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This is a very elementary question but I am no good at math
I am writing a program and I need a function to determine the coordinates of an element in a table given its serial position (counting left to right, top to bottom) and the number of rows and columns. to visualise, I might have a grid like this.
1234 5678
I might want to look up the row and column of 6, which would be column 2 row 2. element 4 would have column 4, row 1. I have a feeling I should use mod somehow but I can't figure out how. Thanks! -- 71.86.121.200 ( talk) 02:21, 30 January 2008 (UTC)
0123 4567
The PVC figure like this one, requires manual panting.
http://www.entertainmentearth.com/images/%5CAUTOIMAGES%5CORG40601.jpg
Suppose it has 3 sections that require painting, A , B and C. And suppose each section has a painting failure rate. If a painting failure occurs, the whole thing would need to be discarded resulting in a loss of effort.
So the question is, in what sequence should the painting occurs? My gut feeling is that the section with the highest failure rate should be painted first, and then the second highest and so on until the section of the lowest failure rate is painted last.
Now if each section requires a different amount of effort, then this will change the problem. So what algorithm should be followed to find the sequence to minimize the loss of effort.
For a figure with only 3 section, the search is trivial and brute force can be used. But if a figure has 17 painting sections, each section with an effort rate and a failure rate then finding the optimum sequence is required.
Example:
Is this type of problem a hard problem to solve?
202.168.50.40 ( talk) 02:30, 30 January 2008 (UTC)
Start - A passes - B passes - C passes (effort 26, prob = .8 * .2 * .9) \ \ \ C fails (effort 26 + E, prob = .8 * .2 * .1) \ \ B fails (effort 6 +E, prob = .8 * .8) A fails (effort 5+E, prob = .2)
Wow! I'm really impressed! I wonder how many anime otaku has actually considered the mathematics of painting anime PVC figures before they start painting their garage kits. I guess mathematics applies everywhere even where you would have least expected it. Thanks. Oh by the way, here is another nice figure http://www.tmpanime.com/images/medium/4560228201475_MED.jpg 122.107.141.142 ( talk) 10:40, 31 January 2008 (UTC)
I know that you can put a polynomial through any finite set of points, no two of which are directly vertical. Can you do the same thing with a power series and an infinite set of points? Does it matter if there are accumulation points? Black Carrot ( talk) 04:18, 30 January 2008 (UTC)
The purpose behinde this post is an attempt to expand integration space to include functions with undesirable points where the function could be undefined.I need help to understand if this post makes sense.The main question here, is this way could be more general than using the measurement set theory to expand the integration?
Assume the function F(x),
F:[a,b]→R
We will put the interval[a,b] as acombination of subsets, UGk,
UGk=GNUGQUGQ̀....etc.(U,here stands for combination symbol) which stands for natural set,rational set ,irrational set,….etc.
Lets define the subsets,
өi={gk:F(xi)≥gk≥0},
xiЄ[a,b],
Let,(pi) be apartitoning of Gk in [a,b] and (pөi) apartitioning of(өi),
(Mөi=SUPөi), and (mөi=infөi),
(UFөi,pi)= Σi Mөi(xi-xi-1), (UFөi,pi)=upper darboux sum.
(LFөi,pi)= Σi mөi(xi-xi-1),(LFөi,pi)=lowe darboux sum.
(UFөi,pi,Pөi)=inf{UFөi:piPөi,(Pөi) partioning of(өi), (pi) partitioning of (Gk)},
(LFөi,pi,Pөi)=sup{LFөi:piPөi,(Pөi) partioning of(өi),(pi) partitioning (Gk)},
now, we put the integration in the form,
∫Fөi,over,Gk={0,UFөi}={0,LFөi}=the subsets,sөi,
∫F,over,[a,b]=Usөi
for example;
f(x):[0,1]→R,
f(x)=x , xЄ irrational numbers=Q̀, within[0,1],
f(x)=1,xЄ rational numbers=Q, within[0,1],
∫F,over,[o,1]={0,1\2}Q̀,U{0,1}Q={0,1\2}R,U,{1\2,1}Q ,(Q̀,Q AND,R,ARE SUFFIXES HERE AND U STANDS FOR COMBINATION symbol.)
