3theta) /4, ( 1 + 3theta)/4, (1+ theta)/4,
where theta is a real constant. Find the range of values for which this is a valid probability mass function.
i think the answer is -1/3 =< theta =< 1/3
BUT HOW DO PROVE IT!!!!!!!
help!!
• Mar 11th 2013, 01:21 PM
ILikeSerena
Re: Discrete Random Variables - pmf
Quote:
Originally Posted by Matt1993
The discrete random variable R takes the values in S = { -3, -1, 1, 3 } with probabilities respectively,
( 1 - theta )/4, ( 1 - 3theta) /4, ( 1 + 3theta)/4, (1+ theta)/4,
where theta is a real constant. Find the range of values for which this is a valid probability mass function.
i think the answer is -1/3 =< theta =< 1/3
BUT HOW DO PROVE IT!!!!!!!
help!!
Hi Matt1993! :)
What is the reason you think that -1/3 =< theta =< 1/3?
That is likely the key to the proof...
• Mar 11th 2013, 01:36 PM
Matt1993
Re: Discrete Random Variables - pmf
See all I did was let the pmfs equal zero and then I just tried values until I came up with that answer. That's the problem
• Mar 11th 2013, 01:47 PM
ILikeSerena
Re: Discrete Random Variables - pmf
Quote:
Originally Posted by Matt1993
See all I did was let the pmfs equal zero and then I just tried values until I came up with that answer. That's the problem
That's close.
The axioms of probability (see wiki) require in particular 2 things from the probabilities:
1. Each probability is at least 0: $\displaystyle p \ge 0$.
.
2. The sum of all probabilities is 1: $\displaystyle \sum p = 1$.
$\displaystyle ( 1 - \theta )/4 \ge 0$
$\displaystyle ( 1 - 3\theta) /4 \ge 0$
$\displaystyle ( 1 + 3\theta)/4 \ge 0$
$\displaystyle (1+ \theta)/4 \ge 0$
$\displaystyle ( 1 -\theta )/4 + ( 1 - 3\theta) /4 + ( 1 + 3\theta)/4 + (1+ \theta)/4 = 1$ | 656 | 1,968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.817437 |
https://nursesgoalstatement.com/30400/italy_of_law.aspx | 1,674,947,249,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00868.warc.gz | 468,477,409 | 15,051 | ##### Why cars are of law acceleration is ever lived
Previous Post Jan Additional force between acceleration of law holds it accelerate at what that.
# Law Of Acceleration Examples
Email Signup Laws around a law.
Force due to law example, laws being designed systems of.
Newton's Second Law of Motion Force Mass & Acceleration. Psychotherapist writing on acceleration of laws of appeals decision must cancel. Have students to fall increasingly being thrown to the fuel mixture to look is. What kind of examples of the mass is the discussion with the universe, and roll this hypothetical system allows the vocabulary. Each law example of laws of mass by applying a conclusion of forces are. Find it accelerate upwards to discussion about. How few laws play, joy and bridges to law? Earth is a law example is a horizontal. Your acceleration and accelerate a law example, laws of examples of forces exerted by doing so quickly and worksheet to acknowledge the accelerator. You can accelerate in acceleration of law enters again, find that these sites for general relativity theory of physics is astronomy and in? The state university, cos is made of acceleration, you can calculate its acceleration of. The acceleration of examples accelerate a suitable electrical interactions between acceleration: what happened to a straight down and forklifts, a black background. You can accelerate to acceleration and long charging periods of examples of different conditions of motion that test your reality mirrors what type of an accelerator. Most excellent example of examples of an essay about how anobject in its direction of any of force directly proportional. If you can see that example, laws and stop without breaking of law does not required to a pulley. Definitions and the laws and for? It measures speed of law apply mathematical equation has been deduced and weather driving your palms pressed against each. Also examples accelerate? The acceleration of examples.
### If you back of examples
You might change if you look at regular brake is acceleration. In the examples that include people in the discussions about forces the role of the. Dummies are examples for example, laws to law of the exhaust valve allows for? The laws of examples include equations from above were developed to show motion where students participated in an asymmetrical tires. First law is acceleration vector, laws and accelerate in real world. In the law mathematically describes the engine? The laws are examples of physical activity is needed. Student safely on acceleration is accelerating it accelerate the law actually happens if you are examples, a flat surface with a toothpick partly into your spam folder. Tell them to law example to an arc upwards. Physics laws are examples in terms. These laws of the objects up high school does tell whether the law of acceleration examples of change is. Another is directly proportional to push forward as they do people in a middle of astronomical size of government and notice with only certain time of our local stage supply shops or features causing them. Choose up a double the same answer makes an ever lived just an engine drives the law of acceleration of the vertical forces. The newtonÕs first, because of velocity tangent to a frame of a straight which tends to time taken to live science standards were stationary lift. Opposing forces acting on top ball downward force depends solely on it until something wrong. How acceleration or law? What is called law example of examples for it exerts a number of a car accidents and physiological effects. The law is thrown or subtracted from our examples of a magnetic interactions are external force to have been created in different types of. It is considered to change its side of examples of rest or, and inertia and due to newton school does the example. Students played prior to law of acceleration examples that is perhaps the bottle tends to reinsert the time. Find out jerkiness of laws?
### The bottle lies behind the design of examples of
Long as the law of examples accelerate in applications. And acceleration of examples of electric fields, and forces in the accelerator of. An example of laws of elastic cord supporting a certain radius, he is applying more. Each law to accelerate at a climber shows us, laws are examples include it follows from being during compression and one turns. What will travel a law says that examples did not used to net force. We should be acceleration is a law unto herself away. Note that examples of laws are some unknown quantities are going around earth are three laws and parallel to slide the indianapolis public submits written across and shape. Horizontal force acts upon by the weight is exerted by a an infinite number of the vertical acceleration of law examples accelerate in the presence of mass of a rotating at thestarting line. Add it accelerate if you double that acceleration due to it makes use for instance, laws simply means that acceleration? If these techniques for example of an object will happen for understanding by an opportunity to prevent oil in this. This is pushing or the cheese is easier or gradually through it related to divide into a bullet back of examples of the brakes to modify this work in the velocities and the willful application of. See if these laws. This law reproducible and accelerating spaceship between physics laws are examples, theories of molecular motors that. This shows that examples of. Calculus to law describes how does not allowed to illustrate known repulsive force upwards you. It accelerate continually until an acceleration once it is needed for solving force and narrower blocks cut two examples accelerate it. This law of examples accelerate very short distance and then make personal challenge and can be uploaded because really quickly, what ror without loss. Note that there are in a podcast on a lover might not accelerate because it is traveling cart in your hand straight up ws an accelerator.
Acceleration occurs when there is a net force on an object. Physics laws of examples of a close together with comparable mass a change. The frictional forces so in vermont is of law acceleration of them towards the situation where only one speeds are the motion? Laws are staying still equal magnitude of acceleration in half of water. When it changes by laws are examples in time. We have conceived of acceleration of a car for? This law of examples of toothpicks that. Felipe watched over each other. You might surprise you will accelerate it is acceleration and rules based on it is what thisexperiment shows no other times are examples did some time. The acceleration and accelerate the above examples accelerate an important to do not produce zero since moving object already happened to be able to ancient greece and apply. The screen to law to the shutter just in mind of the small changes by the second law does happen. The acceleration if these sites. If you see in this special can. What kind used to be able to avoid losing mass and requires that every action is rather than a flat road is. While acceleration is accelerating downward as a law example, laws of examples accelerate very high velocity and motionthat we treat all the accelerator. For this law of force required. The law of examples of an object positions and perpendicular to remove dirt from this universe can be covered by an intrinsic property of. The result of examples that the plane, pointing towards different mass is varied, green light suspended on. Determine the law equation, as in to this under the time intervals on the object or lessons, the difference leads to push away from which we still?
An object and inertia of examples of law of rubber squashes. Or law example, laws and we can be represented by equations to a thick piece of. These laws of examples accelerate very soon after it accelerates very well. To the class into groups and power and row in your local fire an object, mass of car engine pushes on the pedals would happen to fall. Kids learn about the science behind the basic three Laws of Motion Forces. Make about law apply to accelerate upwards to work? This system as positive direction of examples of. Calculate an infinite universe can check to gravity pulls in direction opposite reaction force makes it is accelerating, since different place. So in acceleration, laws and accelerate. When two parts of the force and we are more mass experiences greater is an external force arrives at rest will imply more. As being involved in this law of laws of plastic cup with fundamental point on earth is? Then swings up old holes in the whole width of examples of law acceleration as the trolley is? This law actually describes inertia: action of laws are examples accelerate it takes a basket, resting on a rocket that force which car to make things. The law enters again. This law example, laws in the accelerator too hard pill to accelerate upwards equal force upwards. Consider a ruler to know and exponential functions, without permission to consider a bit harder to exceptions to turn. Some examples of law of work a function of motion and is very hard and to mass is a transparent acetate sheet! In problem at once it becomes great concepts of law acceleration examples that case, if the car turns out over a greater. So that examples did not assume that a variety of laws are not necessarily to be expressed by acceleration are separated from linear and gender.
### Start the liquid
Researchers can be transcended or changes in rocketry was. The scientist that passes through the person stands on the force or to record data. Examples take turns round a road, laws of those forces on any force acting on. EarthÕs center of acceleration and accelerating versus non directional tread patterns on your name: sive corporum duorum actiones in? Second law example, laws of examples of acceleration should be at one. Legislatures and acceleration and tools float? And draw a law example can be asked to slide on. Use scientific laws of examples accelerate it accelerates very important to be accelerating versus non directional tires have not been added or used to see that example. If you wish to law for? This means that a car compare prices and has engineered a law of the pivot point here. The acceleration is common sense mechanically to accelerate in this system, several possibilities here, sometimes referred to cut or in? The related to use the bed to keep an impactor with the acceleration of law of net force between the constitution is? In general a scientific law is the description of an observed phenomenon It doesn't explain why the phenomenon exists or what causes it The explanation of a phenomenon is called a scientific theory It is a misconception that theories turn into laws with enough research. Move may seem like fuelthat pushed occured at rest until everyone unconsiously knows these laws of law example. All this law is not processing if we perceive from the laws which it to do with which we double the fuel cell is applied to reinforce concepts. In acceleration is accelerating from ma we can accelerate more examples accelerate, accelerates both marks in velocity rather than i had him lifting himself. Did the acceleration is ok, the mass and accelerate the resultant force to the cannon idea of examples in the bottle tends to follow from deceleration. The laws of examples of three learning for the book and latin and horizontal motion, draw pictures theyÕvedrawn in? There is that examples accelerate more force acting when the forces present in your spam folder. The properties of law the accelerator earlier ones were stationary and their assumed to allow researchers to maneuver.
Invitestudents to law example to local stage booster rockets. The motive force is the smallest applied force can a lesser than the deflection of. Every motion described by laws of examples of static friction on the example, intuition develops gradually reduces the weight. This law of laws may seem like your rocket through the fishing line. Make a law example to accelerate toward its location. How adding and accelerates. It takes time intervals on a car more mass of each other end until the greater than to be hurt if the book was. Continued before and colliding into laws and cart moved on human subjects at a law is extremely high speeds creates a certain force is. Let us that law of. In acceleration and accelerate the law of examples of the redirect does a brick wall with. Examples that your neet is high enough to complete body when it sliding relative motion will imply more complicated machines like a force were. Just enough to get maximum speed of motion and only when a change by nasa and provides an objectÕs state of different. The acceleration will accelerate very little awareness that examples take their eyes followed by three. This video learn some cylinders inside of three laws in some; her fingers to tighten a constant. Suppose the law is needed to turn and its direction of examples of the directions parallel, nails would be directly. Classical advancement they are examples accelerate upwards in acceleration experiment is known force on an accelerator. Will generate a gun fires a diagram below on the accelerator earlier section on the fastest line passing through in motion unless a full of.
When they will accelerate it is acceleration is because it was shown in turn which law. Of Obligations.
##### The board begins sliding with the car moving forward and quickly, of examples of the fundamental constituents of
We think it is the surface because the elevator moves on it should gain energy. Cover Best Letter Architects
##### Lex i want the pressure
You slide of law example, and exploit as shown in the class. With a law example, laws of examples of a reaction force can easily with the first. The accelerator too is gone causing it forward on by an incredible coincidence that a single force multiplication gained by eötvös. Blocks instead of mass of the moment for? Find acceleration are examples accelerate it stays at one will always sags a law example with teacher will contribute to complete brake petal to modify the accelerator. Mass times acceleration of law example, accelerating spaceship they have read the accelerator. Which is applied on your fingers from our use arrowsto show center of this slide because they stay in? We have enough fluid without this? We refer to accelerate more examples; the accelerating it and accelerates toward its acceleration of electric forces involved, as a conclusion of the motion. If given point between a car accident, separate from rest will decrease in inverse square law of getting dry ice unless a jar by changing. Gravity can be reduced by doing so logically and acceleration of a larger force while the mass? How two examples and opposite directions parallel to law will have been sent too tight between motion. In acceleration of law example. Mendel knew why moving with a law example, accelerates towards the accelerator earlier but in the one place to accelerate, the slope of examples.
You to law of laws simply a copy of each student on it was at this. Death All Notices. | 2,887 | 15,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.92991 |
https://www.geogebra.org/m/GvEUjyAu | 1,716,128,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00076.warc.gz | 701,992,341 | 28,421 | GeoGebra Classroom
# RECTANGLE
## Follow the instruction in the above fig. As you drag the vertex, answer the following questions.
1. How are the lengths of the sides related?
2. How are measures of the angles related?
3. The diagonals are of equal length.
Select all that apply
• A
• B
• C
4. A rectangle is a parallelogram.
Select all that apply
• A
• B
• C
5. A rectangle is a Rhombus.
Select all that apply
• A
• B
• C
6. Diagonals of a rectangle are perpendicular.
Select all that apply
• A
• B
• C | 147 | 515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.806296 |
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Great introduction to Curve Sketching for your students applying what they have learned about the first and second derivatives. Great for visual learners. This innovative activity is designed for Calculus 1, AP Calculus , and Calculus Honors. It is an introduction to Unit 3, Applications to the D
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Calculus PreCalculus Limits from a Graph with Notes task cards and HW/QuizThis lesson is designed for Calculus 1, AP Calculus AB, and PreCalculus for some curricula, and is a great beginning activity in the Limits Unit. Included: ✓ Guided student notes with four examples. There are two versions,
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Calculus Continuity Foldable Engaging Activity in Trifold format This resource is designed for first semester Calculus students and is from Unit 2, Limits and Continuity. The trifold contains three activities in one, is paper friendly, and can be used in Interactive Notebooks if desired.In section
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Partial Fractions Decomposition - Guided Student Notes Worksheet designed for PreCalculus and AP Calculus BC, and College Calculus 2 This great handout with worksheet or quiz explains decomposing a rational expression into partial fractions. It uses the substitution method which eliminates the nee
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Volume of Revolution Disk and Washer Methods These engaging rigorous activities are designed for Calculus AB, BC, Honors, and Calculus 2. It is part of the unit Applications of Integration. Included: ☑ 12 Task Cards which are both number and color coded. #1 – 3 are disk about the x-axis, #4
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Calculus: Volume of Solids of Known Cross Section Task Cards, and way more. This lesson is designed for AP Calculus AB, BC or College Calculus 2. It is from the Applications of Integration Unit. Included: ✓ In class guided notes and examples on Volumes Solids of Known Cross Sections, with a
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AP Calculus BC Calculus 2 Taylor and MacLaurin Polynomials - Taylor Series Infinite Series -Organizers, Guided Notes, plus Practice Problems with FULL SOLUTIONSAmaze your AP Calculus BC, and College Calculus 2 students with the visualization of Taylor Polynomials included in this product. Inclu
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Calculus Introduction to Riemann Sums This activity is designed for AP Calculus AB, AP Calculus BC, Honors Calculus and College Calculus 1. It emphasizes the visualization of the Riemann sums, and how curves are partitioned. Included: ✓ Task Cards : There are 10 task cards with partitione
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Calculus Slope Fields Differential Equations with Guided Notes and Matching Task Cards. This lesson is designed for AP Calculus AB, AP Calculus BC and College Calculus 1 or 2 classes. These activities will help your students thoroughly understand and master the concepts in this section. Activity B
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This Calculus resource on Particle Motion is designed for the Unit on Derivatives. It will give your students a solid foundation on a difficult and very important topic. Included in the Lesson: ✓ Guided Notes with a completed multi-part example ✓ 18 color coded Task Cards with two blank car
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Calculus Definite Integration Interactive Digital Maze GOOGLE® Edition with HW I am excited to offer this 1:1 paperless digital version of my very popular printable version of the Calculus Integration Fun Maze.This innovative, engaging, fun product is designed for AP Calculus AB, BC, Honors Calcul
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Limits at Infinity Horizontal Asymptotes Task cards HW/Quiz This activity is designed for Calculus 1 or AP Calculus. It is usually part of the third unit, Applications of the Derivative. Students are to find the limits of various functions as x approaches infinity. Determining these limits wi
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Calculus, Separation of Variables Differential Equations Guided Notes, Task Card Activity, plus HW and optional QR This lesson is designed for AP Calculus AB, AP Calculus BC and College Calculus 1 or 2 classes. These activities are designed to help students thoroughly understand and master the con
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PreCalculus, Calculus: Finding Limits Analytically This is the second lesson in a five-lesson unit on INTRODUCTION TO CALCULUS designed for PreCalculus Honors students who have a working understanding of function behavior. Students will be able to: ★ Evaluate limits using substitution ★ Evaluate l
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Activity Based Learning with Task Cards really works to help reinforce your lessons. Task Cards and Station cards get your students engaged and keep them motivated. This set of 16 task cards will help students practice finding the derivative of Trigonometric Functions using the Chain Rule. This l
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Calculus, AP Calculus AB, AP Calculus BC, College Calculus, Calculus Honors, Area Under a Curve This is the first lesson in a ten-lesson unit on Integration for students enrolled in AP Calculus AB or BC, Calculus Honors, or College Calculus. Every lesson includes: ✎ A set of Guided Student Note
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This is the final lesson in a thirteen-lesson unit on Differentiation for students enrolled in AP Calculus AB or BC, Calculus Honors, or College Calculus. Every lesson includes:✎ A set of Guided Student Notes✎ A daily homework assignment✎ Four forms of a daily homework quiz or exit ticket✎ T
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Activity Based Learning with Task Cards really works to help reinforce your lessons. Task & Station cards get your students engaged and keep them motivated. This product is designed for AP Calculus AB, BC or College Calculus 1. It is from Unit 1. Included: ☑ 15 task cards. The questio
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This lesson is designed for AP Calculus AB, AP Calculus BC, Calculus 1, and Calculus Honors. This lesson is related to Integration and is part of an older bundle. It is a stand alone lesson for teachers who need to teach this lesson separately. The lesson includes: ♦ Guided Notes handout ♦
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This is a four-lesson unit on Differential Equations for students enrolled in AP Calculus AB or BC, Calculus Honors, or College Calculus. Every lesson includes ✎ A set of Guided Student Notes ✎ A daily homework assignment ✎ Four forms of a daily homework quiz or exit ticket ✎ Teachers also h
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PreCalculus, Calculus: The Derivative and Tangent Lines This is the fourth lesson in a five-lesson unit on INTRODUCTION TO CALCULUS designed for PreCalculus Honors students who have a working understanding of function behavior. Students will be able to: ★ Find slope using the limit process ★ Find
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This is the coolest activity I have developed for calculus. You have to see the color on the activity to appreciate it--check out the preview! This activity introduces the concepts of curve sketching--when f (x) is increasing, f '(x) is positive. Students will use the same color to mark those int
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Calculus Curve Sketching This packet contains 5 worksheets that you can use to help students work on the concept of curve sketching. The worksheets in this packet focus on the sketching of graphs. If your students need practice with the algebraic portion of the curve sketching process, see my Cal
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Calculus, AP Calculus AB, AP Calculus BC, College Calculus, Calculus Honors, Antiderivatives and Indefinite Integration This is the third lesson in a ten-lesson unit on Integration for students enrolled in AP Calculus AB or BC, Calculus Honors, or College Calculus. Every lesson includes: ✎ A se
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,424 | 9,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-43 | latest | en | 0.873541 |
http://www.answers.com/Q/How_many_square_feet_of_land_is_one_cent_in_India | 1,550,797,331,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247511573.67/warc/CC-MAIN-20190221233437-20190222015437-00229.warc.gz | 318,859,930 | 49,684 | # How many square feet of land is one cent in India?
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1Cent->435.5sqft.
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# How many square feet of land is one cent?
One cent is 1/100th of an acre. In square feet, one cent is 435.540069686 square feet.
# One cent is how many square feet?
Unless you are speaking a different language, there is no relationship between cents and square feet. Cents is money (unless you have a different meaning) and square feet is a
# How many square feet equals one cent?
In India(especially south india), land will be measured in cents. 1 acre = 100 cents. and 1 acre = 4840 square yards => 100 cents = 4840 square yards => 1 cent = 48.4 squa
# One cent equal to how many square feet?
If you are asking how many square feet is in a centimeter, it cannot be converted. A centimeter is a measure of length and square feet is a measure of area.
# One cent is equal to how many square feet?
You need to measure the diameter of the cent. Then divide the diameter in half to get the radius. The formula for area of a circle = ?r 2 , where ? is 3.14159. I think you wou
# How many square meters of land are contained in one cent?
See the conversion chart at the related link that states 1 cent equals 40.50522648083624 meters.
