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Another qual problem:
Let , and suppose is a continuous linear functional on such that for every , . Show that is the zero functional. Show that this is false if .
I'm not sure how to do this, obviously, or I wouldn't ask. But I have gotten somewhere, whether it is helpful or not, I do not know. First, by the Riesz Representation Theorem, there exists such that
for all . In particular, let for any real a. Then, using the condition on and using a change of variables gives
for any real a. Here I am stuck. My thought was perhaps I could show this implies g is periodic... I could use instead maybe. I don't know. Then, if I get that g is periodic, perhaps I could pick another choice of f that gives a contradiction. Any suggestions? Thanks. StatisticsMan ( talk) 02:20, 30 June 2009 (UTC)
What is the theoretical relationship between the unemployment rate and the number of applicants per job opening? Neon Merlin 03:17, 30 June 2009 (UTC)
Please could you check that my thinking on this problem is mathematically sound. The company that I work for tenders for various products and services. As part of the evaluation process we come up with a "scoring" matrix, which scores each company against a number of categories. The points in each category are set to give a weighting. A simplified example might be:
COMPANY 1 | COMPANY 2 | COMPANY 3 | |
---|---|---|---|
PRICE (0-10) | |||
FUNCTIONALITY (0-20) | |||
SUPPORT (0-10) | |||
COMPANY STABILITY (0-20) |
In practice there would be many functional areas and criteria, i.e. many rows. Each scorer would potentially put a score in the range given against each company. If everyone filled in the table entirely then scoring would be easy - just totalling the scores for each company. In practice two things happen:
Some people only score certain areas This is expected, technical people might only be able to answer questions about functionality and not (for example) company stability. The thing is that if we just add up the scores some areas would not get the weighting they need; for example only three people might score company stability but ten functionality. I figure that the way to cope with this is to give the people who have not scored in an area a score equal to the average of the scores that have been given.
Some people do not score all companies Ideally this would not happen, but due to ongoing work, unexpected calls, sickness, etc. some people may miss some of the presentations. Obviously it would be wrong to mark a company down because fewer people attended their presentation, so I figure that a solution to this is to give companies they missed an average of the scores they gave to ones they attended. I thought that this is better than giving an average of the scores given by other people because some people seem to mark high and some low.
Is this mathematically sound? I can see no reason why I should apply the correction for missing rows before I apply the correction for missing columns, but it feels right. In practice we have used these corrections and they do tend to come out with results that match the "general feelings" that people had. Thanks in advance . -- Q Chris ( talk) 09:36, 30 June 2009 (UTC)
I think about it this way. The strength of a category of a company should be a number between 0 and 1, like a probability, P, of success. This number P is unknown, but some knowledge of P is gained by scoring. The number of times it is scored is n, and the number of successes in scoring is i. So the score i is an integer between 0 and n, while the strength P is a real between 0 and 1. The likelihood function of P is the beta distribution having mean value m = (i+1)/(n+2) and variance s2 = m(1-m)/(n+3). So if a category is not scored, simply put n = i = 0 in the above formula and get m = 1/2 and s2 = 1/12. Bo Jacoby ( talk) 06:27, 1 July 2009 (UTC).
Our greatest common divisor article says that the expected value E(k) of the gcd of k integers is E(k) = ζ(k-1)/ζ(k) for k > 2. Does anyone have a reference for this? (While you're at it, if you're knowledgeable, you can try to fix up the sentences following this statement in the article which currently don't make any sense to me.) My real question: Is a similar formula known for the expected value of the reciprocal of the gcd? Staecker ( talk) 14:46, 30 June 2009 (UTC)
Hi. I just made an edit to the section of Residue (complex analysis) on calculating residues. I wonder if someone could have a look at it to double-check that I didn't say anything false, or otherwise break the article. Thanks in advance. - GTBacchus( talk) 15:27, 30 June 2009 (UTC)
Suppose f(x, y) is a bounded measurable function on . Show that if for every a < b and c < d
then f = 0 a.e.
In our study group, we looked at a similar problem we have yet to figure out, which is that the integral over each open disk in R^2 is 0 and we are supposed show f = 0 a.e. [as far as the Lebesgues measure on R^2].