this way enable us to exclude the undesired points from the integration where f(x)may be undefined. 88.116.163.226 ( talk) 11:02, 30 January 2008 (UTC)husseinshimaljasim
well,i dont think so,in Lebesgue integration,the above example would be, μ(0,1\2)inQ`set+μ(0,1)inQset=1\2+0=1\2,iam suggesting to keep integration in form of combination of subsets regardless they were countable or not.wouldnot this be more genral? 210.5.236.35 ( talk) 14:36, 31 January 2008 (UTC)husseinshimaljasim
If ,N1,N2,s and k,represent 4 natural numbers,then why the equation, [(N1)^s]+2=[(N2)^k)],where,N1,N2,s,and k Є(N-SET) has only one solution in N-SET,where, N1=2,s=1,N2=2,k=2? i guess. 88.116.163.226 ( talk) 11:19, 30 January 2008 (UTC)husseinshimaljasim
what i was thinking!?i must be needing some sleep. 210.5.236.35 ( talk) 15:56, 30 January 2008 (UTC)husseinshimaljasim
From the article Summation: Mathematical notation has a special representation for compactly representing summation of many similar terms: the summation symbol, a large upright capital Sigma. This is defined thus:
The subscript gives the symbol for an index variable, i. Here, i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation. Here i = m under the summation symbol means that the index i starts out equal to m. Successive values of i are found by adding 1 to the previous value of i, stopping when i = n.
I have been reading on how to calculate correlation coefficient using spearman formula But I got confused on the way when ranking the scores of this nature. According to the procedure of ranking the arbitrary ranks of 64,64 and 64 of x if 9th 10th and 11th respectively. Now is it true that the ranks of each of the numbers above is 10? X Y 22 36 24 24 30 25 40 20 45 48 50 44 50 40 52 56 64 62 64 68 64 56 72 32 78 78 78 68 84 68 90 58 —Preceding unsigned comment added by Nkomali ( talk • contribs) 14:31, 30 January 2008 (UTC)
Hello,
I've been looking at the article on dual curves, but it wasn't very informative.
Am I right when I say this is the construction for the dual curve :
For each point on the curve, draw the tangent. Then take the perpendicular to that tangent that passes through the origin. Labelling the point at the intersection of those two lines P, take the point H that is at distance 1/OP from the origin, that lies on the perpendicular line. The set of all such points H forms the dual curve.
If that's the case, doesn't that give rise to two different curves : one when we choose H to be the same side of the origin, one when H is the other side ?
Finally, why does this rather unlikely combination of constructions lead to a curve of any interest ? Seeing as the position of the origin has an importance for the resulting curve...
Thanks. -- Xedi ( talk) 23:31, 30 January 2008 (UTC)
Mathematics desk | ||
---|---|---|
< January 29 | << Dec | January | Feb >> | January 31 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
This is a very elementary question but I am no good at math
I am writing a program and I need a function to determine the coordinates of an element in a table given its serial position (counting left to right, top to bottom) and the number of rows and columns. to visualise, I might have a grid like this.
1234 5678
I might want to look up the row and column of 6, which would be column 2 row 2. element 4 would have column 4, row 1. I have a feeling I should use mod somehow but I can't figure out how. Thanks! -- 71.86.121.200 ( talk) 02:21, 30 January 2008 (UTC)
0123 4567
The PVC figure like this one, requires manual panting.
http://www.entertainmentearth.com/images/%5CAUTOIMAGES%5CORG40601.jpg
Suppose it has 3 sections that require painting, A , B and C. And suppose each section has a painting failure rate. If a painting failure occurs, the whole thing would need to be discarded resulting in a loss of effort.
So the question is, in what sequence should the painting occurs? My gut feeling is that the section with the highest failure rate should be painted first, and then the second highest and so on until the section of the lowest failure rate is painted last.
Now if each section requires a different amount of effort, then this will change the problem. So what algorithm should be followed to find the sequence to minimize the loss of effort.
For a figure with only 3 section, the search is trivial and brute force can be used. But if a figure has 17 painting sections, each section with an effort rate and a failure rate then finding the optimum sequence is required.
Example:
Is this type of problem a hard problem to solve?
202.168.50.40 ( talk) 02:30, 30 January 2008 (UTC)
Start - A passes - B passes - C passes (effort 26, prob = .8 * .2 * .9) \ \ \ C fails (effort 26 + E, prob = .8 * .2 * .1) \ \ B fails (effort 6 +E, prob = .8 * .8) A fails (effort 5+E, prob = .2)
Wow! I'm really impressed! I wonder how many anime otaku has actually considered the mathematics of painting anime PVC figures before they start painting their garage kits. I guess mathematics applies everywhere even where you would have least expected it. Thanks. Oh by the way, here is another nice figure http://www.tmpanime.com/images/medium/4560228201475_MED.jpg 122.107.141.142 ( talk) 10:40, 31 January 2008 (UTC)
I know that you can put a polynomial through any finite set of points, no two of which are directly vertical. Can you do the same thing with a power series and an infinite set of points? Does it matter if there are accumulation points? Black Carrot ( talk) 04:18, 30 January 2008 (UTC)
The purpose behinde this post is an attempt to expand integration space to include functions with undesirable points where the function could be undefined.I need help to understand if this post makes sense.The main question here, is this way could be more general than using the measurement set theory to expand the integration?