# One cent land how many square feet?
Land Measurement in South India . This question is relating to land measurements in India, especially in south India.. In India, one cent is 1/100th of an acre. In square
40.5 meters
# How many square feet of land is one decimal in India?
One decimal of land in India equals approx 436 sq. feet (435.35 sq. feet).
# One square feet is how many cents?
1 cent = 435.60 sq.feet (ft 2 ) = 1/100 acre
# How many square feet of land is one dismal in India?
Dismil is a unit of measuring land in Orissa, Bihar district in India 1 dismil = approx 435 sqft 100 dismil = 1 acre
# How many meter of land is one cent in india?
10
In Math and Arithmetic
# How many square feet a cent have?
435.6 squar feet
In Math and Arithmetic
# How many square feet of land is 5 cent in India?
5 U.S. cents (square feet) = 0.004645152 m 2 U.S. Dollars
In Math and Arithmetic
# How many links is one cent of land in India?
1 cent = 1000 square links. That is all.
In Area
# How many square feet of land is one katha in India?
Approx. 1360 sq. feet
In Math and Arithmetic
# How many Square feet of land in 1 cent?
1 cent=432 sq.ft. | 691 | 2,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-09 | latest | en | 0.946212 |
http://www.diychatroom.com/f97/rated-windows-119072-print/ | 1,498,308,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320261.6/warc/CC-MAIN-20170624115542-20170624135542-00370.warc.gz | 517,606,256 | 4,308 | DIY Chatroom Home Improvement Forum (http://www.diychatroom.com/)
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amyevans 10-03-2011 04:21 AM
A rated windows
I've been looking into these recently http://www.stormclad.co.uk/arated.php
Apparently they reduce the energy lost through the windows to zero; is this even possible? And does anyone know how much this might save on heating bills?
ralphf 10-03-2011 04:56 AM
Zero does not seem possible
Quote:
Originally Posted by amyevans (Post 740983) I've been looking into these recently http://www.stormclad.co.uk/arated.php Does anyone know anymore about this type of window? Apparently they reduce the energy lost through the windows to zero; is this even possible? And does anyone know how much this might save on heating bills?
They State zero none but there graph shows A has 1.2 rating. For zero loss I think the rating would have to be 0. Even a vacuum thermos bottle has heat loss. My opinion not possible to have zero heat loss if the outside temperature is lower than inside temperature. Maybe contact them for clarification of the zero none statement
amyevans 10-03-2011 10:48 AM
Quote:
Originally Posted by ralphf (Post 740985) They State zero none but there graph shows A has 1.2 rating. For zero loss I think the rating would have to be 0. Even a vacuum thermos bottle has heat loss. My opinion not possible to have zero heat loss if the outside temperature is lower than inside temperature. Maybe contact them for clarification of the zero none statement
Thanks Ralph, that's what I was thinking, except you've actually managed to explain why I was thinking it!
I just might give them a call...
AGWhitehouse 10-03-2011 01:43 PM
Well that's false advertisement if I've every seen it. It says an "A" rating is the highest a double pane window can achieve. I can tell you that a double-pane window is not the most efficient window available. When you call them, ask what the U-value is. U-value is the inverse of the R-value (thermal resistance value).
The industry "Energy Star" efficiency windows right now are R-5 (U-0.20). These are triple pane. There are more efficient windows but they can get really pricey and sometimes totally custom.
SPS-1 10-05-2011 06:38 PM
Quote:
Originally Posted by AGWhitehouse (Post 741305) The industry "Energy Star" efficiency windows right now are R-5 (U-0.20).
Now the OP is really going to be confused. His location is England. U value in Europe is measured in Watts......., and in the US they are measured in BTU........ (too many units to list them all).
So they are different. I read where European U value is 5.678 times bigger than US U value. If this is accurate than a US 0.2 equals a European 1.14 U value. These guys were advertising 1.2. So That would make these rather good windows, but not extraordinary windows.
AGWhitehouse 10-06-2011 09:53 AM
Quote:
Originally Posted by SPS-1 (Post 742927) Now the OP is really going to be confused. His location is England. U value in Europe is measured in Watts......., and in the US they are measured in BTU........ (too many units to list them all). So they are different. I read where European U value is 5.678 times bigger than US U value. If this is accurate than a US 0.2 equals a European 1.14 U value. These guys were advertising 1.2. So That would make these rather good windows, but not extraordinary windows.
Oooohhh...Thanks! Maybe someday, as technology brings us closer together, we'll have common systems and measurements.
operagost 11-16-2011 12:09 PM
Quote:
Originally Posted by AGWhitehouse (Post 743242) Oooohhh...Thanks! Maybe someday, as technology brings us closer together, we'll have common systems and measurements.
I don't think I'll get used to telling people I weigh about 11 1/2 stone.
DangerMouse 11-16-2011 02:05 PM
Quote:
Originally Posted by SPS-1 (Post 742927) His location is England. U value in Europe is measured in Watts......., and in the US they are measured in BTU........
Thank you for the laugh....
BTU stands for WHAT again? :laughing:
Yeah, it's confusing all right. Maybe someday we'll all speak the same language too.....?
DM
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https://physics.stackexchange.com/questions/617911/electric-and-magnetic-charge-quantization-v-s-the-radius-of-the-u1-or-mat/617965 | 1,620,500,234,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00294.warc.gz | 470,120,634 | 40,390 | # Electric and magnetic charge quantization v.s. the radius of the $U(1)$ or $\mathbb{R}$ gauge parameter
We know (say from Griffiths E&M Problem 8.12) that the electric $$q_e$$ and magnetic charge $$q_m$$ (with a distance $$\vec{z}$$ apart) can store the angular momentum in the space: $$\vec{L}=\int d^3 V (\vec{r} \times (\epsilon_0 \vec{E} \times \vec{B}))=\frac{\mu_0}{4 \pi} q_e q_m \vec{z} = n \frac{\hbar}{2}, \quad n \in \text{integers}.$$ Here the $$(\epsilon_0 \vec{E} \times \vec{B})$$ is the EM momentum obtained from the Poynting vector. This value in the quantum theory should be quantized $$n \frac{\hbar}{2}$$ because the angular momentum is quantized in Planck units. This result is general true regardless which $$q_e$$ or $$q_m$$ charges we take; thus we should better have both $$q_e$$ or $$q_m$$ charges quantized also in integers. (Any loop holes?)
The above approach to derive is based on
a classical EM derivation + a quantum theory constraint on angular momentum $$\vec{L}=n \frac{\hbar}{2}, \quad n \in \text{integers}$$.
This is also known as Dirac quantization.
This result implicitly assumes the underlying gauge theory is a $$U(1)$$ gauge theory instead of the $$\mathbb{R}$$ gauge theory. This means that the underlying $$U(1)$$ gauge transformation demands an identification $$\theta \in U(1)$$: $$\theta = \theta + 2 \pi$$ where $$\theta$$ is the parameter in front of the 2-dimensional operator $$\exp(i \theta \int * dA)$$.
My questions here are that:
1. Based on the above derivation $$\frac{\mu_0}{4 \pi} q_e q_m \vec{z} = n \frac{\hbar}{2}$$, how do we see the quantization of $$q_e$$ or $$q_m$$ charges go away to un-quantized, if we change the compact $$U(1)$$ gauge theory to a $$\mathbb{R}$$ gauge theory? So $$\theta \in \mathbb{R}$$ is no longer identified with other values of $$\theta$$.
2. how do the quantization of $$q_e$$ or $$q_m$$ charges go, if we change the compact radius of the $$U(1)$$ gauge theory by $$k$$ times? So $$\theta = \theta + 2 k \pi.$$
• There are no magnetic charges in $\mathbb{R}$ gauge theory because there is no possible non-trivial winding of the gauge group around non-trivial (in the deRham sense) cocycles. – Richard Myers Mar 1 at 19:58
First of all, let me just point out that the radius of the $$U(1)$$ is related to the gauge coupling, so rescaling the radius is essentially the same as rescaling the gauge coupling of the theory. The only place I've seen this pointed out is in references dealing with $$S$$-duality in electrodynamics since part of the transformation is on the gauge coupling, and hence it matters there because the radius of the $$U(1)$$ also gets changed as a result under the $$S$$-transformation. Basically the idea is if you choose the normalization of the gauge field $$A$$ such that the magnetic charges are integers (this does require a normalization choice and is not implied, as you conjecture, by Dirac's condition), the normalization factor then also multiplies any gauge transformation $$d\Lambda$$. From this the radius can be deduced, though I don't remember the specifics off the top of my head, just that the result is a factor of $$1/e^2$$ where $$e$$ is the gauge coupling.
The more interesting point to make in this answer is that there are no magnetic charges in $$\mathbb{R}$$ gauge theory. As a result, the Dirac quantization condition is trivial in that case. The difference is a matter of the gauge group's topology, so there isn't a smooth deformation between the two as the original question seems like one might hope. As I mentioned in a comment, the difference is essentially down to the fact that in an $$\mathbb{R}$$ gauge theory, there can be no non-trivial winding around non-trivial 2-cycles of the gauge parameter.
This is simplest to see in the canonical example of a monopole sitting stationary at the origin. Topologically, the space is $$\mathbb{R}^3/\{\boldsymbol 0\}\times\mathbb{R}$$. That is, $$\mathbb{R}^4$$ with the origin in a spatial slice excised. This is the basic example of how a monopole can exist without breaking gauge invariance (Gauss' laws). Because we have deleted a point, we can no longer cover all of spacetime in a single coordinate chart. Usually the way we handle this is to look at just one spatial slice (this is a static situation so the temporal factor can largely be ignored) and in particular a 2-sphere enclosing the deleted point. The space can be covered by two coordinate charts, one valid over the north pole of the sphere (and for all radii) and one valid over the south pole which overlap at the equator.
Since the magnetic charge is defined to be (I have made some normalization choices here for simplicity, these won't matter to the point I'm looking to make) $$q_m=\int_\Sigma F^{(2)}$$ where $$\Sigma$$ is any closed 2-dimensional surface and $$F^{(2)}$$ is the 2-form field strength, we can choose $$\Sigma=S^2$$ to be the sphere enclosing the deleted point. But we can write this sphere as being the north hemisphere plus the south hemisphere, $$S^2=N\cup S$$ and write $$q_m=\int_{N\cup S}d A^{(1)}=\int_N dA^{(1)}+\int_S dA^{(1)}.$$ Applying Stokes' theorem here and noting that $$\partial N=E$$ and $$\partial S=-E$$ where $$E$$ is the equator of the 2-sphere (the important part here really is that the orientations are opposite, the exact signs depend on how you orient $$E$$), the above may be written $$q_m=\int_E(A^{(1)}_N-A^{(1)}_S)$$ where $$A^{(1)}_N$$ is the vector potential in the northern patch and $$A^{(1)}_S$$ is the potential in the southern patch. From one coordinate patch to another, the vector potential can only ever differ up to a gauge transformation, so there exists some gauge parameter $$\Lambda$$ such that $$A^{(1)}_N-A^{(1)}=d\Lambda$$. In terms of this, the above integral may be written (again, Stokes theorem) as $$q_m=\Lambda(0)-\Lambda(2\pi)$$ where I have written $$\Lambda$$ as a function of the angular coordinate around the equator.
Now, this is all a lot of words to get to a final expression, $$\Lambda(0)-\Lambda(2\pi)$$, which at face value should just be zero. The reason it may not be zero, however, is because $$\Lambda$$ may not be single-valued. This is where the gauge group we are working with finally enters the picture. You see, what we really demand is that the gauge parameter be a smooth function on spacetime valued in the gauge group. If that gauge group is $$\mathbb{R}$$, then that means that since $$0$$ and $$2\pi$$ represent the same point in spacetime, they must also represent the same point in $$\mathbb{R}$$. For $$\mathbb{R}$$, that means they are literally the same, so $$\Lambda(0)=\Lambda(2\pi)$$ and hence $$q_m=0$$. This is precisely the statement that there are no magnetic monopoles in $$\mathbb{R}$$ gauge theory.
Just to drive this point home, let's see what were to happen if we took the group to be $$U(1)$$ instead. Here, because $$0$$, and $$2\pi$$ represent the same spacetime point, $$\Lambda(0)$$ and $$\Lambda(2\pi)$$ must again still represent the same point, but this time in $$U(1)$$. Within $$U(1)$$, we have the identification $$\Lambda=\Lambda+2\pi$$, and hence for all $$n\in\mathbb{Z}$$, $$\Lambda+2\pi n$$ represent precisely the same point. Hence, we then must have that $$\Lambda(0)=\Lambda(2\pi)+2\pi n$$ for any integer $$n$$, this integer being precisely the winding around the $$U(1)$$.
Putting this into our expression for the magnetic charge, we immediately obtain $$q_m=2\pi n$$ (again, remember, I made some normalization choices from the beginning for simplicity). So this is how in $$U(1)$$ we are able to obtain magnetic monopoles. The key is the difference in topology between $$U(1)$$ and the $$\mathbb{R}$$. | 2,092 | 7,720 | {"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": 86, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.701973 |
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# 2.6: Truth Tables
Difficulty Level: At Grade Created by: CK-12
### Truth Tables
So far we know these symbols for logic:
• \begin{align*}\sim\end{align*} not (negation)
• \begin{align*}\rightarrow\end{align*} if-then
• \begin{align*}\therefore\end{align*} therefore
Two more symbols are:
• \begin{align*}\land\end{align*} and
• \begin{align*}\lor\end{align*} or
We would write “\begin{align*}p\end{align*} and \begin{align*}q\end{align*}” as \begin{align*}p \land q\end{align*} and “\begin{align*}p\end{align*} or \begin{align*}q\end{align*}” as \begin{align*}p \lor q\end{align*}.
Truth tables use these symbols and are another way to analyze logic. First, let’s relate \begin{align*}p\end{align*} and \begin{align*}\sim p\end{align*}. To make it easier, set \begin{align*}p\end{align*} as: An even number. Therefore, \begin{align*}\sim p\end{align*} is An odd number. Make a truth table to find out if they are both true. Begin with all the “truths” of \begin{align*}p\end{align*}, true (T) or false (F).
\begin{align*}p\end{align*}
\begin{align*}T\end{align*}
\begin{align*}F\end{align*}
Next we write the corresponding truth values for \begin{align*}\sim p\end{align*}. \begin{align*}\sim p\end{align*} has the opposite truth values of \begin{align*}p\end{align*}. So, if \begin{align*}p\end{align*} is true, then \begin{align*}\sim p\end{align*} is false and vise versa.
\begin{align*}p\end{align*} \begin{align*}\sim p\end{align*}
T F
F T
To Recap:
• Start truth tables with all the possible combinations of truths. For 2 variables there are 4 combinations for 3 variables there are 8. You always start a truth table this way.
• Do any negations on the any of the variables.
• Do any combinations in parenthesis.
• Finish with completing what the problem was asking for.
#### Drawing a Truth Table
1. Draw a truth table for \begin{align*}p, q\end{align*} and \begin{align*}p \land q\end{align*}.
First, make columns for \begin{align*}p\end{align*} and \begin{align*}q\end{align*}. Fill the columns with all the possible true and false combinations for the two.
\begin{align*}p\end{align*} \begin{align*}q\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*}
Notice all the combinations of \begin{align*}p\end{align*} and \begin{align*}q\end{align*}. Anytime we have truth tables with two variables, this is always how we fill out the first two columns.
Next, we need to figure out when \begin{align*}p \land q\end{align*} is true, based upon the first two columns. \begin{align*}p \land q\end{align*} can only be true if BOTH \begin{align*}p\end{align*} and \begin{align*}q\end{align*} are true. So, the completed table looks like this:
This is how a truth table with two variables and their “and” column is always filled out.
2. Draw a truth table for \begin{align*}p, q\end{align*} and \begin{align*}p \lor q\end{align*}.
First, make columns for \begin{align*}p\end{align*} and \begin{align*}q\end{align*}, just like Example A.
\begin{align*}p\end{align*} \begin{align*}q\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*}
Next, we need to figure out when \begin{align*}p \lor q\end{align*} is true, based upon the first two columns. \begin{align*}p \lor q\end{align*} is true if \begin{align*}p\end{align*} OR \begin{align*}q\end{align*} are true, or both are true. So, the completed table looks like this:
The difference between \begin{align*}p \land q\end{align*} and \begin{align*}p \lor q\end{align*} is the second and third rows. For “and” both \begin{align*}p\end{align*} and \begin{align*}q\end{align*} have to be true, but for “or” only one has to be true.
#### Determining the Truths of Variables
Determine the truths for \begin{align*}p \land (\sim q \lor r)\end{align*}.
First, there are three variables, so we are going to need all the combinations of their truths. For three variables, there are always 8 possible combinations.
\begin{align*}p\end{align*} \begin{align*}q\end{align*} \begin{align*}r\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
Next, address the \begin{align*}\sim q\end{align*}. It will just be the opposites of the \begin{align*}q\end{align*} column.
\begin{align*}p\end{align*} \begin{align*}q\end{align*} \begin{align*}r\end{align*} \begin{align*}\sim q\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
Now, let’s do what’s in the parenthesis, \begin{align*}\sim q \lor r\end{align*}. Remember, for “or” only \begin{align*}\sim q\end{align*} OR \begin{align*}r\end{align*} has to be true. Only use the \begin{align*}\sim q\end{align*} and \begin{align*}r\end{align*} columns to determine the values in this column.
\begin{align*}p\end{align*} \begin{align*}q\end{align*} \begin{align*}r\end{align*} \begin{align*}\sim q\end{align*} \begin{align*}\sim q \lor r\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
Finally, we can address the entire problem, \begin{align*}p \land (\sim q \lor r)\end{align*}. Use the \begin{align*}p\end{align*} and \begin{align*}\sim q \lor r\end{align*} to determine the values. Remember, for “and” both \begin{align*}p\end{align*} and \begin{align*}\sim q \lor r\end{align*} must be true.
\begin{align*}p\end{align*} \begin{align*}q\end{align*} \begin{align*}r\end{align*} \begin{align*}\sim q\end{align*} \begin{align*}\sim q \lor r\end{align*} \begin{align*}p \land(\sim q \lor r)\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
### Examples
Write a truth table for the following variables.
#### Example 1
\begin{align*}p \land \sim p\end{align*}
First, make columns for \begin{align*}p\end{align*}, then add in \begin{align*}\sim p\end{align*} and finally, evaluate \begin{align*} p \land \sim p\end{align*}.
\begin{align*}p\end{align*} \begin{align*}\sim p\end{align*} \begin{align*}p\land \sim p\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
#### Example 2
\begin{align*}\sim p \lor \sim q\end{align*}
First, make columns for \begin{align*}p\end{align*} and \begin{align*}q\end{align*}, then add in \begin{align*}\sim p\end{align*} and \begin{align*}\sim q\end{align*}. Finally, evaluate \begin{align*}\sim p \lor \sim q\end{align*}.
\begin{align*}p\end{align*} \begin{align*}q\end{align*} \begin{align*}\sim p\end{align*} \begin{align*}\sim q\end{align*} \begin{align*}\sim p \lor \sim q\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
#### Example 3
\begin{align*}p \land (q \lor \sim q)\end{align*}
First, make columns for \begin{align*}p\end{align*} and \begin{align*}q\end{align*}, then add in \begin{align*}\sim q\end{align*} and \begin{align*}q \lor \sim q\end{align*}. Finally, evaluate \begin{align*}p \land (q \lor \sim q)\end{align*}.
\begin{align*}p\end{align*} \begin{align*}q\end{align*} \begin{align*}\sim q\end{align*} \begin{align*}(q \lor \sim q)\end{align*} \begin{align*}p \land (q \lor \sim q)\end{align*}
\begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*}
\begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
\begin{align*}F\end{align*} \begin{align*}F\end{align*} \begin{align*}T\end{align*} \begin{align*}T\end{align*} \begin{align*}F\end{align*}
### Review
Write a truth table for the following variables.
1. \begin{align*}(p \land q) \lor \sim r\end{align*}
2. \begin{align*}p \lor (\sim q \lor r)\end{align*}
3. \begin{align*}p \land (q \lor \sim r)\end{align*}
4. The only difference between #1 and #3 is the placement of the parenthesis. How do the truth tables differ?
5. When is \begin{align*}p \lor q \lor r\end{align*} true?
6. \begin{align*}p \lor q \lor r\end{align*}
7. \begin{align*}(p \lor q) \lor \sim r\end{align*}
8. \begin{align*}(\sim p \land \sim q) \land r\end{align*}
9. \begin{align*}(\sim p \lor \sim q) \land r\end{align*}
Is the following a valid argument? If so, what law is being used? HINT: Statements could be out of order.
1. \begin{align*}p \rightarrow q \\ r \rightarrow p\\ \therefore r \rightarrow q\end{align*}
2. \begin{align*}p \rightarrow q\\ r \rightarrow q\\ \therefore p \rightarrow r\end{align*}
3. \begin{align*}p \rightarrow \sim r\\ r\\ \therefore \sim p\end{align*}
4. \begin{align*}\sim q \rightarrow r\\ q\\ \therefore \sim r\end{align*}
5. \begin{align*}p \rightarrow (r \rightarrow s)\\ p\\ \therefore r \rightarrow s\end{align*}
6. \begin{align*}r \rightarrow q\\ r \rightarrow s\\ \therefore q \rightarrow s\end{align*}
To view the Review answers, open this PDF file and look for section 2.6.
