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#23 | |
"Bob Silverman"
Nov 2003
North of Boston
3·5·509 Posts |
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you. I may be condenscending toward you, but I have not been arrogant. It is not arrogance to tell someone that they lack basic understanding of a subject. Based on your level of understanding, I would have guessed that you were a high school freshman at best. You wrote: "Thanks for your replies and good advice. I was not trying to drastically change the power, only by about 5% or so, so I guess I thought I could compensate with the coefficient." The fact that you had these thoughts show a FUNDAMENTAL lack of understanding of basic level algebra; i.e. math you should have had a long time before 11th grade. |
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#24 | |
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11×389 Posts |
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#25 |
Sep 2004
2×5×283 Posts |
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Condescending and arrogant...humm.. I wonder if R.D. Silverman behaves like that with his mother.
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#26 | |
"Bob Silverman"
Nov 2003
North of Boston
1DD316 Posts |
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about very elementary mathematics to attempt to "dabble" in an area of math that requires a fair bit of knowledge, mathematical maturity and sophistication. It is not arrogant for an expert in a subject to tell a newbie that his/her knowledge of a subject is inadequate. |
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#27 | |
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
11·389 Posts |
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Can we just all agree you're being a jerk ("A foolish, rude, or contemptible person.") about the whole thing, and leave it at that? |
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#28 |
Sep 2004
2·5·283 Posts |
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I think we should make a poll.
Last fiddled with by em99010pepe on 2008-07-16 at 17:20 |
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#29 | |
"Richard B. Woods"
Aug 2002
Wisconsin USA
11110000011002 Posts |
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Come on, Dr. Silverman -- not everyone has good innate math ability. This is a public forum, with no "entrance" requirements. Say -- why don't you start your own subforum: "R. D. Silverman's Math" or something like that? With your name on it, none of the rest of us could complain about how you responded to inquiries there, especially if you stated your requirements in the title, like: "R. D. Silverman Answers Clearly-Phrased Questions Asked by Knowledgable Persons" Last fiddled with by cheesehead on 2008-07-16 at 18:09 Reason: various and sundry |
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#30 |
Banned
"Luigi"
Aug 2002
Team Italia
4,871 Posts |
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If we claim for freedom, we should let people free to behave as they like, especially on a free forum. Even if we don't agree with them. We may disagree, but again we cannot call people for having called people or we'd act as those who got our blame.
Everybody here know Mr.Silverman: he is not going to change because of posts. Let's accept him for how he is and what he gives to us. That said, maybe someone could direct our new friend Roger to our "math for non mathematicians" forum, Miscellaneous math threads... Luigi ![]() P.S. I learnt a lot about how to formulate a math question from his answers. Last fiddled with by ET_ on 2008-07-16 at 18:47 |
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#31 |
"Jason Lynch"
Jun 2008
Spokane, WA
22×3 Posts |
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Of course, if you're willing to tolerate some error, all is not lost. Especially if you're dealing with smaller changes in the exponent, you can adjust the coefficient to provide a similar curve for an area of interest (a particular range of x values). As the size of this area of interest increases, however, the second curve will diverge more and more from the first. Depending on your exact application, this may or may not be useful.
And as far as fitting to a particular set of data, there will generally be only one best set of coefficients and exponents for a given equation that will best fit the data. Once you have that, I'm not sure I see the value in trying a different exponent, though again, I know nothing about this particular application. |
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#32 | |
"Bob Silverman"
Nov 2003
North of Boston
3·5·509 Posts |
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You talk about "trying a different exponent" in complete ignorance???? This entire thread is STUPID. The OP was trying to curve fit a function that counts twin primes... But if the OP had bothered to do any READING he would have found out that there is a well established THEORY that tells us what the curve should be. The number of twin primes less than N is uniformly asymptotic to C N/(log^2 N) where C is the twin prime constant. [given by an infinite product]. Trying to fit ANY kind of "power law" curve to this data is MORONIC. It can not possibly fit a curve of that type...... If you want to know more (i.e. the derivation of C) then ask. But for someone at the level of the OP to even attempt work in this area is crazy. |
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#33 |
"Lucan"
Dec 2006
England
647410 Posts |
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