Can any one help us figure these out? Thanks! StatisticsMan ( talk) 20:47, 30 June 2009 (UTC)
Mathematics desk | ||
---|---|---|
< June 29 | << May | June | Jul >> | July 1 > |
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. |
Another qual problem:
Let , and suppose is a continuous linear functional on such that for every , . Show that is the zero functional. Show that this is false if .
I'm not sure how to do this, obviously, or I wouldn't ask. But I have gotten somewhere, whether it is helpful or not, I do not know. First, by the Riesz Representation Theorem, there exists such that
for all . In particular, let for any real a. Then, using the condition on and using a change of variables gives
for any real a. Here I am stuck. My thought was perhaps I could show this implies g is periodic... I could use instead maybe. I don't know. Then, if I get that g is periodic, perhaps I could pick another choice of f that gives a contradiction. Any suggestions? Thanks. StatisticsMan ( talk) 02:20, 30 June 2009 (UTC)
What is the theoretical relationship between the unemployment rate and the number of applicants per job opening? Neon Merlin 03:17, 30 June 2009 (UTC)
Please could you check that my thinking on this problem is mathematically sound. The company that I work for tenders for various products and services. As part of the evaluation process we come up with a "scoring" matrix, which scores each company against a number of categories. The points in each category are set to give a weighting. A simplified example might be:
COMPANY 1 | COMPANY 2 | COMPANY 3 | |
---|---|---|---|
PRICE (0-10) | |||
FUNCTIONALITY (0-20) | |||
SUPPORT (0-10) | |||
COMPANY STABILITY (0-20) |
In practice there would be many functional areas and criteria, i.e. many rows. Each scorer would potentially put a score in the range given against each company. If everyone filled in the table entirely then scoring would be easy - just totalling the scores for each company. In practice two things happen:
Some people only score certain areas This is expected, technical people might only be able to answer questions about functionality and not (for example) company stability. The thing is that if we just add up the scores some areas would not get the weighting they need; for example only three people might score company stability but ten functionality. I figure that the way to cope with this is to give the people who have not scored in an area a score equal to the average of the scores that have been given.
Some people do not score all companies Ideally this would not happen, but due to ongoing work, unexpected calls, sickness, etc. some people may miss some of the presentations. Obviously it would be wrong to mark a company down because fewer people attended their presentation, so I figure that a solution to this is to give companies they missed an average of the scores they gave to ones they attended. I thought that this is better than giving an average of the scores given by other people because some people seem to mark high and some low.
Is this mathematically sound? I can see no reason why I should apply the correction for missing rows before I apply the correction for missing columns, but it feels right. In practice we have used these corrections and they do tend to come out with results that match the "general feelings" that people had. Thanks in advance . -- Q Chris ( talk) 09:36, 30 June 2009 (UTC)
I think about it this way. The strength of a category of a company should be a number between 0 and 1, like a probability, P, of success. This number P is unknown, but some knowledge of P is gained by scoring. The number of times it is scored is n, and the number of successes in scoring is i. So the score i is an integer between 0 and n, while the strength P is a real between 0 and 1. The likelihood function of P is the beta distribution having mean value m = (i+1)/(n+2) and variance s2 = m(1-m)/(n+3). So if a category is not scored, simply put n = i = 0 in the above formula and get m = 1/2 and s2 = 1/12. Bo Jacoby ( talk) 06:27, 1 July 2009 (UTC).
Our greatest common divisor article says that the expected value E(k) of the gcd of k integers is E(k) = ζ(k-1)/ζ(k) for k > 2. Does anyone have a reference for this? (While you're at it, if you're knowledgeable, you can try to fix up the sentences following this statement in the article which currently don't make any sense to me.) My real question: Is a similar formula known for the expected value of the reciprocal of the gcd? Staecker ( talk) 14:46, 30 June 2009 (UTC)
Hi. I just made an edit to the section of Residue (complex analysis) on calculating residues. I wonder if someone could have a look at it to double-check that I didn't say anything false, or otherwise break the article. Thanks in advance. - GTBacchus( talk) 15:27, 30 June 2009 (UTC)
Suppose f(x, y) is a bounded measurable function on . Show that if for every a < b and c < d
then f = 0 a.e.
In our study group, we looked at a similar problem we have yet to figure out, which is that the integral over each open disk in R^2 is 0 and we are supposed show f = 0 a.e. [as far as the Lebesgues measure on R^2].
Can any one help us figure these out? Thanks! StatisticsMan ( talk) 20:47, 30 June 2009 (UTC)