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86 Chapter Three Stoichiometry Thus the mass of 1 mole of CaC0 3 (1 mol Ca 2 + plus 1 mol CO/-) is 100.09 g. This is the molar mass. b. The mass of 1 mole of CaC0 3 is 100.09 g. The sample contains nearly 5 moles, or close to 500 g. The exact amount is determined as follows: 100.09 g CaC0 3 4.86 ~ X = 486 g CaC0 3 1~ To find the mass of carbonate ions (CO/-) present in this sample, we must real- ize that 4.86 moles of CaC0 3 contains 4.86 moles of Ca2+ ions and 4.86 moles of CO/- ions. The mass of 1 mole of C0 3 2 - ions is 1 C: 1 X 12.01 = 12.01 g 30: 3 X 16.00 = 48.00 g Mass of 1 mol CO/- = 60.01 g Thus the mass of 4.86 moles of CO/- ions is 60.01 g C0 3 2 - 2- 4.86 ~ X 1111Ql. .-fB{~ = 292 g C0 3 SEE EXERCISES 3.51 THROUGH 3.54 EXAMPLE 3.8 j Isopentyl acetate is released when a bee stings. Isopentyl acetate Oxygen Carbon Hydrogen Isopentyl acetate (C 7 H I4 0 2 ) is the compound responsible for the scent of bananas. A mo- lecular model of isopentyl acetate is shown in the margin below. Interestingly, bees release about 1 ILg (1 X 10-6 g) of this compound when they sting. The resulting
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Unformatted text preview: scent attracts other bees to join the attack. How many molecules of isopentyl acetate are released in a typical bee sting? How many atoms of carbon are present? Solution Since we are given a mass of isopentyl acetate and want to find the number of molecules, we must first compute the molar mass: 7 mol C X 12.01!r = 84.07 g C 14 mol H X l.008!r = 14.11 g H 2 mol 0 X 16.00!r = 32.00 g 0 130.18 g This means that 1 mole of isopentyl acetate (6.022 X 10 23 molecules) has a mass of 130.18 g. To find the number of molecules released in a sting, we must first determine the num-ber of moles of isopentyl acetate in 1 X 10-6 g:-6 1 mol C 7 H 14 2-9 1 X 10 ~ X = 8 X 10 molC 7 H 14 2 130.18~ Since 1 mole is 6.022 X 10 23 units, we can determine the number of molecules: 6.022 X 10 23 molecules _ 8 X 10-9 ~ X = 5 X 10 1 ) molecules 1~...
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The text below was copied from Daniel Fry's Books and reformatted for the wiki, including new headings for important sections.
## Description of the Craft's Propulsion
"You are seeing the parts of the ship and its mechanism which your mind is capable of grasping. The large drum like structure just above the central bulkhead is the differential accumulator. We can recharge it from the energy banks of our own ship but this is seldom necessary. In your stratosphere, for example, there are several layers of ionized gas which, although they are highly rarefied are also highly charged. By placing the ship in a planetary orbit at this level it is able to collect during each orbit several times the energy required to place it in orbit. It would also, of course, collect a significant number of high energy electrons from the sun."
"By the term 'charging the differential accumulator' I merely mean that a potential difference is created between two poles of the accumulator. The accumulator material has available free electrons in quantities beyond anything of which you could conceive. The control mechanism allows these electrons to flow through various segments of the force rings, which you see at the top and bottom of the craft. The tremendous surge of electrons through the force rings creates a very strong magnetic field. Since the direction and amplitude of the flow can be controlled through either ring and in several paths through a single ring, we can create a field, which oscillates in a pattern of very precisely controlled modes. In this way, we can create magnetic resonance between the two rings or between several segments of a single ring. As you know any magnetic field, which is changing in intensity will create an electric field, which at any given instant is equal in amplitude, opposite in sign and perpendicular to the magnetic field."
### Artificial Gravitational Force Vector
If the two fields become mutually resonant, a vector force will be generated. Unless the amplitude and the frequency of the resonance is quite high, the vector field will be very small and may pass unnoticed. However, the amplitude of the vector field increases at a greater rate than the two fields which generate it and, at high resonance levels, becomes very strong. The vector field, whose direction is perpendicular to each of the other two, creates an effect similar to and in fact identical with a gravitational field.
### Center of Force Field Propulsion
If the center of the field coincides with the craft's center of mass, the only effect will be to increase the inertia or mass of the craft. If the center of mass does not coincide with the center of force the craft will tend to accelerate toward that center. Since the system which creates the field is a part of the ship it will, of course, move with the ship and will continue constantly to generate a field whose center of attraction is just ahead of the ship's center of mass so that the ship will continue to accelerate as long as the field is generated."
## The Non Linearity of Physical Law
For several thousands of years, the more advanced thinkers among the races of earth, have dreamed of the day when earthmen would succeed in breaking the bonds of his terrestrial prison, to soar freely out into space, and to explore at will, the utmost depths of a boundless universe. To most men, however, the dream had seemed to be one that was impossible of fulfillment.
Now we are suddenly awaking to the fact that the dream is becoming a reality, that this generation is going out into space.
The author has stated, in a number of recent lectures, his opinion that some of the young men who are now in their first or second year of high school, will stand upon the surface of Mars or Venus before they reach their thirtieth birthday. While this prediction may appear to be over optimistic to most readers, it is nevertheless a fact, that the development of the physical science progresses at a rate which constantly amazes, even those who are foremost in its pursuit.
Man's attempt to escape from the rather irksome confines of his tiny planet, has always been hampered by his lack of understanding of four of the primary factors of the universe: Gravity, Space, Time and Energy. It has always seemed that there was too much of gravity and space, and not enough of energy or time. About the year nineteen hundred and five, however, it was brought to man's attention that these factors were not the absolute and independent entities that he had always considered them to be, but that they were variable factors, each having a value which depended upon the value of others. Thus the first faint light of understanding began to filter through the dense screen of absolute determinism which had been erected about the physical science.
Unfortunately, our men of science, instead of pursuing this bright gleam of truth, attempted, from force of habit, to mould it into the common pattern of knowledge, by reducing it to a mathematical formula, which could be used without the necessity of understanding it.
The series of mathematical formulae which Albert Einstein gave to the world in 1905, he called "A Theory of Special Relativity." We have attempted to make of it a 'universal law of absolutes'.
We have ignored the foreword with which he prefaced the mathematics, and so have created the very thought blocks which he had hoped to prevent.
### Apparently Different Laws in the Microcosm and Macrocosm
We will refer to this problem later on, but it might be wise first, to devote a little time to the consideration of what we will call 'the non linearity of physical law.' Until a few decades ago the physical laws which govern the universe were considered to be linear. That is: we had, by trial and error, by observation and test, developed a set of laws which apparently held true for all of the small segment of nature which we were able to observe at the time. We assumed therefore that these laws would hold true in any segment of nature, no matter how far removed from our point of observation. When, however, the study of physics moved into the microcosm, that is, when we began to examine the interior of the atom, we found that we were dealing with laws which did not agree with those to which we had been accustomed. These laws too appeared to be linear but followed a different ratio of response. The same disturbing situation was discovered in our examination of the macrocosm. When our astronomers developed the giant telescope capable of peering many millions of light years into space, they found that here too, the physical laws appeared to undergo a definite change.
For a time we attempted to accustom ourselves to the existence of three sets of physical laws, each set linear within its own range of observation, but having different fundamental characteristics. Then with the development of the principles of relativity, we began to understand; or at least, we should have understood, that these different sets of linear laws, were not actually linear, nor were they different sets of laws. They were simply three widely separated segments of the one great curve of natural law.
As long as we were dealing with quantities which could be observed with the unaided eye, or with simple instruments we were unable to detect the curvature, because the segment which we were observing constituted such a tiny portion of the curve that its deviation from linearity was too slight to be detected.
For most practical purposes connected with the ordinary mechanics of our daily lives, these laws are still considered to be linear. Calculations are simpler when they are so considered, and the resulting error is negligible. For the same reason, a surveyor who is surveying a small residence lot, does not find it necessary to take into consideration the curvature of the earth, because the error resulting from this neglect has no significant effect upon the placing of his stakes. If, however, the surveyor is to make accurate measurements of large areas such as a state or continent, it does become imperative to consider the curvature of the earth's surface, and to do this, of course, it is necessary to have a reasonably accurate knowledge of the radius of that curvature.
### Radius of Curvature of Natural Law
The necessity of an accurate determination of the radius of curvature of natural law was first realized perhaps, by Dr. Albert Einstein, who devoted a large part of his life and his work to this problem. The results which he obtained have filled a number of textbooks, and have proved to be of inestimable value in the progress of the physical science. They pointed the way to the utilization of nuclear energy, and have many other implications which are sensed, but are not yet completely understood.
The difficulty with our present mathematical approach to the problems of relativity lies not in any error of the mathematics themselves, but in the fact that the methods and terms used in the attempt to explain them often lead to incorrect thinking and assumptions.
In the theories of relativity given to the world by Dr. Einstein, the natural laws, in general, are assumed to be linear, but the space in which they operate is considered to be 'curved.' This concept offers the simplest mathematical presentation, since all of the deviations from linearity can thus be explained by a single postulate. Unfortunately, like most of our mathematical presentations, the concept offers but little for the mind to grasp. A curved space cannot be pictured mentally, nor can it be drawn upon paper. The question always arises, if space is inside the curve, what is outside?
### Sine Wave & Zero Point
We have discovered that the linear mathematics which we commonly apply to the 'laws' or rules of nature, do not hold true when carried to an extent which permits the error to be measured, because they do not follow a straight line reaching to infinity, but a curve of finite radii. In a timeless universe, this curve, in any given plane, would be represented by a circle, but since the laws operate through time as well as space, the curve may be more readily understood if represented by a 'sine curve' or 'wave.' The 'base' (which is the center line of the curve) represents zero, while the portions above and below the zero line represent the positive and negative aspects of the law.
Thus we see that there are positions and conditions in which the effect of a natural law will reach zero value with respect to a given reference point, and that beyond these positions and conditions, the law will become negative, reversing its effect with respect to the observer. (The constant repetition of the term 'reference point' or ‘observer' is necessary to emphasize the frequently forgotten fact that none of the basic factors of nature have any reality or significance except when considered from a specified position, or condition.)
If, therefore, we exchange the existing mathematical postulate of linear laws operating in a curved space, for a concept based upon the curvature of natural law, we will find that we have not invalidated or changed any of the presently accepted mathematics which we apply to these concepts. They can still be applied in the same way, and will give the same results. By the exchange, however, we will have achieved a position from which the operation of the natural laws can be pictured by the mind, and can be charted upon paper. Thus we will have taken a great step in the direction of understanding.
## Gravity
Perhaps the greatest obstacle to man's achievement of his dream of space travel has been a factor which has been given the name of Gravity. Its 'discovery' is usually credited, in elementary school text books, to a seventeenth century mathematician and physicist, Sir Isaac Newton. Actually, of course, every man 'discovers' gravity soon after birth; and the stone age man who first rolled a boulder down upon the head of the cave bear who was attempting to scramble up the cliff after him, was making a practical application of this force. It was, however, Sir Isaac Newton who first made a complete mathematical analysis of the subject. His conclusions were compatible with subsequent observation and test, and were virtually unchallenged until the dawn of the era of relativity.
In brief, his conclusions were that gravity is a quality which is inherent in all matter, and that it manifests itself as a mutual attraction between all bodies of matter. The value of this attraction between any two given bodies was said to be directly proportionate to the product of their mass and inversely proportionate to the square of the distance between them.
The attraction between the earth and an object near its surface is an example of this force, although it is usually described as being the 'weight’ of the object.
The difficulty with the statement that the force varies inversely as the square of the distance lies in the implication that if the distance becomes zero, the force should become infinite. Thus it would at first seem that a man standing or lying upon the surface of the earth would be one of two bodies between which the distance was zero, therefore, the weight of the man should be infinitely great. The reply to this assumption is that the force acts as though it originated at the center of the mass, called the 'center of gravity, and that the man on the surface of the earth is still some four thousand miles from its center of gravity. This explanation, however, creates a new problem in that, if we accept it literally, we must assume that if there were a well or shaft extending to the center of the earth, and if a man descended this shaft, his weight would increase as he approached the center of gravity, becoming infinite as he reached it. Actually, of course, his weight would decrease, becoming zero when his center of gravity coincided with that of the earth. So we are forced to the further explanation that gravity is inherent, not in 'bodies’, but in particles of matter, and since a man at the center of the earth would have an equal number of particles attracting him from every direction, the resultant of the forces would be zero.
### Gravity in the Atom
If we assume the gravity to reside independently within each atom, our problem is solved as far as the man and the earth are concerned, but if we look within the atom itself in the attempt to find the point where the distance becomes zero, and the force infinite, we find that the same problem again confronts us. We have not solved it, we have only changed our scale of observation. There is conclusive evidence that the attraction, called the binding energy, which exists between the Newtonian particles, (the protons and the neutrons) is intense almost beyond our ability to describe. This force, however, does not increase uniformly with increasing mass, but at certain points not only reaches zero but actually becomes negative.
We can demonstrate this fact by adding a single unit of Newtonian mass, a neutron, to the nucleus of an atom of Uranium 235. When this is done, we find that the gravitational force within the nucleus, instead of increasing becomes negative, that is, the attraction between its parts becomes a repulsion, and the parts begin to separate with considerable brisance. During the expansion, however, several new centers of gravity are formed, which, because of the smaller amount of mass involved in each, are strongly positive. The result is that two or more simpler atoms are formed.
### Atoms Not Split
In most text books, this phenomenon is described as the 'splitting' of the atom. There is an implication that it is the 'impact, or the kinetic energy of the neutron which causes the atom to split. If this were true, then obviously, a high speed neutron would split the atom more easily and surely than one with much lower speed. Actually, the opposite situation is true. The high speed neutron will not split the uranium atom at all. It must be slowed to thermal velocity so that it can settle into the nucleus before fission occurs.
Occasionally a neutron will be captured by a uranium atom, without falling directly into the nucleus. The neutron may orbit the nucleus for a very long time (as time is counted in nuclear physics), perhaps several seconds or even minutes. Eventually the neutron drops into the nucleus, and 'delayed fission' occurs, again demonstrating the fact that it is not the impact of the neutron, but its presence in the nucleus, which results in its expansion.
The expansion and subsequent condensation into several simpler atoms is a completely random process. Many simpler types of atom can, and do result from the condensation, in each case however, the smaller atoms cannot contain as many neutrons in proportion to the number of protons as the larger atom, so there are always several neutrons left over.
This phenomenon, if carefully examined and considered, will furnish several strong clues to the nature of gravity itself, but let us for the moment, content ourselves with the observation that it demonstrates that a gravitational field can, under certain conditions, become negative.
### Gravity Can Go Negative
Because of the manner in which our gravitational laws have been expressed, it has commonly been assumed that a gravitational force can manifest itself only as an attraction between two bodies of matter. This is not, however, a necessity of thought, since there is no logical reason why it should necessarily be true. In fact if it were true, it would set gravitational fields apart as the only force fields with which we are familiar which could not produce a repulsion, as well as an attraction between bodies of matter. The reason for the assumption of a universal attraction is simply that all of our early and limited observations seemed to indicate that this was true. However, as we have already mentioned, any number of observations, if made on a sufficiently limited scale, will tend to indicate that the earth is flat, rather than spherical.
### Origins of Levity Explained
For many years a school of thought existed which recognized that gravitational fields, like all other fields, must possess a dual polarity. They called these poles, gravity and levity. They assumed that some objects and materials normally possessed the quality of gravity, while others normally possessed the quality of levity. An object possessing levity would be repelled by all objects possessing gravity. The theory eventually became discredited, and was almost universally discarded, not because it was ever disproved, but because so many attempts had been made to assign this quality of levity to objects and materials which did not actually possess it. For instance it was, for a time, assumed that gases such as hydrogen and helium possessed levity because when they were contained in a light bag or envelope, they were observed to rise against the gravitational field. It was soon demonstrated, however, that their rise was not caused by any quality of levity, but simply because their specific gravity was less than that of the air they displaced. After a number of unsuccessful attempts to assign the quality of levity to specific materials and objects, the theory fell into disrepute to the extent that the very word levity has become synonymous with humorous nonsense. Nevertheless, the philosophers who developed the theory were perfectly correct in their primary postulate. They erred only in failing to realize that gravity and levity are not properties of specific materials but are conditions under which all matter may come.
We have now observed negative gravitation in the microcosm (the interior of the atom), we also observe it in the macrocosm (between the galaxies).
### Expanding Universe
Many technical articles have been written in recent years concerning "Our Expanding Universe," yet where, in any of them, can we find any logical explanation or reason why it should expand at all? Under the theory of universal attraction, all of the matter in the universe should be rapidly coalescing into one gigantic lump. Instead, we find that every one of the large groups of stars which we call 'galaxies' or 'galactic clusters' are retreating from every other group, at velocities which increase with their distance from the observer. Velocities of recession approaching that of light have been calculated for those which are most distant from us.
A number of interesting but hardly convincing theories have been advanced in the attempt to reconcile the observed state of the universe with the existing concept of universal attraction. Some of our cosmic theorists have proposed that at one time all of the matter in the universe was contained in a single tremendous star, or 'atom'. For some reason, which is not given, this atom exploded, hurling outward the matter which has become the star clusters, and imparting to them the motion which we now observe, several billions of years later. This theory, first propounded by Abbe Lemaitre, has become known in colloquial parlance, as "The big bang theory." It was popular for a time, but as knowledge of the size and nature of the universe increased, it became obvious that such a theory would not stand up if examined under the existing concept of linear natural laws.
### Why Black Holes Don't Exist
In the first place, such an inconceivably huge mass of matter, even at the very great temperature which was assumed for it, would, under Newtonian laws, produce a gravitational field so intense that no velocity less than that of light itself would be an 'escape' velocity. In fact it has been calculated that even the light emitted by this huge sun would not escape completely, but would circle in a comparatively small orbit around it. Through the concept of the curvature of physical law, however, we see that the addition of mass to an existing body does not, necessarily, increase the force of attraction between its parts, but may, under certain conditions, cause the field to become negative, and the attraction to become a repulsion. We can explain the observed actions of the present universe by postulating that an attraction exists between the individual bodies within a galaxy, because their total mass and distance is such that they are within the positive portion of the gravitational curve with respect to each other. In the vast spaces between the galaxies however, the curve dips below the zero line, with the result that a repulsion exists between
### Parry Moon Paper
In July 1958. Parry Moon, of the Massachusetts Institute of Technology, and Domina Eberle Spencer of the University of Connecticut, published an excellent paper in the Journal of the Franklin Institute, titled "The Cosmological Principle and the Cosmological Constant." This paper demonstrates, logically and mathematically, that the assumption of a positive gravitational force within galaxies or galactic clusters, and a negative gravitational force between the clusters, offers the only practical means of explaining the observed actions of these bodies.
### Zwicky Reference
In the January 1959 issue of the magazine "Astronautics', Fritz Zwicky of the California Institute of Technology, published an article headed, “Is Newton's Law of Gravitation Really Universal?" In this article Zwicky pointed out that present observations indicate rather conclusively that the gravitational fields of the galactic clusters reach zero value between the galaxies themselves. This also explains why matter, although rather evenly distributed throughout the known universe, is not distributed uniformly, but is found in quite similar concentrations at comparatively regular distances.
### How Not to Control Gravity
At this point we hear someone say, "These explanations may be very interesting to the astronomer or to the theoretical physicist, but how can they help us in achieving space travel?" The answer is, of course, that we must have some understanding of the physical laws before we can make the proper use of them in attaining our own personal ambitions.
In his dream of space travel, man has generally considered only three possibilities of escaping from the earth. First, gravity must be destroyed. That is, the operation of the gravitational field must cease between the space craft and the earth, so that it will not hinder the departure of the craft. While a number of highly imaginative stories have been written along this line of thought, no theory has ever been evolved, or test conducted which could give us any hope that such a condition can be achieved.
Despairing of the first possibility, we pass on to the second. Gravity must be shielded. Some type of screening material must be interposed between the craft and the earth to cut off or absorb the gravitational field so that while it still exists, it will no longer act upon the craft.
Here again we have found imagination raising our hopes, and reality disappointing, for no material has been discovered which shows any promise of fulfilling such a function. With our hopes considerably subdued, we pass on to the third possibility. Gravity must be overcome. We must apply a greater force, so that we can rise against the pull of gravity, even though we must pay an exorbitant tribute of energy for each foot of progress. In this last plan, we have achieved a certain degree of success. Instrument packages have been placed in orbit about the earth, one has been dispatched to the moon, and several have been placed in independent orbits about the sun.
It does not appear however, that the proper solution has yet been achieved.
When man attempts to attain his ends by pitting one natural law against another, he usually finds that it is a wasteful and laborious process. While it is true that it is perfectly possible to propel a rowboat by throwing rocks from the stern, it is not a method which an intelligent man would choose if he were aware of other possibilities. In the first place, the thrown rock must accelerate, not only the boat, bur all the rocks which remain to be thrown. If a long journey were planned, the greatest problem would be to find enough room in the boat to store the required number of rocks. Since the thrust produced is equal to the mass of the rock multiplied by the velocity of its ejection, it is obvious that there are three limiting factors.
First, there is the total mass of the available rocks, which is limited by the size of the boat which contains them. Second, there is the total amount of energy available. (This is a factor only because we have so little understanding of the true nature of energy.) The third, and at the present time the most serious factor, is the limited mechanical strength of the throwing arm.
