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arc length is measured in m or cm or other units of length. How can we measure a curved length with straight scale .why in formaula arclength = radius . (angle in radian) units of both sides are not equal. —Preceding unsigned comment added by 119.154.76.143 ( talk) 00:55, 29 October 2008 (UTC)
I asked this question here a few days back and was at that point satisfied with the response. The question was : How many finite k-term Arithmetic Progressions are possible in the set {1,2,3,...m}. After deducing that the answer was the person who was kind enough to answer the question also suggested that the next step cound be to find this number equal to some asymptotic (I presume by using small o/big O notation). At that point I was merely interested in bounding the function (or finding a suitable condition within which the function could be bound) and so I didn't bother with finding the asymptotic. Another reason is that other then the basic definitions I don't know much about how to use asymptotics.
Now however I need to show that the above expression is equal to . How do I go about doing it is my main problem. What I can do is to show the given expression is greater then by using the properties of the floor function. (These properties force the greater then sign which is the source of my problem here.) This I presume is the same as that as goes to zero with m going to infinity. But I need to show equality not greater then. Can someone comment or explain how to handle big O and small o with floor functions. Thanks (I don't understand a similar proof using big 'O' in Lemma 10 here.)-- Shahab ( talk) 06:33, 29 October 2008 (UTC)
Hi Shahab... I think last day I gave you quite a complete answer... why don't you tell us what was not clear in the answer? Your number N of arithmetic progressions is bounded form below and above with:
(As I said, these bounds are also sharp, because the left, respectively, the right inequality is an equality provided is an integer, respectively, a half-integer number). However, they tell you
as m goes to infinity, which is an even more precise asymptotics. For the bounds the key point was that N, as given by your formula, is computed in an (integer) point close to the maximum point. -- PMajer ( talk) 09:45, 29 October 2008 (UTC)
Yes, correct. As to and , they mean that f(n) is above an infinitesimal sequence, respectively, below an (other) infinitesimal sequence, so you can well conclude as you do. It's the sandwich theorem. -- PMajer ( talk) 16:21, 29 October 2008 (UTC)
Just by the way, this would be much easier to read if you typeset it as (for example)
-- Trovatore ( talk) 02:55, 30 October 2008 (UTC)
By the way - II :) the value of the above expression coincides with
(for the first one came as the sum of all positive terms of the form m-(k-1)h over integer h>0; the second one came as the sum of all nonnegative terms of the form m-(k-1)h over integer h>0); the second is actually a bit better to deal with in Shahab's problem -- PMajer ( talk) 11:50, 30 October 2008 (UTC)
Mathematics desk | ||
---|---|---|
< October 28 | << Sep | October | Nov >> | October 30 > |
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. |
arc length is measured in m or cm or other units of length. How can we measure a curved length with straight scale .why in formaula arclength = radius . (angle in radian) units of both sides are not equal. —Preceding unsigned comment added by 119.154.76.143 ( talk) 00:55, 29 October 2008 (UTC)
I asked this question here a few days back and was at that point satisfied with the response. The question was : How many finite k-term Arithmetic Progressions are possible in the set {1,2,3,...m}. After deducing that the answer was the person who was kind enough to answer the question also suggested that the next step cound be to find this number equal to some asymptotic (I presume by using small o/big O notation). At that point I was merely interested in bounding the function (or finding a suitable condition within which the function could be bound) and so I didn't bother with finding the asymptotic. Another reason is that other then the basic definitions I don't know much about how to use asymptotics.
Now however I need to show that the above expression is equal to . How do I go about doing it is my main problem. What I can do is to show the given expression is greater then by using the properties of the floor function. (These properties force the greater then sign which is the source of my problem here.) This I presume is the same as that as goes to zero with m going to infinity. But I need to show equality not greater then. Can someone comment or explain how to handle big O and small o with floor functions. Thanks (I don't understand a similar proof using big 'O' in Lemma 10 here.)-- Shahab ( talk) 06:33, 29 October 2008 (UTC)
Hi Shahab... I think last day I gave you quite a complete answer... why don't you tell us what was not clear in the answer? Your number N of arithmetic progressions is bounded form below and above with:
(As I said, these bounds are also sharp, because the left, respectively, the right inequality is an equality provided is an integer, respectively, a half-integer number). However, they tell you
as m goes to infinity, which is an even more precise asymptotics. For the bounds the key point was that N, as given by your formula, is computed in an (integer) point close to the maximum point. -- PMajer ( talk) 09:45, 29 October 2008 (UTC)
Yes, correct. As to and , they mean that f(n) is above an infinitesimal sequence, respectively, below an (other) infinitesimal sequence, so you can well conclude as you do. It's the sandwich theorem. -- PMajer ( talk) 16:21, 29 October 2008 (UTC)
Just by the way, this would be much easier to read if you typeset it as (for example)
-- Trovatore ( talk) 02:55, 30 October 2008 (UTC)
By the way - II :) the value of the above expression coincides with
(for the first one came as the sum of all positive terms of the form m-(k-1)h over integer h>0; the second one came as the sum of all nonnegative terms of the form m-(k-1)h over integer h>0); the second is actually a bit better to deal with in Shahab's problem -- PMajer ( talk) 11:50, 30 October 2008 (UTC)