Assume the function F(x),
F:[a,b]→R
We will put the interval[a,b] as acombination of subsets, UGk,
UGk=GNUGQUGQ̀....etc.(U,here stands for combination symbol) which stands for natural set,rational set ,irrational set,….etc.
Lets define the subsets,
өi={gk:F(xi)≥gk≥0},
xiЄ[a,b],
Let,(pi) be apartitoning of Gk in [a,b] and (pөi) apartitioning of(өi),
(Mөi=SUPөi), and (mөi=infөi),
(UFөi,pi)= Σi Mөi(xi-xi-1), (UFөi,pi)=upper darboux sum.
(LFөi,pi)= Σi mөi(xi-xi-1),(LFөi,pi)=lowe darboux sum.
(UFөi,pi,Pөi)=inf{UFөi:piPөi,(Pөi) partioning of(өi), (pi) partitioning of (Gk)},
(LFөi,pi,Pөi)=sup{LFөi:piPөi,(Pөi) partioning of(өi),(pi) partitioning (Gk)},
now, we put the integration in the form,
∫Fөi,over,Gk={0,UFөi}={0,LFөi}=the subsets,sөi,
∫F,over,[a,b]=Usөi
for example;
f(x):[0,1]→R,
f(x)=x , xЄ irrational numbers=Q̀, within[0,1],
f(x)=1,xЄ rational numbers=Q, within[0,1],
∫F,over,[o,1]={0,1\2}Q̀,U{0,1}Q={0,1\2}R,U,{1\2,1}Q ,(Q̀,Q AND,R,ARE SUFFIXES HERE AND U STANDS FOR COMBINATION symbol.)
this way enable us to exclude the undesired points from the integration where f(x)may be undefined. 88.116.163.226 ( talk) 11:02, 30 January 2008 (UTC)husseinshimaljasim
well,i dont think so,in Lebesgue integration,the above example would be, μ(0,1\2)inQ`set+μ(0,1)inQset=1\2+0=1\2,iam suggesting to keep integration in form of combination of subsets regardless they were countable or not.wouldnot this be more genral? 210.5.236.35 ( talk) 14:36, 31 January 2008 (UTC)husseinshimaljasim
If ,N1,N2,s and k,represent 4 natural numbers,then why the equation, [(N1)^s]+2=[(N2)^k)],where,N1,N2,s,and k Є(N-SET) has only one solution in N-SET,where, N1=2,s=1,N2=2,k=2? i guess. 88.116.163.226 ( talk) 11:19, 30 January 2008 (UTC)husseinshimaljasim
what i was thinking!?i must be needing some sleep. 210.5.236.35 ( talk) 15:56, 30 January 2008 (UTC)husseinshimaljasim
From the article Summation: Mathematical notation has a special representation for compactly representing summation of many similar terms: the summation symbol, a large upright capital Sigma. This is defined thus:
The subscript gives the symbol for an index variable, i. Here, i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation. Here i = m under the summation symbol means that the index i starts out equal to m. Successive values of i are found by adding 1 to the previous value of i, stopping when i = n.
I have been reading on how to calculate correlation coefficient using spearman formula But I got confused on the way when ranking the scores of this nature. According to the procedure of ranking the arbitrary ranks of 64,64 and 64 of x if 9th 10th and 11th respectively. Now is it true that the ranks of each of the numbers above is 10? X Y 22 36 24 24 30 25 40 20 45 48 50 44 50 40 52 56 64 62 64 68 64 56 72 32 78 78 78 68 84 68 90 58 —Preceding unsigned comment added by Nkomali ( talk • contribs) 14:31, 30 January 2008 (UTC)
Hello,
I've been looking at the article on dual curves, but it wasn't very informative.
Am I right when I say this is the construction for the dual curve :
For each point on the curve, draw the tangent. Then take the perpendicular to that tangent that passes through the origin. Labelling the point at the intersection of those two lines P, take the point H that is at distance 1/OP from the origin, that lies on the perpendicular line. The set of all such points H forms the dual curve.
If that's the case, doesn't that give rise to two different curves : one when we choose H to be the same side of the origin, one when H is the other side ?
Finally, why does this rather unlikely combination of constructions lead to a curve of any interest ? Seeing as the position of the origin has an importance for the resulting curve...
Thanks. -- Xedi ( talk) 23:31, 30 January 2008 (UTC)