In a rocket motor, the 'rocks' are represented by a gas produced by combining or 'burning' the fuels within the combustion chamber, the gas, at a high temperature and pressure, is expelled through an opening or 'venturi' in the stern. Since the amount of fuel is limited by the size of the rocket, the only means of increasing the total thrust is to increase the velocity of ejection, but this can only be accomplished by increasing the temperature and pressure of the gas within the combustion chamber. Regardless of the amount of energy which is available, the amount of thrust which can be produced is limited by the ability of the chamber to withstand the temperatures and pressures involved. Since these limits are reached (and often exceeded) by ordinary chemical energies, it is clear that the vastly greater energies available in nuclear reactions, are, at the present time at least, of academic interest only to the rocker engineer. In the case of craft which remained in our atmosphere, of course, more 'rocks' could be taken aboard while in flight, by scooping up the atmosphere through which the ship was traveling, and allowing the surplus energy to act upon it. In space flight, however, this is not practical.
Attempts are being made to overcome this problem through the concepts of the "Ionic" or the 'Photonic' drive, in which ions or photons are used as the 'rocks' to be thrown overboard. Ions and photons have a basic advantage over atoms or molecules in that they achieve much higher velocities without the necessity of high temperatures or pressure.
There are, however, great obstacles to the embodiment of these concepts in practical devices, and it appears unlikely that either will lead to economical space travel in the near future.
It is time to re-examine our position to see if there is not something we have overlooked. Have we forgotten the old saying, "If you can't lick 'em join 'em?"
We have tried for centuries to 'lick' the force of gravity. We have tried to destroy it, and failed. We have searched for some method of shielding ourselves from its effect. We have not discovered it. We have attempted to overcome it by opposing it with superior force, and found it a wasteful and cumbersome process. Isn't it about time we gave up the idea of fighting the force of gravity, and began to consider the possibilities of making use of it?
### Negative Gravity
We have learned that gravity, like all natural factors, has a negative, as well as a positive value. If after building our space craft, we could arrange conditions so that the ship was in the negative portion of the gravitational curve, it would fall away from the earth as easily and as naturally as a stone dropped from a tower falls toward the earth.
Of course, we hear at once the objection that, while negative gravitational fields have been shown to exist, they have been found only within the atom and at inter-galactic distances. How can we place a space ship within the negative portion of the curve, with respect to the earth? The answer to this question lies in the fact that, as we have already learned, the natural laws are not absolute, but relative. That is, the size and shape of the curve of one law is dependent upon the value and position of the others. We have seen that the nucleus of the atom of uranium 235 dips below the zero line with the addition of only one mass unit, making a total of 236, yet the nucleus of the atom of uranium 238, although close to the zero line is still on the positive side of the curve because of the fact that the shape of the gravitational curve is modified not only by the mass present but also by the number and position of the electrical charges.
When we acquire a better understanding of the laws, we will be able to produce any shape of curve we desire, with the earth as one reference point and the spacecraft as the other.
Suppose you were to hand a bar magnet and a similar bar of soft iron to a man who was intelligent, but uneducated, with the request that he examine and test the two objects in order to determine their properties. One of the properties which the researcher would be certain to list would be the 'inherent' property of mutual attraction between the two objects. He would observe that when either end of one bar approached either end of the other bar, a condition of attraction was observed. He would probably conclude that the attraction was an inherent quality of these objects, and that it would continue to persist regard-less of anything which could be done.
We know, of course, that if a length of insulated wire were wound around the soft iron bar, and a flow of electrons were induced in the winding, the two bars could be made to exhibit a repulsion as readily as an attraction. Note that in this case we have not destroyed the field of permanent magnet, we have not shielded the field, nor have we overcome it. We have simply produced a field which is in opposition to it, and the two objects now tend to separate rather than to come together.
The same possibility exists with respect to gravitational fields. While the results will probably not be achieved in the same way, it should not be too difficult to work out means of polarizing a gravitational field, once we discard the old assumption that it is impossible.
## Matter and Mass
Much of the confusion which exists in our scientific concepts today is brought about by our failure to distinguish carefully between matter and mass. Until a comparatively few years ago, it was assumed that mass was a property which was exhibited only by matter. Upon closer examination, however, it appeared that energy also possessed mass, since when energy was added to a body of matter, the mass of the body was increased.
We have defined mass as being resistance to change in the existing state of motion. It is measured by the amount of the energy which is required to produce a given change in velocity. All matter has the property of mass, but not all mass has the properties of matter. For the purposes of this discussion, we will postulate that there are two types of mass, inertial mass, which is simply the property of resistance to change in a state of motion, and the mass inherent in matter, which we will call Newtonian mass, because it includes all mass which obeys the original laws laid down by Sir Isaac Newton. Since the reader may be under the impression that all mass obeys the Newtonian laws, let us pause here long enough to examine the facts and to point out the differences in the properties of inertial and Newtonian mass.
All physicists of today are agreed that the electron has mass. Yet if it were possible for us to hold an electron between two of our fingers and then suddenly release it, we would find that there was not the slightest tendency for the electron to fall to the earth (unless the surface happened to be positively charged at the moment). The electron is nor in the least affected by the gravitational field of the earth, so long as it is at rest with respect to that field (if the electron is moving through the field, however, the direction of the motion will be affected).
The electron has mass only because it has an electric charge. As we know, when an electric charge is accelerated in space, a magnetic field is produced, and energy is required to produce this field. The energy 'spent in producing this field, is said to be the 'mass' of the electron, since it is the entire cause of its resistance to acceleration. The greater the degree of acceleration, of course, the more intense the field, and the greater the amount of energy required to produce it. So we say that the electron gains 'mass' with every increase in its velocity. If an electron could be accelerated to the velocity C (commonly called the velocity of light), it would have acquired the maximum velocity with which energy can be propagated.
It is obvious, therefore, that no amount of energy could further accelerate this electron (with respect to its original reference point), so it would be considered to have acquired 'infinite' mass.
Let us take time to examine this statement carefully, since it is a point upon which there is much confusion. The electron would have acquired infinite mass only in reference to its original energy level. If observed from a reference point which had itself received the same degree of acceleration, the mass of the electron would not have changed a particle. This increase of inertial mass with increasing velocity, is simply the measure of the kinetic energy differential between the observer and the point which he is observing.
### Infinite Energy Example - Gun & Cork
We will attempt a simple analogy, in the hope of making this more readily understood. An observer is stationed in 'free space' far from any gravitational or other fields which might affect the results of the experiment which he proposes to make. He has, in one hand, a sphere of cork or other light material which has a mass of 10 grams. In the other hand he has a pistol which fires bullets also having a mass of 10 grams and a velocity of 1000 feet per second. The man holds the ball out at arms length, and fires a bullet from the gun into it. We will postulate that the bullet is not absorbed by the cork, but shares its kinetic energy with it, so that after the impact, the bullet and the cork ball each have a velocity of 500 feet per second. The observer now fires a second bullet at the cork. This bullet also has a velocity of 1000 feet per second with respect to the observer, but now the target has a velocity of 500 feet per second in the same direction, so that there is a differential of only 500 feet per second which the bullet can share with its target. After this impact, the bullet and the ball each have a velocity of 750 feet per second. When the observer fires the third bullet, he finds that now there is a differential of only 250 feet per second between it and the target, so that the velocity of the target is raised by only 125 feet per second, and so on.
The observer notes that each succeeding bullet, although it has the same energy with respect to him, produces a smaller and smaller acceleration in the target. He would observe that the 'mass of the target' (its resistance to acceleration) appears to increase with its velocity. If he made mathematical calculations based upon his observations, they would show that the greatest velocity which he could ever induce in the target would be 1000 feet per second (the velocity of the bullets), and that to produce this velocity it would be necessary to fire an infinite number of bullets. His experiment demonstrates conclusively that as the velocity of the target approaches 1000 feet per second, his ability to further accelerate it approaches zero. Persons with lesser intelligence or insight than our observer might be convinced that this figure of 1000 feet per second was an absolute and inescapable limit.
The observer, however, as we said, has greater understanding. After he has accelerated his target to the 'limiting' velocity of 1000 feet per second (by firing an infinite number of bullets), he steps aboard a small space ship (with which he has thoughtfully provided himself), and takes off in the direction of the target. He accelerates his ship to a velocity of 1000 feet per second, with respect to his starting point, and now finds that he is back upon exactly the same energy level as his target. If there were no other bodies of matter in the universe, there would be no way in which he could determine that either he or the target were in motion, since there would be no relative motion between them, and no other reference points from which motion might be determined. In fact, he finds that the situation is exactly the same as it was before he fired the first shot, and he can now begin his shooting all over again. He does so and observes that his first bullet accelerates the target to a velocity of 500 feet per second with respect to his new reference point, and he notes that the 'infinite mass' of the target returns to its original 10 grams, as soon as he reaches the same energy level. He realizes then that the 'increasing mass' of the target is only the measure of the kinetic energy differential which exists between them. The mass approaches infinity only as the energy level approaches that of the accelerating force. (In this case it is 1000 feet per second.) In the case of the quantity C, usually called the velocity of light, the differential is equal to 3 x 1010 centimeters per second, or if we convert this velocity to its equivalent energy we would have 9 x 1020 ergs per gram of mass.
In our discussion of non-linearity of physical law, it was pointed out that the energy inherent in a gram, or any other quantity of matter is precisely the quantity of energy necessary to accelerate its mass to a velocity equal to the quantity C by energy conversion. This statement may be hotly disputed by some students who have not yet learned to distinguish between matter and mass. Their argument is to the effect that no mass can ever be accelerated to the velocity of light since the mass would then be `infinite' and consequently the energy required to produce the velocity would also be `infinite.' The incorrectness of this assumption can be demonstrated simply by pressing the button of a pocket flashlight. A beam of light will be produced which any physicist will agree has mass and which, by its very definition, is moving at the velocity of light. Yet all the energy required is released by a small amount of chemical change taking place within the cells of a battery.
## Space
Among all of the great basic factors of the Universe, perhaps the most difficult to define or explain is that which we call space. While many of our greatest philosophers and scientists have attempted definitions, few have succeeded in offering anything which the average mind could readily grasp. The German mathematician Leibnitz said, "Space is simply the order or relation of things among themselves." Several centuries afterwards, the late Dr. Einstein used almost identical terms. "Space has no objective reality except as an order or arrangement of the objects we perceive in it."
The average man's definition of space is: "That in which matter can be placed" or "that which matter occupies." This last definition is subject to dispute by those who maintain that matter does not occupy space, but is itself, only a warp or distortion in space. Another school of thought insists with equal vigor, that while matter does occupy space, it creates a warp or distortion in the space surrounding it. Since both of these concepts are subject to the same set of mathematical laws, the same laws can be offered in support of either. There is little, however, in either of these postulates which seems to furnish a good foundation for understanding and it is understanding rather than algebraic formulae that we are seeking in this discussion.
For our purpose, a simple definition will suffice. Space is that which separates bodies of matter, whether these bodies be atoms, galaxies or any component part of either. We can extend this definition by stating that the degree of separation which exists between any two bodies is determined by the degree of curvature of the natural laws which exist between them. In making observations, of course, we must remember that, since the natural laws are relative, the mass of the body itself influences the degree of curvature. In the theories of relativity given to the world by Dr. Einstein, the natural laws, in general, retain their linearity, but the space in which they operate is considered to be curved. This concept offers the simplest mathematical presentation, since all of the observed deviations from linearity can thus be explained by a single postulate. Unfortunately, like most of our mathematical presentations, the concept offers but little for the mind to grasp. A curved space cannot be pictured mentally, nor can it be drawn upon paper. There is always something remaining outside the curve. Furthermore, attempts to rationalize this concept lead to many paradoxical statements which become more and more evident, the greater the effort to explain.
One of the best efforts to bring to the average mind an understanding of the principles of relativity, was made by Lincoln Barnett in his well known book, "The Universe and Dr. Einstein." Because of its careful preparation and its explicit presentation of present theory however, it brings out very clearly the paradox which must exist between successive assumptions. For instance: reference was made, as has already been noted, to the theory of Abbe Lemaitre, which supposed that at one time all the matter in the universe was contained in one huge lump or star. Since the curvature of space is considered to be determined by the amount or density of the matter present in it, at that time the universe was very small. That is, it had a very high degree of curvature. Light and other forms of energy do not move outward from this curve, but follow the circumference, so that the light emitted by this body, after a comparatively short journey, returned to its starting point.
No attempt was made to speculate upon the length of time in which this body had existed, or the origin of the matter and energy of which it was composed. The theory merely supposed that, after perhaps an infinity of quiescence, this body suddenly exploded. Portions of the mass moved outward in all directions and thereby enlarged the radius of space. If the radius of space was increased, it is obvious that the matter did not follow the curvature of space, but actually moved perpendicularly to it, (or perhaps at a tangent). At any rate, we see that while the radiated energy followed the 'curvature' of space whose radius was determined by the mass and density of the matter, when the matter itself expanded, instead of following the curve, its motion increased the radius.
It is interesting to note that the statement is repeatedly made that this sudden expansion began about two billion years ago, yet in the preceding paragraphs it has been stated that the calculated radius of the universe is now about 35 billion light years. Simple calculation would indicate then that the universe, or at least that portion which we call space, must have moved outward at an average velocity equal to about seventeen times the velocity of light. Either this velocity of expansion is still maintained or at some period in the past it must have been even greater.*
These statements raise some perplexing questions. In our theories of relativity it is assumed that light follows the 'curve' of space. Yet it is difficult to picture a photon following a curve whose radius is expanding at a rate equal to seventeen times the velocity of the particle.
In the book "The Universe and Dr. Einstein" it is also stated that: while space is expanding rapidly, the matter of the universe, which is likened to "inelastic patches on the surface of an expanding balloon," is not expanding with the space, since if it were, we could not detect the expansion.
If it is space that is expanding, it is difficult to understand why we have never detected the increasing distance between the earth and the moon or the sun. No attempt was made to explain why the space which exists between the individual atoms, and between the component parts of those atoms, should not expand also.
None of these difficulties, of course, invalidate any of the mathematical laws from which the concepts have been derived, but they do emphasize the great need for explanations which are more compatible with reason and understanding. For instance, in the above case would it not be simpler to assume that the degree of separation which exists between the Galaxies, when considered as individual bodies, is apparently increasing because they occupy opposite portions of the sine curve of natural law?
If we exchange our postulate of linear laws and a 'curved space' for a concept which incorporates the curvature of natural law, we find that we have not thereby destroyed or invalidated any of our present mathematics, but we have achieved a position from which the operation of the natural laws can be pictured by the mind, and can be charted upon paper. Thus we have taken a great stride in the direction of understanding.
### Space Summary
In summing up our discussion of space we should recall
1. Our definition- Space is that which separates bodies of matter. This separation is a vector function of the time, energy and mass differentials.
2. The degree of separation which exists between any two bodies, or reference points, determines the degree of curvature of the natural laws between them.
3. The natural laws are relative. That is, the value of one can be altered between any two reference points by altering the value or relationship of the other. This last fact should always be borne in mind when we hear some dogmatist solemnly declare that we are forever barred from reaching the stars by the hopelessly great degree of separation which exists between us.
## Quantity C
We have seen that the factor known as the quantity C has a greater significance than is usually credited to it. It is not merely the velocity with which light and other forms of energy are propagated in a vacuum. The quantity C is a degree of energy differential. We can define it as the maximum differential which can exist between two reference points in the factor which we call matter. We can also define it as the minimum differential which can exist between a reference point in matter, and one in energy. This is only true, however, when the reference point in matter is at the same energy level as the observer.
One of the postulates of the theory of relativity is that as a body of matter accelerates and approaches the velocity of light, or a kinetic energy differential equal to the quantity C with respect to a given observer, the body loses dimension in the direction of motion. If the velocity reaches the velocity of light it will appear to have lost all of its dimension in this direction. To this observer it would no longer be matter, since matter, by definition, requires three dimensions. The matter would have become energy insofar as the original observer was concerned since it would now exhibit a kinetic energy differential equal to the total energy inherent in the original matter.
### Three Craft Example
This statement, however, seems to produce a misconception in the minds of many students of physics. We will therefore attempt to clarify the concept by the use of a simple analogy. We will assume that we have three space ships assembled at a given point upon the surface of the earth, (or at a given point in space.) For the purpose of this analogy we will assume that the ships are capable of any desired degree of acceleration. We will dispatch two of these ships into space, flying side by side in a given direction. We will launch the remaining craft in the opposite direction in space. We have an observer upon each of the three craft and a fourth observer who remains at the point from which they departed. We will designate the ships which departed together as A and B, the ship which is moving in the opposite direction as C, and the observer at the starting point as D. When we have accelerated all three of the ships to a velocity equal to one half that of light, (with respect to the starting point) we pause to determine what changes, if any, have taken place. To the observer at the starting point D, the three ships have become slightly shorter in the direction of their motion, and have gained a small amount of `mass,' but are otherwise unchanged. The observer upon the ship C, however, discovers that while he and his own ship appear to be unchanged, the ships A and B have lost all dimension in the line of motion, because they have reached the velocity C with respect to his reference point. They have ceased to exist as matter and have entered the plane of energy. The two observers upon the ships A and B also note that C has ceased to exist as a material object, but when they examine themselves and each other, they find that no change whatever has occurred to them or to their ships since they are all upon exactly the same energy level and no differential exists between them.
We will now accelerate all three ships to the velocity C with respect to their starting point D. At this velocity the three ships cease to exist materially insofar as the observer at D is concerned, since they have entered the plane of energy, and are also at the zero point of the curve of time with respect to him. The observer upon the ship C would note that the ships A and B were again in existence but that they were now in the negative portion of the curve. Since this concept may prove somewhat difficult to grasp at the first attempt, it will be explained further and a simple analogy given in the chapter on Time.
The foregoing analogy also demonstrates that the term velocity has no meaning or significance except as an observed kinetic energy differential between two specified points of reference.
### The Only True Constant C
If we examine this analogy carefully, we will find that we have demonstrated the most important aspect of the factor which we have named the quantity C. C is a constant, the only true constant in the universe, because it is the pivotal point about which the natural laws become manifest. It is the factor for which many great physicists have spent years of search, even though they had it constantly in their possession. In short, the quantity C is the measure of the radius of curvature of natural law. It is the factor which will enable us to determine precisely the degree of change in the curvature of one law which will be brought about by a specified change in the application of the others. It is the factor which will eventually tell us how to place our spacecraft in either the positive or negative portion of the gravitational curve with respect to the earth or any other planet which we may choose to visit.
When we state that the quantity C is the radius of the curvature of natural law, we mean simply that if a differential of energy equal to this quantity exists between the observer and the point which he is observing, the natural laws will be suspended. If the energy differential is in excess of the quantity C, the laws will appear to operate in reverse at that point. As we stated earlier, this effect will be demonstrated by a simple analogy in our discussion of the factor called time.
### Consider C as Frequency
While we have repeatedly referred to the quantity C as an energy differential, we have heretofore considered it only in terms of kinetic energy. Some may believe that it can be reached only when there is a rate of increase or decrease in the degree of spatial separation between the reference points, equal to 3x1010 centimeters per second, or in simpler terms, a velocity equal to that of light. It is necessary therefore to point out the fact that an energy differential does not necessarily manifest itself as a velocity. It can also exist as a frequency. Our present laws of physics state that the energy level upon which an electron, a photon, or other particle exists is proportionate to its frequency. The mathematical rule is E equals Fh where E is the energy, F is the frequency and h is a factor called Planck's constant.
We can now see that a frequency differential which by the above formula is equal to 9x1020 ergs per gram also represents the quantity C. When such a frequency differential exists between the observer and the point which he is observing, we again find that the natural laws at the observed point reach zero value with respect to the observer. If the frequency differential exceeds this value, the action of the laws will become negative. A material object such as a spacecraft upon or near the surface of the earth would cease to exist as matter and would enter the plane of energy insofar as the observer on earth was concerned, but as we have previously pointed out, an observer upon or within the object, whose frequency or energy level had been raised to the same degree as that of the craft, would be unable to detect any change.
We must clear our minds of the thought block produced by the assumption that the quantity C is a factor of absolute limit. We must realize that it is a limiting factor only with respect to two given reference points, and that it is perfectly possible to conceive of a series of consecutive reference points between each two of which a differential equal to the quantity C may exist.
## Time
In his examination of the natural laws or facts of the Universe, man is greatly handicapped by the fact that insofar as time is concerned, he has never progressed beyond a uni-dimensional perception. Those who are familiar with the analogies used to explain some portions of the theory of relativity, will recall that in attempting to achieve a concept of a four dimensional continuum, the reader is asked first to imagine a man who is conscious of only one dimension in space. His entire universe consists of a single line. If a dot were placed on the line in front of him, and one behind, he would be completely imprisoned, since he would not be able to conceive of going over or around them. As his intelligence and consciousness developed, he would eventually become aware of a second dimension, and to imprison him then, it would be necessary to enclose him in a circle. With further development, he would become aware of a third dimension in which a sphere would be a prison, and so on.
We are now conscious of three dimensions of space, and have done considerable mathematical reasoning in regard to a fourth. Unfortunately, insofar as time is concerned, our consciousness has never progressed beyond the first dimension. We are confined to a single line in time. We have no concept of lateral motion, nor can we even turn around upon that line. We can only go forward. Many of the difficulties which we encounter in our attempt to understand the operation of the natural laws arise because of our severely restricted concept of the nature of time.
Time has often been referred to as the `fourth dimension' by those who attempt to explain our present concept of relativity. It is usually pointed out that, since all known bodies of matter in the Universe are constantly in motion with respect to each other, if we wish to describe the position of any body, it is necessary to give a point in time as well as a spatial relationship to any other body or bodies. There is, however, a more convincing method of demonstrating that time is a dimension, although we believe it would be more precise to consider it as the first dimension rather than the fourth since it is the one dimension in which all motion must take place. We are at the present, conscious of three dimensions of space, and we know that motion can take place in any one of the three, but whichever dimension of space is involved, the motion must also take place in time. Our term for the rate of motion is the word velocity, which is defined as being the degree of change in location per unit of time. If an object has a velocity of 1000 feet per second, with respect to our point of observation, we will see that in one thousandth of a second the object will have moved one foot. In one millionth of a second it will have moved only one thousandth of a foot, and so on. We can easily see that if the time becomes zero the motion must also become zero.
