Wikipedia:Reference desk/Archives/Mathematics/2008 October 29

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October 29

arc length

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)

See the article titled arc length. Michael Hardy (talk) 02:50, 30 October 2008 (UTC)

The convention is that 1 radian = 1, so yes, the units are the same. Without this convention, you would need an extra factor in the formula, such as "arclength = radius . angle . 1 radian^-1", just like you must have "arclength = radius . angle . pi/180°" if you measure the angle in degrees (of course, pi/180° = 1 radian^-1, and with this handy convention, both are equal to 1). As for the more philosophical question, it's a bit tricky. One way to measure the length of (for instance) a circle would be to define it as the length that is longer than the circumference of any convex polygon contained within it, but shorter than the circumference of any convex polygon that contains it. Proving that such a length exists uniquely is not the easiest thing, but it should be clear that it can be measured with the same units. -- Jao (talk) 01:08, 29 October 2008 (UTC)

(I see that this has just been answered, but now I've written this I may as well post it)

1. Conceptually, the (curved) arc is split into very many straight-line segments, each of which can be measured with a straight scale, and then summed. The arc length is then the limit of this summation as the number of straight-line segments tends to infinty and their individual lengths tend to zero.

2. The units in "arclength = radius . (angle in radian)" do match. Radians are unitless -- they are just a number. So, both sides have units of length (metres, centimetres, or whatever you choose). 01:15, 29 October 2008 (UTC) —Preceding unsigned comment added by 86.142.111.10 (talk)

Returning to the question, how can a curved length be measured with a straight scale, if the scale is rigid the operation can be done by rolling it without slipping round the curve. If the curvature changes direction, of course, the scale would have to be transferred to the other side of the curve. Another way of looking at it is to consider the scale to be curved without distortion so as to follow exactly the line of the curve, then the one can be placed against the other. …86.132.232.36 (talk) 22:03, 29 October 2008 (UTC)

Asympotitic number of arithmetic progressions

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 $m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}$ 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 ${\frac {m^{2}}{2(k-1)}}(1+o(1))$. How do I go about doing it is my main problem. What I can do is to show the given expression is greater then ${\frac {m^{2}}{2(k-1)}}+{\frac {m-mk-k}{2(k-1)}}$ 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 ${\frac {m^{2}}{2(k-1)}}(1+o(1))$ as ${\frac {\frac {m-mk-k}{2(k-1)}}{\frac {m^{2}}{2(k-1)}}}$ 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)

Maybe this is obvious, but you do know you can get a less than using the other half of $x-1<\lfloor x\rfloor \leq x$, right? Dragons flight (talk) 06:55, 29 October 2008 (UTC)

That's how I get $m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}>{\frac {m^{2}}{2(k-1)}}+{\frac {m-mk-k}{2(k-1)}}$. From here I can conclude that $m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}>{\frac {m^{2}}{2(k-1)}}(1+o(1))$. What I need is the equality not the greater then sign. --Shahab (talk) 07:14, 29 October 2008 (UTC)

$m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}>m({\frac {m-1}{k-1}}-1)-{\frac {k-1}{2}}{\frac {m-1}{k-1}}{\Big (}{\frac {m-1}{k-1}}+1{\Big )}$

$m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}<m{\frac {m-1}{k-1}}-{\frac {k-1}{2}}({\frac {m-1}{k-1}}-1){\Big (}({\frac {m-1}{k-1}}-1)+1{\Big )}$

Use both parts of the inequality relation to get upper and lower bounds. Dragons flight (talk) 07:38, 29 October 2008 (UTC)

Thanks. No small part of my difficulty was that I was doing a careless mistake. I have now shown that if $X=m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}$ then $X>{\frac {m^{2}}{2(k-1)}}(1+o(1))$ and also $X<{\frac {m^{2}}{2(k-1)}}(1+o(1))$. I see a technical glitch preventing me from claiming equality though, that is shouldn't the inequalities be non-strict.--Shahab (talk) 08:56, 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:

${\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}\leq N\leq {\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}+{\frac {k-1}{8}}$.

(As I said, these bounds are also sharp, because the left, respectively, the right inequality is an equality provided ${\frac {m}{k-1}}$ is an integer, respectively, a half-integer number). However, they tell you

$N={\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}+O(1)$

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 $\textstyle mx-{\frac {k-1}{2}}x(x+1)$ computed in an (integer) point $\textstyle x$ close to the maximum point. --PMajer (talk) 09:45, 29 October 2008 (UTC)

Hi. What you said earlier was pretty clear to me other then the fact that $p(a)=p(b)\leq N$ which I reasoned was because p(x) increases at a, there is no other turning point in [a,b] and p(b)=p(a). (Is this reasoning correct?). Aside from that the main issue was to handle small o notation for me. Your asymptotic $N={\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}+O(1)$ implies $N={\frac {m^{2}}{2(k-1)}}(1+o(1))$ is also clear to me. But I was wondering if $f(n)>o(1)$ and $f(n)<o(1)$ imply $f(n)=o(1)$. Can you please elaborate on this point. Thanks for all the help.--Shahab (talk) 15:57, 29 October 2008 (UTC)

Yes, correct. As to $f(n)>o(1)$ and $f(n)<o(1)$, 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)

$m\left\lfloor {\frac {m-1}{k-1}}\right\rfloor -{\frac {k-1}{2}}\left\lfloor {\frac {m-1}{k-1}}\right\rfloor \left(\left\lfloor {\frac {m-1}{k-1}}\right\rfloor +1\right)$

--Trovatore (talk) 02:55, 30 October 2008 (UTC)

By the way - II :) the value of the above expression coincides with

$m\left\lfloor {\frac {m}{k-1}}\right\rfloor -{\frac {k-1}{2}}\left\lfloor {\frac {m}{k-1}}\right\rfloor \left(\left\lfloor {\frac {m}{k-1}}\right\rfloor +1\right)$

(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)