The science of photography has reached a state of development which permits us to take photographs with very short exposure times. By the stroboscopic method of photography, which is now being superceded by an even faster method, we were able to take several hundred thousand consecutive pictures in one second. In these pictures even the fastest projectile seems frozen into immobility. We have taken pictures of a rifle bullet penetrating an ordinary electric light bulb, in which three complete and consecutive pictures have been made between the time the bullet first touched the bulb and the time that the first crack appeared in the glass. In these pictures, the bullet appears to be completely motionless. Of course the taking of the pictures actually did involve a very small elapse of time, and so a very small amount of motion did occur during the taking, but it again illustrates the fact that no motion which we can perceive, can take place except within that dimension of time of which we are conscious.
Having pointed out the limitations of our consciousness concerning this factor which we call time, let us now go back and examine it as best we can, with that degree of consciousness and understanding which we have.
### What Does Simultaneous Mean?
We will again attempt to choose the simplest possible definition. We defined space as ‘that which separates bodies of matter,' so we will define time as 'that which separates events.' (If there is no discernible separation in this respect, the events are said to be simultaneous.) Of course we immediately hear the objection that events may be separated by space as well as by time, or that they may be separated by apace without being separated by time. This statement, while usually considered to be true, yet forms a stumbling block which has precipitated many a philosopher into the quagmire of misunderstanding and paradox. The difficulty arises in our attempt to define the term simultaneous. If two events are separated by space, how shall we determine whether or not they are separated by time? The observer cannot be present at the site of both events, and so must observe one or both of them through the separation of space, and therefore through the curvature of natural law which the separation represents. In referring to this problem in the introduction to his first book on relativity, Dr. Einstein pointed out that since our only contact with the world about us is through our senses, and since all of the knowledge which we have concerning the universe has come to us through them, if we are to formulate mathematical rules based upon our observations, we must begin with the postulate that the things which our senses tell us, are true. If we should observe, through a large telescope, the creation of a nova in a remote galaxy, and at the same time observe the eruption of a volcano upon our own earth, we must assume, for the purpose of our mathematics, that the two events are simultaneous. This a postulate which is difficult to accept because the faculty which we call reason immediately interposes the objection that a separation in space involves an elapse of time between the event and our perception of it. However, Dr. Einstein points out that if we allow our reason to modify our observations, we will be evolving a concept whose value is based only upon the validity of our reason rather than upon the accuracy of our observations. We must postulate that events which are observed simultaneously, occur simultaneously insofar as that observer is concerned, and that, therefore, the simultaneity of events is a condition which depends entirely upon the position of the observer with respect to those events.
### Flight to Proxima Centauri Example
If we examine this concept carefully, we find that time follows the same curve of natural law which is apparent in the operation of all the basic factors of nature, and again the radius of that curvature is measured by the quantity C. A simple analogy may serve to make this statement more readily understood. Suppose we were to start today to build a space ship. We will postulate that the ship will require one year of our time to build, and that when completed, it will be capable of infinite acceleration. We will assume that a continuous supply of energy is available from an outside source, and that the craft will continue to accelerate so long as this energy acts upon it. During the year which we spend in building the craft, light is being reflected from us into space, so that an observer with a telescope stationed at some other point in space could follow the course of its construction. When we have completed the construction of our craft we will enter it and take off for a destination which we will assume to be a planet orbiting about Alpha or Proxima Centauri, our next nearest suns, about four light years distant. We have a telescope of unlimited power in the rear of the craft pointed toward the earth which we are leaving, and another telescope at the front, focused upon the planet which is our destination. We will set the field strength for a constant acceleration, and seat ourselves at our telescopes to observe the result. After we have risen a few miles from the surface, we will, for the purpose of furnishing an additional reference point, eject from the craft and its field, a cannon ball or other sphere of metal which has been specially painted so that it can readily be observed from any distance with the aid of our unlimited telescopes. Since we had not yet reached escape velocity when the ball was ejected, we will observe that it soon begins to fall back to earth.
As we continue to accelerate, we will observe that the kinetic energy differential which we are producing between ourselves and our points of observation is producing exactly the effect upon time which is predicted by our postulate of the curvature of natural law. Since the distance or degree of separation between ourselves and the earth is increasing with time, the energy differential is negative, which means that the natural laws at the observed point will be displaced towards the base or zero line of the sine curve, insofar as our observations are concerned. If we reach a velocity equal to one half that of light, and then observe a clock on earth through our telescope, we will see that in ten hours of our time, only five hours have been recorded by the earth clock. If we observe the test sphere which we ejected during our take off, (assuming that it has not yet reached the ground) we will see that it is not falling at the rate predicted by our laws of gravitation, but at a rate only half as great. We will also observe that the sphere is not accelerating at the rate predicted by our laws, nor even at half that rate. Since we ourselves are still accelerating, the observed acceleration of the sphere is diminished by a factor which is proportionate to ours. We must remember that we can only observe events through the light which is emitted or reflected by the objects concerned with those events, and if we ourselves have a motion equal to one half that velocity in the direction in which the light is moving, then a column or sequence of light impulses which were emitted from the earth during a five hour period, would require ten hours to pass our point of observation.
When the velocity of our craft reaches that of light with respect to the earth, there will be a negative energy differential, equal to the quantity C, existing between us and our point of observation. We will observe that all natural laws upon the earth have reached zero value with respect to us. All motion and all changes have ceased. If we observe our test sphere we will see that gravity is no longer acting upon it, since it has ceased to fall. All laws of motion are in abeyance and the factor which we call time has ceased to have any significance.
To make these observations, of course, we would require one of the new telescopes which operates on the retention of vision principle, where the last image to arrive remains upon the viewing screen until a new light image arrives to change it. When we reach the velocity C, no new light will arrive, hence the picture will not change until we change our velocity.
Since we postulated at the beginning of this analogy that our craft was capable of unlimited acceleration, and since the postulated force continues to act, our velocity will continue to increase and we will have between ourselves and the earth, a rate of increase in the degree of separation which is greater than that specified by the quantity C. We can do this from our point of reference although, as will be explained later, we cannot do it from the point of view of an observer upon the earth. When we have passed through the velocity C, a startling change occurs in our observations. We no longer observe the earth from the telescope at the rear of the craft. The earth now appears in the telescope at the front, and we are no longer leaving the earth, we are now approaching it. We will see a craft which is identical to ours, and which is indeed our own craft, detach itself from us and move back toward earth ahead of us at a rate which is proportionate to our excess over the velocity C. If we observe the earth, we discover that all natural laws are operating in reverse. If we observe the test sphere we will see that it is now falling away from the earth rather than towards it. Gravity between the earth and the sphere has become negative with respect to our point of reference as have all the natural laws. We observe this through the forward telescope rather than that at the rear, because we are now overtaking the light which had passed us before we had reached the velocity C, and since we are now overtaking it, we encounter first the light which had passed us last. All events occur in reverse, just as would the scenes in a motion picture film which is being run backwards.
If we complete our journey to the planet which is our destination, at an average velocity equal to 4 times C, we will arrive with an elapsed time of one year as measured by the clocks on our own craft. During the journey, however, we will observe the elapse of five years of time upon the planet which we are approaching, and the elapse of three years of negative time upon the earth which we are leaving. In other words we will arrive at our destination three years before we left the earth. If immediately upon our arrival we seat ourselves at a telescope of sufficient power to observe the earth at close range, we will see ourselves going about the daily tasks which we performed two years before we began to build the space craft in which we made the journey. If we then focus the telescope upon the proper point in space we will see ourselves in our space craft, flying backwards toward the earth.
### Position to Observe the Sine Wave
We are now in a position from which we can observe the sine curve nature of all natural law, and to measure precisely the radius of the curvature. If we observe the earth, we see that time there is positive. That is: it is moving in the direction which we consider normal. Since there is no significant energy differential, the time rate is essentially the same, but because of the degree of spatial separation there will be a displacement along the time curve, between the observer and the point which he is observing. According to our theory of the curvature of natural law, this displacement should be equal to D divided by C, where D is the distance and C is our basic factor. In the case of our present observation the distance is equal to 4C Years, which if divided by C will equal 4 years, which is precisely the degree of displacement which we observe. If we now turn our attention to the space craft, we find that we are observing it through an energy differential which exceeds the quantity C and therefore the craft is within the negative portion of the curve, and all natural laws will be operating in reverse at that point. We are now in a unique position, in that we now can, from a single point in time or at least from a single point in the only dimension of time of which we are conscious, observe ourselves occupying three rather widely separated positions in space, First: our position at the telescope as the observer. At this point time is positive. Second, our position on the surface of the earth. Here time is also positive but has a negative displacement upon the time curve which is equal to four years. Third, our position in the space craft: here time is negative, as demonstrated by the fact that we observe it flying backwards toward the earth, and all actions taking place within it occur in reverse order. This is, of course, due to the fact that the craft had a velocity greater than that of C and so was constantly leaving behind the light which was emitted or reflected from it. As we observe the craft from our new reference point, the last light which it emitted arrives first.
If we continue to observe for several years, we will eventually see ourselves build the craft and take off into space. At the same time we can see ourselves in the same craft hurtling backward through space toward the inevitable meeting point where the past and the future join to become the present. Since we are observing ourselves simultaneously occupying three different positions in space, we can readily see that we are forced to a concept of time which includes more than one dimension. If we continue to observe the two craft, we will see that the one which is moving away from us is constantly slowing down, while the one coming toward us from the earth is accelerating. At the instant in which the velocity of the receding craft reaches zero, the approaching craft will reach it, coincide with it, and both craft will disappear completely from our view. Our lateral excursion into time has completed its curve and we have returned to the starting point of our uni-dimensional concept.
There is only one thing left to do. We immediately leap into our space craft and begin our return journey to earth. As before, we achieve an average or mean velocity equal to 4C. We land our craft near the observatory of an astronomer who is a friend of ours, and rush in to tell him of our return. We find him seated at his telescope observing our landing upon the planet which we had set out to visit. When we inform him that we achieved an average velocity of 4C, his reply is that this is impossible since the laws of relativity clearly state that no object can achieve a velocity in excess of C (with respect to a given reference point.). He will also point out that he has been observing us constantly since our take off from the earth and that only now, today, five years later, were we observed to have reached our destination. Since the journey required five years of earth time, our average velocity was only four fifths that of light.
According to the primary postulate of relativity, that for mathematical purposes we must accept the results of our observations as valid, the astronomer is perfectly correct in his statement that we did not, and could not have exceeded the velocity C. The mere fact that we may have returned, be seated at his side, and may perhaps be assisting him in his work, does not in any way affect the validity of his observations nor the mathematics of relativity which he applies thereto. He can only state that our arrival upon the distant planet, and the moment of our return to earth were in fact simultaneous.
We can see that, even if our energy level bad been so close to infinite that the outward trip had required only one second, if during the one second trip we had emitted enough light to make observation possible, the astronomer upon the earth would note that the trip required four years and one second, and so would have undeniable proof of the mathematics which postulate that only with infinite energy may the velocity C be achieved.
## New Galaxies
At this point in our progress of understanding, we shall embark upon a most ambitious journey. We are going out into space. Into the remotest depths of inter galactic space, so that we may observe, at close range, the birth processes of a new star cluster or 'Galaxy.' We will take along our consciousness, our ability to observe, and our understanding. We must, of course, leave our bodies behind, since they would not fare well in space, and also because their mass would create a gravitational field which would tend to alter the natural conditions at our point of observation. We will seek a spot which is at least a few million light years distant from any other galaxy or accumulation of matter; for it is only within these remote areas that we may observe the birth process of a new galaxy.
In the first part of this book, we discussed the almost inconceivably large number of particles which are found in each cubic inch of our atmosphere at sea level. As we move outward from the earth's surface we find that the number of particles diminishes rapidly, but still remains surprisingly large. When we have reached a height of one hundred miles we find that there are only about one millionth as many particles per cubic inch as we found at the surface, this is a density of matter so minute that we require very sensitive instruments, even to detect its existence. Yet, if we count the individual particles, we will find that there are still about 400 million, million particles in each cubic inch of space. At a few hundred miles elevation the density has diminished another million times, and we say that we have entered 'space', yet there are still many millions of particles per cubic inch.
We come to the startling realization that there simply is no such thing as 'empty space.' Astronomers have estimated that even in the remotest depths of intergalactic space, (which is our destination on this trip) there will still be found from twenty five to seventy five or more nuclear or atomic particles per cubic inch. Most of these particles are protons, or simple atoms which have attained escape velocity from the surfaces of some star, and which may have been wandering aimlessly about, perhaps for billions of years, coming into occasional collision with ocher particles, but usually with sufficient relative velocity so that mutual capture could not take place.
In the vicinity of existing galaxies, the gravitational fields created by the innumerable stars within those galaxies, tend to draw in the random particles, many of which eventually fall into one or another of the stars, and thereby assist somewhat in replenishing the mass which each star is constantly converting into energy.
We must, therefore, seek a spot which is remote from any of the existing galaxies, and approximately equidistant from the nearer ones. Even in this remote area of space we will find countless numbers of particles of matter, anti units of charge; electrons, protons or simple atoms, which have achieved escape velocity from some star, or which have been formed in space by random approach and capture. In short, we have all of the building blocks of nature, present in an exceedingly tenuous and diffuse state.
Since each of the particles of matter has mass, each has a force of attraction existing between it and ever other particle of matter in the area.
### Null Balance
If we accept the concept of the non linearity of natural law as previously outlined in this text, we find that each of these particles is also being repelled slightly by the surrounding galaxies or galactic clusters. These forces are almost inconceivably small, yet the net result of their action is to create a tendency upon the part of each randomly moving particle to move ever closer to the center of the area of attraction, which is also approximately but not exactly the center or 'null balance' point of the repulsion of the surrounding galaxies.
We will assume that we have now reached the point from which we will observe the birth of our new galaxy.
This point is at the center of a sphere of space, perhaps thirty thousand light years in diameter, within which the final concentration of matter will take place.
We must be prepared to exercise a great deal of patience, because the forces involved, and the resulting accelerations are so minute that many millions of years will probably elapse before we can detect any significant increase in the number of particles per unit of volume. Nevertheless, all of the particles within several hundreds of thousands of light years are slowly but surely acquiring a velocity in our direction.
As the concentration of matter at the center of our system increases, the intensity of its field will also increase and will add, not only to the velocity, but also to the acceleration of the inward moving particles. We are observing the condensation of a tremendously large volume of exceedingly ratified gas into a relatively small volume.
Let us assume that one hundred million years have passed since we first occupied our point of observation at the center of the newly forming galaxy. All of the particles within some thousands of light years have now acquired a very respectable velocity in our direction, and the density of the gas surrounding us is increasing with comparative rapidity. We observe however, that the particles are not falling directly toward the central point of the condensation.
We can understand this if we realize that the center or null point of the force of repulsion is determined only by the distribution and the distance of the surrounding galaxies, while the center of the force of attraction is determined by the distribution of matter within the area of condensation. Since the center of 'push' is not at the same point as the center of 'pull', there is a tendency toward the creation of an angular velocity. That is: the particles, instead of falling directly toward the center, will tend to spiral inward. Eventually this rotational motion will become general throughout the mass.
The plane in which this spin begins is determined by the location of the existing galaxies and the relative density of particles in different parts of the condensing mass, but once begun, the motion tends constantly to increase as the condensation proceeds.
The particles which are upon either side of the central plane of spin tend to fall toward the plane as well as toward the center, while those particles which are nearly perpendicular to the center of the plane of spin rend to fall inward more rapidly because of their smaller rotational velocities.
Our gas cloud now begins to take on the shape of a disk with a somewhat oblate sphere at the center. The galaxy has begun to assume its final shape, though as yet, there are no stars within it nor does it emir any light. If we were to direct a large telescope on earth towards this gas cloud, we would not be able to see it at all. Since all of the light coming from the galaxies behind it is now being absorbed, we would see only that there was an unusually large dark area in space. We would probably refer to it as a 'dark nebula,' a tremendous body of gas, still somewhat rarefied according to our usual concept of gas; which emits no light, but which does absorb, and convert to lower frequencies, almost all of the light, and other forms of radiant energy which reach it from the countless radiating stars throughout the universe.
As the nebula continues to contract, areas of comparatively high density will develop in many parts of the mass. Each of these points will become a local center of gravity, and accelerated condensation will occur towards these points.
The gas cloud now becomes broken up into a multitude of individual spheres, each of which continues to condense upon its own center, just as a cloud condenses into myriads of tiny water droplets.
Let us now direct our attention to one of these 'droplets' which is eventually to become a star in our new galaxy. It is still several millions of miles in diameter, but shrinking rapidly.
As the gas cloud condenses, the energy which it contains, becomes concentrated. The particles which while they were drifting about in space, had almost infinitely long 'mean free paths’, now come into more and more frequent and more and more violent collisions.
The temperature of the mass constantly rises. The kinetic energy which the particles have been building up during the millions of years while they were accelerating toward the common center, is now being converted into thermal energy. Eventually the mass begins to emit photons having frequencies in the visible portion of the spectrum.
We can now say that the star has been 'born', although it may still have more resemblance to a nebula, than to a star. A great deal more contraction will take place before the internal pressure of the gas begins to balance the gravitational force.
The star which we have chosen for observation is one of the millions which are forming within the central portion of the nebula. Since the nebula was created by the gradual inward movement of particles from an immense volume of space, it is apparent that it is within the spherical area at the center that the gas will first achieve a density sufficient for the process of condensation into separate stars to begin.
By this time the entire nebula has acquired a fairly uniform rotation about its center of mass. The individual stars, during their condensation, will of course retain this rotation but will also develop a rotational motion about their own center of gravity.
As the gas at the core of the new star becomes denser, the gravitational field becomes more and more intense, and the surrounding matter falls, with ever increasing rapidity toward the center.
Most of the gas which, even during the dark nebula stage, occupied dozens of cubic light years, of space, now is compressed into a sphere only a few million miles in diameter.
Earlier in this text we observed that the temperature of a given gas will be inversely proportionate to the volume which that gas occupies, so long as the total thermal energy contained remains the same.
The gas which we are observing is now billions of times more densely packed than it was when the condensation began, and the temperature has risen from a fraction of a degree absolute, to several millions of degrees. This temperature continues to rise as the high kinetic energy which the incoming particles have acquired during their long fall, is converted into thermal energy as those particles impact the randomly moving particles at the surface of the star.
The condensation of the star, from the dark nebula to its present state of development has been comparatively rapid, only a few million years being required for the process. Most of the matter available to the star has now formed into a fairly compact spheroid, and comparatively little new matter is arriving at the surface.
As the mass continues to contract, the temperature within the body of the star continues to rise, but because of the tremendous amount of radiant energy which is now escaping from the surface, its temperature will remain far below that of the interior.
The star is now a member of the class which Walter Baade, then a member of the Mount Wilson Observatory staff, named Population I, a blue white star with a surface temperature of the order of 30,000 degrees absolute, and an internal temperature of several millions of degrees. It is emitting light and heat energy at a rate much greater than can be replaced by the comparatively small amount of material which is still falling into it from the nebular cloud.
If the life process of the star ended here, its period of luminescence would be very short. Within a few thousands of years, the surface temperature would begin to fall below the point of incandescence and the star would appear as a dull red body. The continuing contraction of its mass might maintain the star in this condition for a few thousands of years more, but eventually the surface would become almost entirely dark, and a liquid or solid crust would probably begin to form.
We know, however, even from our relatively short history of astronomical observations, that the active period of a star is much greater than this. Let us, therefore, return to our nuclear scale of observation to determine the source from which the star receives its continuing supply of energy.
We must remember that much of the matter which forms our new star, consists of atoms which, eons ago, escaped from the surface of some other star. Since the atom of normal hydrogen (1H1) is the lightest of the atom family, it will acquire, at a given energy level, a greater velocity than any other atom, and since velocity is the principal factor in the escape of atoms from the gravitational field of a star, we would assume that most of the particles to be found in open space would be hydrogen atoms.
The new, star, which is simply a condensation of these particles, would also be assumed to consist principally of hydrogen.
This fact, which we can predict from our simple study of the behavior of atomic particles, has been verified many times by spectrographic analysis of the newer stars in presently existing galaxies.
Let us examine the interior of the star, to see if we can discover the source of its great energy supply. (Since we left our bodies at home when we embarked upon this extra-galactic tour, we will not be unduly inconvenienced by the high temperatures and pressures which exist in the regions in which we must conduct our observations.)
As we approach the star, we first pass through a region which, in the case of our sun, we call the corona. It is the area about a star where the incoming particles first meet resistance in their long fall. The corona is a belt of exceedingly tenuous gas whose particles have random motions. This layer of gas is much like the upper layers of the earth's atmosphere except that its temperature is very much higher. We must remember that the tremendous gravitational field of the star is attracting particles from all parts of the space surrounding it, and that they acquire very high velocities. As they fall through the star's outer layer of gas, sooner or later, each falling particle comes into direct collision with a particle of the corona gas. The linear kinetic energy is converted to radiant energy of high intensity. We observe temperatures of one trillion degrees Fahrenheit and more. The gas is, however, so ratified that the total amount of heat created per unit volume of space is small compared to the much greater quantities of energy which are being radiated from lower levels.
After we have descended through the corona, we encounter another layer of gas, much denser than the gas of the corona. This layer we will call the photosphere, because it is within this layer that most of the visible light which the star radiates, is created.
Here the temperature, as measured by the activity of the particles, is much lower, only about 11,000 degrees F, yet the gas is so much denser that the energy contained per unit volume, is many times greater than that of the corona.
The photosphere is essentially the receiving and shipping department of the star, receiving great quantities of energy from deeper levels, and radiating that energy into space in a never ending stream.
As we descend deeper into the body of the star, we find that the temperature and the pressure constantly increase. This means, of course, that as the gas becomes denser, the mean free path of the particles is becoming shorter, and their velocity is ever increasing. The frequency and violence with which the particles impact each other becomes almost impossible to describe or imagine.
As we approach the central core of the star, we find temperatures upward of twenty millions of degrees, and pressures in the billions of pounds per square inch.
Although the material is still technically a gas, because all of the particles have velocities greater than their escape velocity from each other, its density is now about ten times that of solid steel.
If we remember that in our atmosphere at 32°F and only 14.7 lbs. per square inch, the average particle has a velocity of 1760 feet per second, and undergoes five billion collisions per second, it may give us some faint comprehension of the number and violence of the collisions which take place between the particles deep within the body of a star.
We see that the shell of force which the planetary electrons create about the nucleus, is not sufficient to withstand impacts of this order, and the nucleus is soon stripped of its planetary electrons. When the bare nuclei impact other bare nuclei at this energy level we see that fusion of the two may, and frequently does take place.
The fusion of two nuclei results in the formation of a single nucleus which has a mass slightly smaller than that of the two parts from which it was created. The mass which is lost, appears as a tremendous burst of radiant energy, most of which subsequently is converted to heat. We note that this fusion or joining together of nuclear particles may occur in a number of ways, but in every case where the resultant nucleus has a mass smaller than the mass of the atom of silver, large quantities of heat will be released as a result of the combination.
We also observe that when the mass of the resultant nucleus is greater than the mass of an atom of silver, a large quantity of energy is absorbed rather than radiated, but this event occurs so infrequently that only an insignificant amount of energy is thus subtracted from the total. It is this energy of fusion which constantly replaces that which is being radiated into space from the surface of the star.
The process of fusion also gradually builds up heavier elements from the hydrogen building blocks which were the principal material of the new star. Consequently we would assume that the life expectancy of a given star is determined largely by the amount of hydrogen which it has available for fusion.
If the principal subject of our study were astronomy rather than the larger field of cosmology, we might devote several chapters to the examination of the inherent stabilities and instabilities which affect the process of fusion within a star. If we had a few billion years to spare, we might watch the infant as it changed slowly from a medium sized blue white star, to a somewhat smaller and denser white, until the ever increasing instabilities of the nuclear reactions within it finally overcame the stabilizing factors, and the entire star suddenly erupted in the tremendous blast of inconceivable energy which we call a nova.
After a few months we would see all of the material which had not been blasted irretrievably into space, slowly settle back into a very small and exceedingly dense core which we would describe as a red dwarf.
Since we have already spent many millions of years in this observational expedition, perhaps it is time for us to consider returning to earth. After all, there are many interesting things going on there too!
Before we leave, however, there is one more pattern of development which we should observe because it is, to our own egos at least, the most important of all.
In the star which we have been observing, the condensation took place in a symmetrical manner, with the result that a single sphere was formed. If we had been able to observe all of the stellar condensations simultaneously, we would have observed that in approximately one our of four or five cases, the condensation did not proceed symmetrically. The reason for this is found in the position and size of neighboring condensations. As in the case of the galactic nebula, the stellar gas cloud also begins to rotate as it condenses, and again a plane of spin is created. The particles outside this plane of spin tend to fall toward the plane as well as toward the center. As the rate of spin increases, the gas at some distance from the center, approaches orbital velocity with respect to that center. In simpler words, the centrifugal force tends to balance the gravitational pull of the central mass, and secondary centers of condensation are formed which are in orbit about the principal mass. These secondary condensations are usually very, small in proportion to the main mass, just as the main mass is small in proportion to the galaxy.
(In extreme cases, the condensing cloud may divide into two or more roughly equal parts, each of which becomes a separate star, but which then arc in rotation about a common center of gravity.
It is in the smaller condensations however, that we are particularly interested at this point.)
These smaller bodies which, in the case of our solar system, we have named 'planets,' will always be found to contain a much larger proportion of the heavier atoms, than will be found in the body of the star.
The reasons for this fact become obvious from our previous examination of atomic behavior. In the first place, we have seen that the lighter atom has a higher velocity at a given temperature, and so will reach escape velocity from a given body at a lower temperature. The condensations which result in planetary bodies, being comparatively small, do not reach the very high temperatures found in the stars, but they do reach temperatures sufficiently high to cause most of the lighter particles to reach escape velocity from the relatively small gravitational field.
Because the body is small, and the temperature low, such nuclear reactions as may occur under these circumstances do not furnish sufficient energy to replace that which is radiated, and the planet soon begins to cool.
A solid crust forms upon the surface, and the elements begin to combine in countless molecular patterns. When the surface has reached a sufficiently low temperature, the stage is set for the creation of the amino-acids which are generally conceded to be the starting point in the development of the organic forms to which we refer collectively as 'life'. The process is a delicate one, and only a small percentage of the planets may develop conditions suitable for this type of synthesis. It is also possible that the process may take place upon only a small percentage of those planets which do have suitable conditions. Yet, among the tens of billions of planets in a single galaxy, it is a virtual certainty, from a statistical standpoint, that synthesis will occur upon at least a few hundred, or perhaps a few thousand planets. (If we assume that the creation of life is directed by Divine Will, then the number might be much larger.) If we wished to follow the development of these first life forms through all of the stages of evolution required to produce a sentient being, we might have to wait for a period of time as long as that required for the formation of the galaxy, but eventually such a genus would appear. A race of beings capable of originating complex thought patterns, followed by equally complex actions.
Sooner or later, such a race would tire of its confinement upon a single planet, and would seek means to broaden the scope of its investigations, and of its movements.
Having achieved space travel, the race would proceed to radiate in all directions from its point of origin, investigating many planets, and perhaps colonizing some of those which were suitable for life but upon which life had not yet developed.
We must recall at this point, that it is the central spheroid of the galaxy which is formed first. It is in the central portion, that planets would first reach conditions suitable for life, and it is upon these planets that life would first achieve a high degree of development. Intelligent life might therefore be said to radiate from the center of a galaxy outward toward the periphery. A process which might take place over a period of several millions of years after the first race had achieved space travel.
It is with this thought, and in a very humble frame of mind that we begin our return journey to our tiny planet earth; located almost on the extreme outer edge of our own galaxy. | 19,523 | 98,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-49 | longest | en | 0.955689 |
https://www.esaral.com/q/in-a-abc-ad-is-the-bisector-of-bac-if-ab-6-cm-ac-5-cm-and-bd-3-cm-then-dc-39898 | 1,723,065,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00481.warc.gz | 598,627,781 | 11,696 | In a ∆ABC, AD is the bisector of ∠BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC =
Question:
In a ∆ABC, AD is the bisector of ∠BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC =
(a) 11.3 cm
(b) 2.5 cm
(c) 3 : 5 cm
(d) None of these
Solution:
Given: In a $\triangle \mathrm{ABC}, \mathrm{AD}$ is the bisector of $\angle \mathrm{BAC} . \mathrm{AB}=6 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$ and $\mathrm{BD}=3 \mathrm{~cm}$.
To find: DC
We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Hence,
$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$
$\frac{6}{5}=\frac{3}{\mathrm{DC}}$
$\mathrm{DC}=\frac{5 \times 3}{6}$
$\mathrm{DC}=2.5 \mathrm{~cm}$
Hence we got the result $(b)$ | 302 | 809 | {"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} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.544359 |
http://yogi--anand-consulting.blogspot.com/2012/12/ | 1,521,664,330,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647692.51/warc/CC-MAIN-20180321195830-20180321215830-00458.warc.gz | 493,332,745 | 30,837 | ## Monday, December 31, 2012
### yogi_Find 1st 2nd 3rd ... Largest Non-Unique And Unique Numbers In Range A2 to F2
Google Spreadsheet Post #948
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 31, 2012
user Ron01 said:(http://productforums.google.com/forum/?zx=7swwh6ge4uwz#!category-topic/docs/spreadsheets/HMfOCW2JO7M)
How can i find the 2nd largest number in an array?
Hi,
I know that i can use, lets say:
=MAX(A2:F2)
to find the the biggest number in the array.
Is there a function, or any other good way to find the 2nd biggest number in that array?
Thanks.
Ron.
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following is a solution to a bit more generalized problem
in addition to a further question by user Ron01 in regard to including corresponding cell references, I have added Sheet2 and Sheet3 wherein the enhanced solution includes cell references as well.
## Sunday, December 30, 2012
### yogi_Facilitate A Student To Enter StudentID Via Form And Read The Score in A Published Spreadsheet
Google Spreadsheet Post #947
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 30, 2012
user AndrewHill (thread) and Uwe Waldmann said:(https://productforums.google.com/forum/#!mydiscussions/docs/D5ekeKRi9PI)
Hi Yogi!
I've just come accross this thread as I was having the same problem like Andrew. Your solution sounds good, but I'm not sure if I got it right. Could you maybe describe the two steps a bit more precisely?
That would be great!
Cheers,
Uwe
---
Thanks for providing me more details. I have actually already tried to follow your recommendations. To make it a bit easier, I have created a form which is similar to mine (but in English):https://docs.google.com/spreadsheet/ccc?key=0Ag80lLrCqkKrdEpBbHpRSk1iclhiNjNkdkM4N29UeHc#gid=0
1) Is it possible to embed a "dynamic spreadsheet" (one in which the participant can still insert the ID) into a website? I tried to embed my spreadhseet but was only able to embed a "static" (one in which nothing can be changed).
2) In this form you will also see a problem, I have posted under another thread (https://productforums.google.com/forum/#!msg/docs/KQkISccbH2k/9RdX9g_E_OoJ) This might also have affected Andrew. After I have seen what you posted under other threads, you might also be able to give an advice on that.
3) For some reason the vlookup-function in sheet2 also didn't work, it did in my original sheet...
Final question for my interest: Do you think it's easier to give feedback via e-mail? Are there possibilities to write nice e-mails (design) via scripts or is that pretty complicated?
Thank you already for your help! You are doing a great job!
Uwe
---
So far I have been able to answer some of my questions myself by checking out some forums and Yogi's blog.
The solution to point no. 2 is an arrayformula like this: =arrayformula(if(C2:C="agree";3;if(C2:C="partly agree";2;if(C2:C="disagree";1;0)))
I have been able to claculate means with a formula like this: =arrayformula(if(row(A:B)="test";0;(H2:H+J2:J)/2)) (I still wonder if there isn't an easier way)
The vlookup-function finally also works: =vlookup(\$C\$4,Sheet1!\$B\$2:\$O\$100,11,FALSE)
There are two things which still don't work:
1) the calculation of standard deviations with an arrayformula (no idea how a solution coluld look like)
2) I still wonder if it's possible to embed a "dynamic" sheet in a website in which the user can enter his ID but change nothing else...
Would be great to get some help on that.
-------------------------------------------------------------------------------------------------------
Uwe has got a whole slew of questions ... he has got his whole project wrapped in here ... normally one can only address a specific technical question via Google Docs help forum, but let me give it a shot
this is the sheet where the student student data is logged in ... then I have computed columns to the right shown with light brown colored background ... this is the main sheet from which the student score would be drawn and published ... this sheet is the work-horse but this will not need to be published.
here is the Form via which a student will s8ubmit his/her ID: once a student has submitted her/his ID, the student can then see her/his score via the following
### yogi_In Summay Sheet SumUp Values For Entities in Row 1 Of Sheets In Column A
Google Spreadsheet Post #946
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 30, 2012
user Nick Wohlfarth said:(http://productforums.google.com/forum/?zx=y1lexgi8u6aw#!mydiscussions/docs/FEU4kEg-318)
Enter a Sheet Name into a Function from a Text Cell
Using this Function on Summary Sheet =SUMIF('agh north'!A:A;"NESE";'agh north'!B:B)
The Function Searches Column A for NESE then Summarize Data from B on Summary Sheet.
The Sheetnames are n Column A of Summary Sheet
Need a Function to Enter SheetName pull name from Column A into Formula.
This Has Been Successful Where A3 is Sheetname > =INDIRECT("'" & A3 & "'!" & C3)
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following is a solution to the problem
## Saturday, December 29, 2012
### yogi_Split Strings in Column A To Numbers Street_Addresses And Names Row By Row As Specified
Google Spreadsheet Post #945
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 29, 2012
user googledocsfan said:(https://productforums.google.com/forum/#!category-topic/docs/spreadsheets/cKhOoPp_8DM)
Is there a formula to separate contents of one cell? Please help! :) An example is listed.
Hi fellow google docs users!
I am currently working on massive data entry. Column A is data I already have. I am trying to achieve the automatic results of Column B and C.
Is there a formula I can place in Column B to achieve the results of B and C? The Last Name in column A (currently Smith) will always be the same.
Is there a formula that would split the contents of Column A after the word "Smith" passes?
A B C
1 Amber Smith 111 First St Amber Smith 111 First St
2 John Smith 15 Cupcake St John Smith 15 Cupcake St
3 Jane Smith 1556 Paradise Rd Jane Smith 1556 Paradise Rd
If this is not possible, I have hours of work to complete. :P
All of your thoughts and suggestions are greatly appreciated!!!!!!!!!!
---
When you can, could you please look at this shared file?
I am trying to produce the results as displayed inside this file. Please feel free to work within this file.
Your assistance is greatly appreciated as always.
Aloha!
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following is a solution to the problem
## Thursday, December 27, 2012
### yogi_Change Multiple Sets Of Data In Column A To Multiple Rows Beginning With Row 1 Skipping Unneeded Rows
Google Spreadsheet Post #944
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 27, 2012
user hshaikh said:(http://productforums.google.com/forum/?zx=7e91x0d3naeu#!category-topic/docs/spreadsheets/xIfao0gKDEc)
Google spreadsheet: how to change multiple sets of data in one column to multiple rows
I have 5 rows of data in one column on google spreadsheets that I would like to be displayed in one row. I have many sets of data like this which are separated by 3 rows. I want each data set to be displayed across a row (5 columns). Is there a formula I can use in order to do that over multiple data sets? I tried transpose function but was unable to loop it for other data sets.
for example, I have one column as follows:
Name
Description
location
rating
review
- 1st row (blank)
- 2nd row (blank)
- 3rd row (blank)
Name2
Description2
location2
rating2
review2
- 1st row (blank)
- 2nd row (blank)
- 3rd row (blank)
Name3
...etc.
This is how I would like this data to be displayed (multiple rows with 5 columns of data):
name description location rating review
name2 description2 location2 rating2 review2
...etc for multiple sets of data
Would love any input as I have been working on this for hours and can't figure it out.
Thanks!
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following is a solution to the problem
## Wednesday, December 26, 2012
### yogi_Combine The Results Of Six Queries In Sheet Comparer Into A Single Function in Sheet yogi_Comparer
Google Spreadsheet Post #943
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 26, 2012
user Gilles-Japon said:(http://productforums.google.com/forum/?zx=potzyfn0cfbg#!category-topic/docs/spreadsheets/5EzIJBtCPik)
Join queries
Hello,
I need to have the results of several queries to appear in a continuous column if possible. I have now 6 independent queries one under the other one. The reason to keep them one under the other, is that the column data are the same for most (ID number, Last name, First name, Spouse name) or can become so (ID Spouse ---> ID, IDchild1 ---> ID, Spouse First Name ---> First Name, Child1 first name ---> First name etc).
One of the problem is that I cannot know how many lines to keep free between queries to allow for the data to be listed without erasing the queries below.
Here are the various queries :
First query :
``` =query('Sur MailChimp'!A:AB;"select G, C, E where not '"&join(";";'Ilot484-11-2012'!A\$2:A)&"' contains G ") ```
Second query :
``` =query('Sur MailChimp'!A:AB;"select N, L, M where not '"&join(";";'Ilot484-11-2012'!A\$2:A)&"' contains N ") ```
Third query :
``` =query('Sur MailChimp'!A:AB;"select S, C, R where not '"&join(";";'Ilot484-11-2012'!A\$2:A)&"' contains S ") ```
Fourth query :
``` =query('Sur MailChimp'!A:AB;"select V, C, U where not '"&join(";";'Ilot484-11-2012'!A\$2:A)&"' contains V ") ```
Fifth query :
``` =query('Sur MailChimp'!A:AB;"select Y, C, X where not '"&join(";";'Ilot484-11-2012'!A\$2:A)&"' contains Y ") ```
Sixth query :
``` =query('Sur MailChimp'!A:AB;"select AB, C, AA where not '"&join(";";'Ilot484-11-2012'!A\$2:A)&"' contains AB ") ```
The sample sheet can be found at : https://docs.google.com/spreadsheet/ccc?key=0AkGTeIoz4AcxdDdDZjdjdWtfaU96eHEwaGx0SEFSVkE
Any idea on how to proceed ?
Thank you in advance for all your help.
Gilles
------------------------------------------------------------------------------------------------
following is a solution using VMERGE custom function written by ahab -- available in script gallery
### yogi_SetUp Formula For Computing Values For Being Early OnTime or Late
Google Spreadsheet Post #942
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 26, 2012
user moses004 said:(http://productforums.google.com/forum/?zx=potzyfn0cfbg#!category-topic/docs/spreadsheets/M6_LewvcuD4)
Can you tell me what's wrong with this formula?
I'm using a Dell laptop, Chrome browser.
Here's the formula:
=IF(C2="Early",100,IF(C2="On time",75,IF(C2="Late",50)))
The only part of it that works is the first option for 100. I have a validation list in C2: Easy, On time, Late. When I choose Early, I get 100 in the correct cell. I get nothing when I select "On Time" or "Late."
Thank you.
Joe Moses
----------------------------------------------------------------------------------------------
the problem was that some entries in user's dataValidation list included a leading space character ... so one way would have been to remove those leading unwanted space characters ... however, since this is a common error that is bound to happen, a passive approach is to account for this in the associated formula using the TRIM function
I have also included in the solution a single arrayformula for an array of values
### yogi_Pull Google Finance Value For Specified Entrity And Attribute For Latest Market Open Date On Or Prior To Specified Date
Google Spreadsheet Post #941
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 26, 2012
user EuDave said:()
Issue while getting historical data through Google finance
I use google spreadsheet on a google chrome browser, and use the google finance function to get the stock symbol's historical data. Since last week, GoogleFinance("NIFTY","High",12/22/2012) is returning the high of the Symbol from 23rd Dec till today, where as It was returning the high from 22nd(mentioned date) to today's date previously.
---------------------------------------------------------------------------------------------
following is a solution to the problem
### yogi_SetUp A Comparison Calculator Where Comparison Is Made For Items From A DropDown List Based On dataSheet
Google Spreadsheet Post #940
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 26, 2012
user donald5000 said:(http://productforums.google.com/forum/?zx=potzyfn0cfbg#!category-topic/docs/spreadsheets/ZmTpkq8kaiM)
Creating a calculator?
Does anyone know how or if I can create a calculator (two drop down lists) with a decision being provided after two items are selected (and a "calculate" button is clicked)?
So essentially, I have this spreadsheet here (Shared):
What I'm wanting to do:
1. Have that above spreadsheet and the values populate two drop down lists
2. Both drop down lists use the same data, so essentially you could choose the same item on each list and the calculator would determine both items are equal
3. Have some sort of end result displayed after both items are selected in each list
4. A calculate button would be nice, but if the calculation happens automatically after the second drop down item is selected, I'm fine with that
5. Even if the spreadsheet just shows the value and doesn't give any sort of direction, I'm fine with that too (meaning, once both drop downs are selected it just gives the number/rating for each next to the drop down box
As you can see, I'm not set on any one single way to do this, any way will work, I just would like my spreadsheet to drive this calculator, so maybe tab 2 has the calculator and tab 1 has the data like i have it?
Thanks for the help!!!!!
-------------------------------------------------------------------------------------------
following is a solution to the problem
## Tuesday, December 25, 2012
### yogi_Compare Lists In Two Different Sheets And Extract Non_Matching Values From Field A D F Of Sheet ilot484-11-2012
Google Spreadsheet Post #939
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 25, 2012
this is an adjunct to the solution in my following blog post
yogi_Compare Lists In Two Different Sheets And Extract in Third Sheet Specified Field Values For Unique Records Only
which was in response to a question by user Gilles-Japon in the following thread in Google Docs Help forum:
http://productforums.google.com/forum/?zx=xqfo7ic9pzt2#!mydiscussions/docs/_4dT8rzAKG8
---------------------------------------------------------------------------------
## Monday, December 24, 2012
### yogi_Compare Lists In Two Different Sheets And Extract in Third Sheet Specified Field Values For Unique Records Only
Google Spreadsheet Post #938
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Dec 25, 2012
user Gilles-Japon said:(http://productforums.google.com/forum/?zx=cmk5k2pk0bbf#!category-topic/docs/spreadsheets/_4dT8rzAKG8)
Compare information from two sheets
Hello,
I have two sheets with information arranged differently.
The first sheet that is sent to me regularly shows one person's information per line, including a unique ID.
The second sheet takes the same information (including unique ID's) and is arranged by family units per row, placing spouses and children in separate columns.
I want to be able to identify quickly the changes made, whenever a new updated sheet is sent to me. It could be that people have been removed from the first sheet, or people have been added.
I therefore need a double check with the results on a third sheet.
First check, looking at column A from the first sheet (one unique ID per row) for the ID and searching in several columns in sheet 2 for their matching field.
The second one, looking at several columns in sheet 2 for the unique ID's and comparing it to the first sheet's column A (where the unique ID are stored).
I have made a query, based on Yogi's blog (super resource) to match the first condition (look into sheet 1 for each unique ID and compare to 6 different columns in sheet 2), but fail to make the second query properly.
Here is the query I could make :
`=query(index('Ilot484-11-2012'!A2:A&"");"select Col1 where not '"&join(";";'Sur MailChimp'!G2:G)&join(";";'Sur MailChimp'!N2:N)&join(";";'Sur MailChimp'!S2:S)&join(";";'Sur MailChimp'!V2:V)&join(";";'Sur MailChimp'!Y2:Y)&join(";";'Sur MailChimp'!AB2:AB)&"' contains Col1 label Col1 'Numics dans Ilot484-11-2012 qui ne sont pas dans Sur MailChimp' ")`
Any idea on how to proceed ?
Thank you in advance for all your help.
Gilles
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following is a solution to the problem | 4,444 | 17,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-13 | longest | en | 0.867994 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/525/2/a/d/ | 1,713,953,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819089.82/warc/CC-MAIN-20240424080812-20240424110812-00610.warc.gz | 770,860,085 | 58,970 | # Properties
Label 525.2.a.d Level $525$ Weight $2$ Character orbit 525.a Self dual yes Analytic conductor $4.192$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [525,2,Mod(1,525)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(525, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("525.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$525 = 3 \cdot 5^{2} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 525.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$4.19214610612$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 21) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q + q^{2} - q^{3} - q^{4} - q^{6} + q^{7} - 3 q^{8} + q^{9}+O(q^{10})$$ q + q^2 - q^3 - q^4 - q^6 + q^7 - 3 * q^8 + q^9 $$q + q^{2} - q^{3} - q^{4} - q^{6} + q^{7} - 3 q^{8} + q^{9} + 4 q^{11} + q^{12} + 2 q^{13} + q^{14} - q^{16} + 6 q^{17} + q^{18} + 4 q^{19} - q^{21} + 4 q^{22} + 3 q^{24} + 2 q^{26} - q^{27} - q^{28} - 2 q^{29} + 5 q^{32} - 4 q^{33} + 6 q^{34} - q^{36} - 6 q^{37} + 4 q^{38} - 2 q^{39} + 2 q^{41} - q^{42} + 4 q^{43} - 4 q^{44} + q^{48} + q^{49} - 6 q^{51} - 2 q^{52} - 6 q^{53} - q^{54} - 3 q^{56} - 4 q^{57} - 2 q^{58} + 12 q^{59} - 2 q^{61} + q^{63} + 7 q^{64} - 4 q^{66} - 4 q^{67} - 6 q^{68} - 3 q^{72} + 6 q^{73} - 6 q^{74} - 4 q^{76} + 4 q^{77} - 2 q^{78} - 16 q^{79} + q^{81} + 2 q^{82} + 12 q^{83} + q^{84} + 4 q^{86} + 2 q^{87} - 12 q^{88} - 14 q^{89} + 2 q^{91} - 5 q^{96} - 18 q^{97} + q^{98} + 4 q^{99}+O(q^{100})$$ q + q^2 - q^3 - q^4 - q^6 + q^7 - 3 * q^8 + q^9 + 4 * q^11 + q^12 + 2 * q^13 + q^14 - q^16 + 6 * q^17 + q^18 + 4 * q^19 - q^21 + 4 * q^22 + 3 * q^24 + 2 * q^26 - q^27 - q^28 - 2 * q^29 + 5 * q^32 - 4 * q^33 + 6 * q^34 - q^36 - 6 * q^37 + 4 * q^38 - 2 * q^39 + 2 * q^41 - q^42 + 4 * q^43 - 4 * q^44 + q^48 + q^49 - 6 * q^51 - 2 * q^52 - 6 * q^53 - q^54 - 3 * q^56 - 4 * q^57 - 2 * q^58 + 12 * q^59 - 2 * q^61 + q^63 + 7 * q^64 - 4 * q^66 - 4 * q^67 - 6 * q^68 - 3 * q^72 + 6 * q^73 - 6 * q^74 - 4 * q^76 + 4 * q^77 - 2 * q^78 - 16 * q^79 + q^81 + 2 * q^82 + 12 * q^83 + q^84 + 4 * q^86 + 2 * q^87 - 12 * q^88 - 14 * q^89 + 2 * q^91 - 5 * q^96 - 18 * q^97 + q^98 + 4 * q^99
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
0
1.00000 −1.00000 −1.00000 0 −1.00000 1.00000 −3.00000 1.00000 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$3$$ $$1$$
$$5$$ $$1$$
$$7$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 525.2.a.d 1
3.b odd 2 1 1575.2.a.c 1
4.b odd 2 1 8400.2.a.bn 1
5.b even 2 1 21.2.a.a 1
5.c odd 4 2 525.2.d.a 2
7.b odd 2 1 3675.2.a.n 1
15.d odd 2 1 63.2.a.a 1
15.e even 4 2 1575.2.d.a 2
20.d odd 2 1 336.2.a.a 1
35.c odd 2 1 147.2.a.a 1
35.i odd 6 2 147.2.e.c 2
35.j even 6 2 147.2.e.b 2
40.e odd 2 1 1344.2.a.s 1
40.f even 2 1 1344.2.a.g 1
45.h odd 6 2 567.2.f.b 2
45.j even 6 2 567.2.f.g 2
55.d odd 2 1 2541.2.a.j 1
60.h even 2 1 1008.2.a.l 1
65.d even 2 1 3549.2.a.c 1
80.k odd 4 2 5376.2.c.l 2
80.q even 4 2 5376.2.c.r 2
85.c even 2 1 6069.2.a.b 1
95.d odd 2 1 7581.2.a.d 1
105.g even 2 1 441.2.a.f 1
105.o odd 6 2 441.2.e.a 2
105.p even 6 2 441.2.e.b 2
120.i odd 2 1 4032.2.a.h 1
120.m even 2 1 4032.2.a.k 1
140.c even 2 1 2352.2.a.v 1
140.p odd 6 2 2352.2.q.x 2
140.s even 6 2 2352.2.q.e 2
165.d even 2 1 7623.2.a.g 1
280.c odd 2 1 9408.2.a.bv 1
280.n even 2 1 9408.2.a.m 1
420.o odd 2 1 7056.2.a.p 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
21.2.a.a 1 5.b even 2 1
63.2.a.a 1 15.d odd 2 1
147.2.a.a 1 35.c odd 2 1
147.2.e.b 2 35.j even 6 2
147.2.e.c 2 35.i odd 6 2
336.2.a.a 1 20.d odd 2 1
441.2.a.f 1 105.g even 2 1
441.2.e.a 2 105.o odd 6 2
441.2.e.b 2 105.p even 6 2
525.2.a.d 1 1.a even 1 1 trivial
525.2.d.a 2 5.c odd 4 2
567.2.f.b 2 45.h odd 6 2
567.2.f.g 2 45.j even 6 2
1008.2.a.l 1 60.h even 2 1
1344.2.a.g 1 40.f even 2 1
1344.2.a.s 1 40.e odd 2 1
1575.2.a.c 1 3.b odd 2 1
1575.2.d.a 2 15.e even 4 2
2352.2.a.v 1 140.c even 2 1
2352.2.q.e 2 140.s even 6 2
2352.2.q.x 2 140.p odd 6 2
2541.2.a.j 1 55.d odd 2 1
3549.2.a.c 1 65.d even 2 1
3675.2.a.n 1 7.b odd 2 1
4032.2.a.h 1 120.i odd 2 1
4032.2.a.k 1 120.m even 2 1
5376.2.c.l 2 80.k odd 4 2
5376.2.c.r 2 80.q even 4 2
6069.2.a.b 1 85.c even 2 1
7056.2.a.p 1 420.o odd 2 1
7581.2.a.d 1 95.d odd 2 1
7623.2.a.g 1 165.d even 2 1
8400.2.a.bn 1 4.b odd 2 1
9408.2.a.m 1 280.n even 2 1
9408.2.a.bv 1 280.c odd 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(525))$$:
$$T_{2} - 1$$ T2 - 1 $$T_{11} - 4$$ T11 - 4
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T - 1$$
$3$ $$T + 1$$
$5$ $$T$$
$7$ $$T - 1$$
$11$ $$T - 4$$
$13$ $$T - 2$$
$17$ $$T - 6$$
$19$ $$T - 4$$
$23$ $$T$$
$29$ $$T + 2$$
$31$ $$T$$
$37$ $$T + 6$$
$41$ $$T - 2$$
$43$ $$T - 4$$
$47$ $$T$$
$53$ $$T + 6$$
$59$ $$T - 12$$
$61$ $$T + 2$$
$67$ $$T + 4$$
$71$ $$T$$
$73$ $$T - 6$$
$79$ $$T + 16$$
$83$ $$T - 12$$
$89$ $$T + 14$$
$97$ $$T + 18$$ | 3,235 | 6,361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.393513 |
https://www.marineinsight.com/naval-architecture/hull-fairing-and-development-why-and-how/ | 1,686,446,779,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00353.warc.gz | 977,772,549 | 80,471 | Hull Fairing And Development: Why And How
Remember how you might have created many things out of paper folding such as animals, flowers etc. back then you had actually learnt the concept of development of 3D dimensional objects starting with 2D laminas (e.g. paper). Now, let us think of something quite complex in three-dimensional terms but quite large in size. Yes, I’m talking about a ship’s hull.
How would you design a ship hull with all its curvatures and characteristic intricacies such that it doesn’t become too complex for production? Obviously, you would search for ways to break it down to smaller manageable pieces which can be easily processed in the workshop as the workshop has to find out practical ways to create a scaffolding/loft for bringing the desired hull shape to reality.
The concept of development of the hull plating for production is a very relevant one even in today’s CAD (Computer Aided Design) era. The ship hull is designed on a software platform which enables the user to visualise the 3D problem (Representation of the three-dimensional hull lines as 2D equivalents in various sections i.e. the sheer plan, half breadth plan and the body plans) in an intuitive manner. By making changes in any one of these sections, the user effectively manipulates the lines in 3D changing the other sections instantaneously.
Given this great advantage, we come to the actual problem of fairing a hullform while keeping the curvature developable with plating. Fairing often tends to get many definitions across literature, but it effectively serves certain purposes:
• Superior hydrodynamics: optimal vessel speed, low resistance, increased fuel efficiency, manoeuvrability, etc.
• Aesthetics: appeal to the human senses, functionality after life cycle (e.g. conversion to resorts, display in museums, etc.)
However well built a vessel might be, it might serve its purpose very well, but the finishing and fairing job done reflects the shipbuilder’s attention to the tiniest of details and is often a matter of pride for them. Finishing and fairing are often considered among the finer arts of shipbuilding.
For a moment, think of you have the job of producing a smooth surface out of a body with a coarse surface, you have the option of removing away the high points on the surface until you reach a low level (something often done with wood), also you have the option of adding material to reach a higher level to maintain evenness (paints and surface applications) you could also take some material from the higher planes and deposit it on the lower planes to reach a degree of smoothness.
Shipbuilding employs paint, surface preparation, and many other techniques to achieve the level of finishing of modern-day standards.
A ship hull is composed of numerous steel plates which are rolled or bent to give the characteristic shape at some section. Take a look here at a ship’s shell expansion plan:
This represents the developed surface of the plating used (at plate level) in the manufacture of a vessel and often gives an estimate of the steel to be used in manufacturing the ship. But since the development of a shell expansion plan comes quite later in to the ship’s design cycle, the designer is faced with the problem of ensuring that the hull surface is fairly developable. This is where software comes in. Advanced ship design productivity suites like MAXSURF or DELFTship provide fairing modules for this. The principle on which these are based is that the curvature of a certain ‘patch’ or element in the hull skin can either stand for single/multiple or positive/negative curvatures for the collection of points within it.
Certain typical representations of such curvature are as follows:
1.) In ship designs, the transverse curvature is particularly higher than the longitudinal curvature and so it might be helpful to exaggerate the transverse curvature in view to aid the designer in finding out the inconsistencies (if any) in the fairness of the model. For this reason, some programs allow this to be represented as one of the rendered forms.
2.) Another important tool allows us to modify particular waterline contours based a graphical representation of the curvature along the contour, often called curvature porcupines. Notice the radially outward emerging lines in grey? The height of these lines are proportional to the amount of curvature inherent at the location. Even a slight change in control point position alters these greatly and hence offer a better intuitive experience of fairing. In the event an outward emerging porcupine ends up inside the waterline, the section indicates a hollow in the hull surface and needs rectification. The porcupine curve which runs along the porcupines should be fairly smooth, if not, then there are options to change the stiffness of the control points used in the model.
3.) One popular and simple way of evaluating the fairness is by using the rendered hull using lighting highlights. This shows how light would bounce off the surface of the hull around the hull and often is useful in creating promotional and non-technical content too. This tool entirely depends upon the user’s perception and can only be used for major bumps/irregularities.
This article is authored by Sudripto Khasnabis and was originally published at the inaugural issue of Navisieger. Navisieger is a magazine published by Learn Ship Design. It is a confluence of insights on the maritime technology sector taken from academic experts, naval architects, industry veterans in this field, in the form of articles and exclusive interviews. Learn Ship Design is the first student initiative in India that works towards enhancing industry-academia engagements in the maritime sector. | 1,144 | 5,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.948324 |
https://www.jiskha.com/search/index.cgi?query=Calculus+%28Answer+Check%29&page=17 | 1,516,739,024,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892238.78/warc/CC-MAIN-20180123191341-20180123211341-00136.warc.gz | 969,402,553 | 13,917 | 87,112 results, page 17
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1. I'll probably go and check out that apple, actually. 2. I'll probably go and check that apple, actually. 3. I'll probably go and check on that apple, actually. ------------------ Which one is correct? Do we have to use 'out' after 'check' here? 4. The waitress who served in...
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same as my question below please help me post your answer on here if you have one on these forms of course h t t p : / / w w w . p h y s i c s f o r u m s . c o m / s h o w t h r e a d . p h p ? t = 4 2 1 7 6 3
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Explain and show how you would estimate 0.5 x 4.62 . Then justify why this is a good choice for arriving at the correct answer.
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### Calculus
ã6(2ã6 - ã5) How do you get 12-ã30? I keep getting a way different answer:\ Btw not sure if the radicals will show up. | 3,892 | 12,674 | {"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} | 3.453125 | 3 | CC-MAIN-2018-05 | longest | en | 0.80789 |
https://inperc.com/blog2/2007/11/30/connectivity-or-lengths-of-digital-curves-part-6.html | 1,624,226,740,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00318.warc.gz | 290,907,794 | 8,265 | November 30, 2007
Connectivity (or Lengths of Digital Curves, part 6)
Yes, part 6! I thought I was done with the topic (Part 1, Part 2, Part 3,…), but a couple of days ago I ran into this blog post: General connectivity (MATLAB Central) . The issue is connectivity in digital images: 4-connectivity, 8-connectivity, and other “connectivities”. The issue (adjacency of pixels) is discussed in the wiki. When I wrote this article though I did not realize that the topic is related to measuring lengths of curves. Indeed, the 8-connectivity produces curves that go only horizontally or vertically while the 4-connectivity allows diagonal edges as well. In the post the curves appear as “perimeters” of objects. More accurately, they should be called contours or boundaries of objects as the perimeter is mathematically the length of the boundary (that’s where “meter” comes from). But bwperim is the name of the standard MATLAB command for finding the boundary and we will have to live with that…
My problem is with this idea: if you are a coder it is important to make the right choice of connectivity. Let’s break this into in two statements:
1. The choice of connectivity is important.
2. The choice of connectivity is up to the coder.
The second statement reflects a general attitude that is very common - be it MATLAB or OpenCV. Important decisions are left to the coder and are hidden from the user. If something is important, I as a user would want software that produces output independent of a particular implementation. The two statements simply contradict each other.
Now, is the choice of connectivity really important?
The change of connectivity changes the boundary of the object and, therefore, its perimeter. This seems important. But we are interested of the perimeter of a “real” object, which should be independent (as much as possible) from the digital representation. This perimeter is the length of a “real” curve – the boundary of the object. We showed (in Part 3) that the relative error of the computation of length does not diminish with the growth of the image resolution. The accuracy is improved only by choosing more and more complex ways to compute the length (roughly, increasing the degree of the approximation of the curve). The choice of connectivity is determined by a 3×3 “matrix” (it’s not a matrix, it’s just a table! – another annoying thing about MATLAB). With finitely many choices the error can’t be reduced to arbitrary low. You may conceivably improve the accuracy if you can choose larger and larger “matrix” (table!), but that seems pointless…
There is another reason to think that this choice isn’t important. About that in the next post.
P.S. To clarify, every matrix is a table but not every table is a matrix (even if it contains only numbers). It is my view that tables should be called matrices only in the context of matrix operations especially multiplication. In particular, a digital image is table not a matrix.
2 Responses to “Connectivity (or Lengths of Digital Curves, part 6)”
1. Steve Eddins Says:
Hi,
MATLAB is widely used for algorithm development and prototyping in the area of image processing. Algorithm developers need to be aware of varying definitions of connectivity, and their tools need to provide appropriate choices. Defining neighbor adjacencies applies to many different image processing algorithms; bwperim was merely the example I chose to illustrate.
It’s true that bwperim does not compute the perimeter, and I understand why you might not like the name. Rather, it computes the set of digital image pixels that lie along the boundary, or perimeter, of a connected set of foreground pixels. Connectivity comes into play in deciding which pixels are on the boundary. Pixels are judged to be on the boundary if they are adjacent to a background pixel. Adjacency depends on the connectivity definition in use. I agree with you that this is not especially relevant to estimating the perimeter of the “real” object, but that is not bwperim’s purpose.
Finally, I think your definition of matrix is unnecessarily restrictive. American Heritage, for example, gives the mathematics definition of matrix as “A rectangular array of numeric or algebraic quantities subject to mathematical operations.” A table is “An orderly arrangement of data, especially one in which the data are arranged in columns and rows in an essentially rectangular form.” Either term seems to apply. Given my own background in discrete systems and signals, I’d really prefer something like “two-dimensional sequence,” but since I work in MATLAB, “matrix” is very convenient to use and I see no definition-based reason not to use it.
Best regards,
Steve | 983 | 4,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-25 | longest | en | 0.943821 |
http://theanalysisofdata.com/probability/B_4.html | 1,701,741,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00751.warc.gz | 47,635,614 | 10,744 | ## Probability
### The Analysis of Data, volume 1
Metric Spaces: The Euclidean Space
## B.4. The Euclidean Space
As described in Chapter A, the Euclidean space $\R^d=R\times\cdots\times\R$ is the set of all ordered $d$-tuples or vectors over the real numbers $\R$ (when $d=1$ we refer to the vectors as scalars). We denote a vector in $\R^d$ when $d>1$ in bold and refer to its scalar components via subscripts. For example the vector $\bb x=(x_1,\ldots,x_d)$ has $d$ scalar components $x_i\in\R$, $i=1,\ldots,d$. Together with the Euclidean distance $d(\bb x,\bb y) = \sqrt{\sum_{i=1}^d (x_i-y_i)^2},$ the Euclidean space is a metric space $(\R,d)$ (we prove later later in this chapter that the Euclidean distance above is a valid distance function).
Definition B.4.1. Multiplication of a vector by a scalar $c\in\R$ and the addition of two vectors of the same dimensionality are defined as \begin{align*} c\bb x &\defeq (c x_1,\ldots,c x_d)\in\R^d\\ \bb x+\bb y &\defeq (x_1+y_1,\ldots,x_d+y_d)\in\R^d. \end{align*}
Example B.4.1. For all vectors $\bb x$, the vector $\bb 0=(0,\ldots,0)$ satisfies $\bb x+\bb 0=\bb x$ and the scalar 1 satisfies $1\bb x=\bb x$.
Definition B.4.2. A function $T:\R^m\to\R^n$ is a linear transformation if $T(\alpha \bb u+\beta \bb v) = \alpha T(\bb u) + \beta T (\bb v)$ for all $\bb u,\bb v\in\R^m$ and all $\alpha,\beta\in\R$.
Example B.4.2. The function $T:\R^d\to\R$, defined by $T(\bb x)=\sum_{i=1}^d c_i x_i$, is a linear transformation.
Definition B.4.3. An inner product is a function $g(\cdot,\cdot):\R^d\times\R^d\to\R$ that satisfies the following for all $\bb u,\bb v, \bb w\in\R^d$ and for all $\alpha,\beta\in\R$.
1. symmetry: $g( \bb v,\bb u) = g(\bb u,\bb v)$
2. bi-linearity: $g(\alpha \bb u+\beta \bb v,\bb w) =\alpha g(\bb u,\bb w) + \beta g( \bb v,\bb w)$
3. positivity: $g(\bb v,\bb v)\geq 0$.
Note that as a consequence of the first two properties above $g(\alpha \bb u+\beta \bb v,\gamma \bb w+\delta \bb z) = \alpha\gamma g(\bb u,\bb w) + \alpha\delta g(\bb u,\bb z)+\beta\gamma g(\bb v,\bb w) + \beta\delta g(\bb v,\bb z).$
Proposition B.4.1 (Cauchy Schwartz Inequality). For any inner product $g$ and for all $\bb x, \bb y\in\R^d$ $g^2(\bb x,\bb y) \leq g(\bb x,\bb x) \, g(\bb y,\bb y).$
Proof. Using the positivity and bi-linearity properties of the inner product, \begin{align*} 0 & \leq g( g( \bb y,\bb y) \bb x-g( \bb x,\bb y) \bb y,g( \bb y,\bb y) \bb x-g( \bb x,\bb y) \bb y)\\ &=g^2(\bb y,\bb y) g(\bb x,\bb x) - 2 g(\bb y,\bb y) g^2( \bb ,\bb y) + g^2( \bb x,\bb y) g(\bb y,\bb y)\\ &=g(\bb y,\bb y)(g(\bb x,\bb x) g(\bb y,\bb y)-g^2(\bb x,\bb y)). \end{align*} Since $g(\bb y,\bb y)\geq 0$, the proposition follows.
Proposition B.4.2 The Euclidean product $\langle \bb x,\bb y\rangle \defeq \sum_{i=1}^d x_iy_i$ is an inner product
Proof. It is easy to verify the three properties above by direct substitution.
Definition B.4.3. A norm is a function $h:\R^d\to \R$ satisfying the following for all $\bb x, \bb y\in\R^d$, and for all $c\in \R$.
1. non-negativity: $h(\bb x)\geq 0$
2. positivity: $h(\bb x)=0$ if and only if $\bb x=\bb 0$
3. homogeneity: $h(c \,\bb x)=|c|\, h(\bb x)$
4. triangle inequality: $h(\bb x+\bb y) \leq h(\bb x) + h(\bb y)$.
Proposition B.4.3 The function $\|\bb x\| = \sqrt{\langle \bb x,\bb x\rangle}=\left(\sum_{i=1}^d x_i^2\right)^{1/2}$ is a norm, commonly known as the Euclidean norm.
Proof. The first three properties are obvious. The fourth property follows from the Cauchy Schwartz inequality applied to the Euclidean inner product \begin{align*} \|\bb x+\bb y\|^2 &= \langle \bb x+\bb y,\bb x+\bb y\rangle \\ &= \|\bb x\|^2 + 2\langle \bb x,\bb y\rangle +\|\bb y\|^2\\ & \leq \|\bb x\|^2 +2 \|\bb x\| \|\bb y\| + \|\bb y\|^2\\ & = (\|\bb x\|+\|\bb y\|)^2. \end{align*}
The Euclidean norm is the most popular norm. The more general $L_p$ norm \begin{align*} \|\bb x\|_p=\left(\sum_{i=1}^d |x_i|^p\right)^{1/p}, \qquad p\geq 1, \end{align*} has the following special cases \begin{align*} \|\bb x\|_2 &=\|\bb x\| \qquad \text{ (the Euclidean norm)}\\ \|\bb x \|_1 &= \sum_{i=1}^d |x_i|\\ \|\bb x\|_{\infty} &= \max\{|x_1|,\ldots,|x_d|\} \qquad \text{ (achieved by letting } p\to\infty). \end{align*}
The $L_p$ norm can be further generalized as follows.
Definition B.4.5. Assuming $W$ is a non-singular matrix (matrices, and the matrix product notation $W\bb x$, are defined at the beginning of Chapter C), the weighted $L_{p,W}$ norm is $\|\bb x\|_{p,W}= \|W \bb x\|_p.$
Weighted norms are convenient for emphasizing some dimensions over others. For example, if $\diag(\bb w)$ is an all zero matrix, except for its diagonal $\diag(\bb w) = \begin{pmatrix} w_1 &0 &\cdots& 0\\ 0 & w_2&\cdots&0\\ \vdots & \vdots & \ddots &\vdots\\ 0 & \cdots & 0& w_d \end{pmatrix},$ the corresponding weighted $L_2$ norm is $\|\bb x\|_{2,\diag(\bb w)} = \sqrt{\sum_{i=1}^d w_i^2 x_i^2}.$
Definition B.4.6. The angle between $\bb x,\bb y\in\R^d$ is the value $\theta$ for which $\cos\theta=\frac{\langle \bb x,\bb y\rangle}{\|\bb x\| \,\cdot\, \|\bb y\|}.$
Definition B.4.7. Vectors ${\bb x}^{(i)}, i=1,\ldots,k$ are said to be orthogonal if $i\neq j\qquad \text{implies} \qquad \langle {\bb x}^{(i)},{\bb x}^{(j)}\rangle=0.$ If, in addition, $\|{\bb x}^{(i)}\|=1$, $i=1,\ldots,k$, the vectors are said to be orthonormal.
Note that the orthogonality condition above corresponds to a a 90 degree angle (perpendicularity) between every two distinct vectors.
Proposition B.4.4. For any norm $h(\cdot)$, the function $d(\bb x,\bb y)=h(\bb x-\bb y)$ is a valid distance function. Accordingly, we may interpret the norm $h(\bb x)$ as the corresponding distance of $\bb x$ from the origin $d(\bb x,\bb 0)$.
Proof. It is straightforward to verify that $d(\bb x,\bb y)=h(\bb x-\bb y)$ satisfies the first three properties of a distance function. The fourth follows from applying the triangle inequality property of norms to $\bb x-\bb y$ and $\bb y-\bb z$: $h(\bb x-\bb z) \leq h(\bb x-\bb y) + h(\bb y-\bb z).$
The propositions above confirm that the Euclidean space $(\R^d,d)$, where $d(\bb x,\bb y) = \sqrt{\sum_{i=1}^d (x_i-y_i)^2},$ is a metric space.
Figure B.4.1 below shows contour lines of four $L_p$ norms in the left column and four $L_{p,W}$ norms in the right column, where $W=\diag(2,1)$. A non-diagonal matrix $W$ would result in rotated versions of the figures in the right column.
The R code below generates the left and right columns of the figure.
s = seq(-1, 1, length.out = 50)
R = expand.grid(x1 = s, x2 = s)
# generate left column (Lp norms)
D = rbind(R, R, R, R)
D$Norm[1:2500] = abs(D$x1[1:2500]) + abs(D$x2[1:2500]) D$Norm[2501:5000] = ((abs(D$x1[2501:5000]))^1.5 + (abs(D$x2[2501:5000]))^1.5)^(2/3)
D$Norm[5001:7500] = ((abs(D$x1[5001:7500]))^2 + (abs(D$x2[5001:7500]))^2)^0.5 D$Norm[7501:10000] = pmax(abs(D$x1[7501:10000]), abs(D$x2[7501:10000]))
D$p = c(rep("Lp norm, p=1", 2500), rep("Lp norm, p=1.5", 2500), rep("Lp norm, p=2", 2500), rep("Lp norm, p = inf", 2500)) ggplot(D, aes(x1, x2, z = Norm)) + facet_grid(p ~ .) + stat_contour(bins = 4) # generate right column (weighted Lp norms) D = rbind(R, R, R, R) D$Norm[1:2500] = abs(2 * D$x1[1:2500]) + abs(D$x2[1:2500])
D$Norm[2501:5000] = ((2 * abs(D$x1[2501:5000]))^1.5 +
(abs(D$x2[2501:5000]))^1.5)^(2/3) D$Norm[5001:7500] = ((2 * abs(D$x1[5001:7500]))^2 + (abs(D$x2[5001:7500]))^2)^(1/2)
D$Norm[7501:10000] = pmax(abs(2 * D$x1[7501:10000]),
abs(D$x2[7501:10000])) D$p = c(rep("weighted Lp norm, p=1", 2500), rep("weighted Lp norm, p=1.5",
2500), rep("weighted Lp norm, p=2", 2500), rep("weighted Lp norm, p=inf",
2500))
ggplot(D, aes(x1, x2, z = Norm)) + facet_grid(p ~ .) +
stat_contour(bins = 4)
Figure B.4.1: Equal height contours of the $L_p$ norm (left column) and weighted $L_{p,W}$ norm with $W=\diag(2,1)$ (right column), in the two dimensional case $d=2$. Each row corresponds to a different value of $p$. As $p$ increases from 1 to $\infty$ the contours change their shape from diamond-shape to square-shape.
Example B.4.3. Consider the set $\R^{\infty}$ defined in Example A.3.2. The proofs of the propositions above can be extended to show that $\langle \bb x,\bb y\rangle=\sum_{n\in\mathbb{N}} x_n y_n$ is an inner product on $\R^{\infty}$, $\|\bb x\|_2=\sqrt{\sum_{n\in\mathbb{N}} x_n^2}$ is a norm on $\R^{\infty}$ and $d(\bb x,\bb y)=\sqrt{\sum_{n\in\mathbb{N}} (x_n-y_n)^2}$ is a distance on $\R^{\infty}$, all provided that the infinite sums converge. In general, however, the infinite sums may not converge, making the generalizations above inappropriate.
Example B.4.4. Consider the set $\R^{\infty}$ defined in Example A.3.2 and the distance function \begin{align} \bar d(\bb x,\bb y) = \sup\left\{ \frac{\min(|x_n-y_n|,1)}{n}\,: \, n\in\mathbb{N}\right\}. \end{align} Note that $\bar d$ is well-defined and finite, is symmetric, is non-negative, and is zero if and only if $\bb x=\bb y$. It also satisfies the triangle inequality \begin{align*} \bar d(\bb x,\bb z) &= \sup\left\{ \frac{\min(|x_n-y_n+y_n-z_n|,1)}{n}\,:\, n\in\mathbb{N}\right\} \\ &\leq \sup\left\{ \frac{\min(|x_n-y_n| + |y_n-z_n|,1)}{n}\,:\, n\in\mathbb{N}\right\}\\ &\leq \sup\left\{ \frac{\min(|x_n-y_n|,1)}{n} + \frac{\min(|y_n-z_n|,1)}{n}\,:\,n\in\mathbb{N} \right\}\\ &\leq \bar d(\bb x,\bb y)+\bar d(\bb y,\bb z) \end{align*} and is therefore a distance function.
If $\bb x\in B_{\epsilon}(\bb 0)$ for a given $\epsilon$ then for all $n\in\mathbb{N}$ we have $\min(|x_n|,1) < n\epsilon$. There exists some $N$ such that $n>N$ corresponds to $n\epsilon > 1$, implying that the components $x_{N+1}, x_{N+2},\ldots$ of vectors $\bb x\in B_{\epsilon}(\bb 0)$ are unrestricted. Similarly, for $\bb x\in B_{\epsilon}(\bb 0)$ the components $x_n$, where $n < N$, satisfy $|x_n| < n\epsilon$. It follows that points in $B_{\epsilon}(\bb 0)$ are an intersection of a finite number of sets of the form \begin{align} \tag{*} \R\times \cdots \times \R\times (a,b) \times \R\times\cdots. \end{align} (we refer to sets of the form (*) as simple cylinders.) Since $\bar d(\bb x+\bb c,\bb z+\bb c)=\bar d(\bb x,\bb z)$, we also have that $B_{\epsilon}(\bb y)$ is an intersection of a finite number of simple cylinders or sets of the form expressed in (*).
On the other hand, let $A$ be an intersection of a finite number of simple cylinders. Then for each $\bb x\in A$ we can construct $B_{\epsilon}(\bb y)$ such that $\bb x\in B_{\epsilon}(\bb y)\subset A$ (taking $\epsilon$ to be sufficiently small). This implies that $A$ is a union of open balls and is therefore an open set. A union of intersections of a finite number of simple cylinders is a union of open sets and therefore is open also.
In summary, we have thus demonstrated that in the space $(\R^{\infty},\bar d)$, the set of open sets is equivalent to the set of unions of intersections of a finite number of simple cylinders.
The convergence problems mentioned in Example~B.4.3 leads to the common practice of defining the metric structure on $\R^{\infty}$ using the distance function $\bar d$ in Example B.4.4 rather than the Euclidean distance. In fact, whenever we refer to the metric structure of $\R^{\infty}$ we will assume the metric structure of $\bar d$ derived above. This metric structure is commonly referred to in the literature as the product topology of $\R^{\infty}$.
The metric structure of the Euclidean space simplifies some of the properties described in the previous chapter.
Proposition B.4.5. The Euclidean space $\R^d$ is second countable, and in particular one choice for $\mathcal{G}$ in Definition B.1.5 is the set of all open balls with rational centers and radii. Similarly, the space $(\R^{\infty},\bar d)$ (see Example B.4.4) is second countable.
Proof. We define $\mathcal{G}$ to be the set of all open balls with rational centers and rational radii. Since $\mathbb{Q}$ and $\mathbb{Q}^d$ are countably infinite sets, the set $\mathcal{G}$ is countably infinite.
Let $G$ be an open set and ${\bb x}\in G$. By Proposition B.1.2 there exists $r>0$ for which $B_r({\bb x})\subset G$. Since for every real number there is a rational number that is arbitrarily close, we can select $B_{r'}({\bb x}')$ where $r',{\bb x}'$ are rationals such that ${\bb x}\in B_{r'}({\bb x}') \subset B_r({\bb x})\subset G$. Repeating this for every ${\bb x}\in G$ and taking the union of the resulting rational balls completes the proof.
In the case of $R^{\infty}$, second countability is demonstrated by taking all simple cylinders (see Example B.4.4) whose base $(a,b)$ has rational endpoints and noting that a countable union of sets that are countably infinite is countably infinite.
Proposition B.4.6. In the Euclidean space ${\bb x}^{(n)}\to \bb x$ if and only if ${ x}^{(n)}_j\to x_j$ for all $j=1,\ldots, d$.
Proof. We recall Proposition B.3.2, which states that continuous functions such as $f(x)=x^2$, $f(x)=\sqrt{x}$, and $f(x,y)=x+y$ preserve limits. It follows that $\|{\bb v}^{(n)}\|\to 0$ if and only if $\|{\bb v}^{(n)}\|^2=\sum_{j=1}^d (v_j^{(n)})^2 \to 0$, which occurs if and only if $(v_j^{(n)})^2\to 0$ for all $j=1,\ldots,d$. This occurs if and only if $v_j^{(n)} \to 0$ for all $j=1,\ldots,d$
Example B.4.5. The sequence of vectors $(2^{-n},3^{-n}), n\in\mathbb{N}$ converges as $n\to\infty$ to $\bb 0=(0,0)$.
Proposition B.4.7. The function $\bb f:\R^d\to\R^k$ defined by $\bb f(\bb x)=(f_1(\bb x),\ldots,f_k(\bb x))$ is continuous if and only if the functions $f_1,\ldots,f_k$ are continuous.
Note that above we are denoting the function using bold-face $\bb f$ to indicate that $\bb f(\bb x)$ is a vector.
Proof. We assume that $f_i$ are continuous for all $i=1,\ldots,k$ and consider an arbitrary $\epsilon>0$. Then for $\epsilon'=\epsilon/\sqrt{k}>0$ there exists $\delta>0$ such that if $\|\bb x-\bb y\|^2<\delta^2$ then $|f_i(\bb x)-f_i(\bb y)|^2<\epsilon^2/k$, $i=1,\ldots,k$. Note that we choose the same $\delta$ for all $i=1,\ldots,k$; for example, by setting $\delta=\min(\delta_1,\ldots,\delta_k)$. It follows that if $\|\bb x-\bb y\|^2<\delta$ then $\|\bb f(\bb x)-\bb f(\bb y)\|^2=\sum_{j=1}^k |f_j(\bb x)-f_j(y)|^2\leq k\epsilon^2/k = \epsilon^2,$ showing that $\bb f$ is continuous.
Conversely, if $\bb f$ is continuous, for all $\epsilon>0$ there exists $\delta>0$ such that whenever $\|\bb x-\bb y\|^2\leq \delta^2$ we have $\|\bb f(\bb x)-\bb f(\bb y)\|^2=\sum_{j=1}^k |f_j(\bb x)-f_j(\bb y)|^2 < \epsilon^2.$ This implies $|f_j(\bb x)-f_j(\bb y)|^2 < \epsilon^2$ and $|f_j(\bb x)-f_j(\bb y)| < \epsilon$, implying the continuity of $f_1,\ldots,f_k$.
Proposition B.4.8. The following functions $f:\R^d\to\R^k$ are continuous: \begin{align*} f(\bb x)&=cx_i\\ f(\bb x,\bb y)&= \bb x+\bb y\\ f(\bb x,\bb y)&=\langle \bb x,\bb y\rangle\\ f(x)&=1/x. \end{align*}
Note that in the first and third cases above, we have $k=1$, and in the last case above, we have $k=d=1$.
Proof. To prove the first assertion, let $\epsilon>0$ be given, and define $\delta=\epsilon/|c|$. Whenever $\|\bb x-\bb y\|^2=\sum_{j=1}^d|x_j-y_j|^2 <\epsilon^2/|c|^2$ we also have $|cx_i-cy_i|\leq |c| \cdot |x_i-y_i|<|c|\epsilon/|c|=\epsilon.$
To prove the second assertion note that whenever $\|(\bb u,\bb w)-(\bb x,\bb y)\|<\epsilon/\sqrt{2d}$, then for all $j=1,\ldots,d$, $|x_j-u_j|< \epsilon/\sqrt{2d}$ and $|y_j-w_j|<\epsilon/\sqrt{2d}$. Using the triangle inequality property of the Euclidean norm, we have \begin{align*} \|(\bb x+\bb y)-(\bb u+\bb w)\| &= \|(\bb x-\bb u)+(\bb y-\bb w)\| \\ &\leq \|\bb x-\bb u\|+\|\bb y-\bb w\| \\ &\leq \sqrt{2d\epsilon^2/(2d)}=\epsilon. \end{align*} The proofs of the other propositions are similar.
Corollary B.4.1. All polynomials are continuous functions.
The concept of compactness (Definition B.1.6) has some important consequences (Propositions B.3.5 and B.3.6). The general definition of a compact space (Definition B.1.6) is hard to verify, but the following simple condition is veru useful for verifying compactness in $\R^d$. A proof is available in (Rudin, 1976).
Proposition B.4.8. If a set $A\subset \R^d$ is closed and bounded, it is compact. | 5,812 | 16,061 | {"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": 2, "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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-50 | latest | en | 0.690981 |
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Online Notes / Algebra (Notes) / Polynomial Functions / Partial Fractions
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Algebra - Notes
## Partial Fractions
This section doesn’t really have a lot to do with the rest of this chapter, but since the subject needs to be covered and this was a fairly short chapter it seemed like as good a place as any to put it.
So, let’s start with the following. Let’s suppose that we want to add the following two rational expressions.
What we want to do in this section is to start with rational expressions and ask what simpler rational expressions did we add and/or subtract to get the original expression. The process of doing this is called partial fractions and the result is often called the partial fraction decomposition.
The process can be a little long and on occasion messy, but it is actually fairly simple. We will start by trying to determine the partial fraction decomposition of,
where both P(x) and Q(x) are polynomials and the degree of P(x) is smaller than the degree of Q(x). Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember.
So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.
Factor in denominator Term in partial fraction decomposition
Notice that the first and third cases are really special cases of the second and fourth cases respectively if we let . Also, it will always be possible to factor any polynomial down into a product of linear factors ( ) and quadratic factors ( ) some of which may be raised to a power.
There are several methods for determining the coefficients for each term and we will go over each of those as we work the examples. Speaking of which, let’s get started on some examples.
Now, we need to do a set of examples with quadratic factors. Note however, that this is where the work often gets fairly messy and in fact we haven’t covered the material yet that will allow us to work many of these problems. We can work some simple examples however, so let’s do that.
Example 2 Determine the partial fraction decomposition of each of the following. (a) [Solution] (b) [Solution] Solution (a) In this case the x that sits in the front is a linear term since we can write it as, and so the form of the partial fraction decomposition is, Now we’ll use the fact that the LCD is and add the two terms together, Next, set the numerators equal. This is where the process changes from the previous set of examples. We could choose to get the value of A, but that’s the only constant that we could get using this method and so it just won’t work all that well here. What we need to do here is multiply the right side out and then collect all the like terms as follows, Now, we need to choose A, B, and C so that these two are equal. That means that the coefficient of the x2 term on the right side will have to be 8 since that is the coefficient of the x2 term on the left side. Likewise, the coefficient of the x term on the right side must be zero since there isn’t an x term on the left side. Finally the constant term on the right side must be -12 since that is the constant on the left side. We generally call this setting coefficients equal and we’ll write down the following equations. Now, we haven’t talked about how to solve systems of equations yet, but this is one that we can do without that knowledge. We can solve the third equation directly for A to get that . We can then plug this into the first two equations to get, So, the partial fraction decomposition for this expression is, (b) Here is the form of the partial fraction decomposition for this part. Adding the two terms up gives, Now, set the numerators equal and we might as well go ahead and multiply the right side out and collect up like terms while we’re at it. Setting coefficients equal gives, In this case we got A and B for free and don’t get excited about the fact that . This is not a problem and in fact when this happens the remaining work is often a little easier. So, plugging the known values of A and B into the remaining two equations gives, The partial fraction decomposition is then,
Algebra - Notes | 1,149 | 5,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2013-20 | latest | en | 0.935667 |
https://math.answers.com/other-math/What_is_the_pattern_rule_for_1_2_4_8 | 1,713,638,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00697.warc.gz | 345,507,688 | 47,842 | 0
# What is the pattern rule for 1 2 4 8?
Updated: 4/28/2022
Wiki User
6y ago
You are doubling the values each time. So it is U{n} = 2ⁿ⁻¹ for n = 1, 2, 3, 4.
------------------------------------
Nothing can be said about values of U{n} beyond n = 4, as that above is only one of infinitely many functions which give the values {1, 2, 4, 8} for n = {1, 2, 3, 4}. For example:
In the above formula, U{5} = 16
But if U{n} = (27n⁴ - 266n³ + 933n² - 1318n + 648)/24
then U{1..4} are also {1, 2, 4, 8}, but U{5} = 42
Wiki User
6y ago | 216 | 539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.802534 |
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