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March 5

semiprime number theorem

so i typed in a random 18 digit number to factorise, and it only had two prime factors:

factor 299129421485828329
299129421485828329: 11 27193583771438939

i was wondering how to work out the probability of this for an arbitrary 18 digit number. i know the prime-number theorem gives the probability of a number x being prime as x/ln(x), but how would i find the equivalent for a semiprime? thanks, 109.246.247.147 (talk) 00:00, 5 March 2010 (UTC)[reply]

The probability of what, specifically? Of being divisible by 11? Of being composite? Of having exactly two prime factors? Of having a prime factor less than 100? Michael Hardy (talk) 01:57, 5 March 2010 (UTC)[reply]

His last sentence says it -- the probability of being semiprime, which our article defines as having exactly two prime factors (not necessarily distinct). (I didn't know the word either.) --Trovatore (talk) 02:01, 5 March 2010 (UTC)[reply]

As well as the title and the first sentence. -- Meni Rosenfeld (talk) 09:58, 5 March 2010 (UTC)[reply]

Almost prime has an asymptotic formula you could use to compute an estimate. I don't know the error term. By the way, x/ln(x) is an approximation of the prime-counting function pi(x), the number of primes below x. The probability of a random x being prime is 1/ln(x). PrimeHunter (talk) 02:42, 5 March 2010 (UTC)[reply]

Prime numbers

If I choose a positive integer at random (with uniform distribution across all positive integers), what is the probability that it's prime? --70.250.214.164 (talk) 03:04, 5 March 2010 (UTC)[reply]

There is no uniform probability distribution on the positive integers. The closest meaningful notion is natural density, which for the primes is zero. Algebraist 03:22, 5 March 2010 (UTC)[reply]

Alternatively, you can say that, by the prime number theorem, the probability of x chosen uniformly from between 1 and N being prime is about 1/log(N) for sufficiently large N. --Tango (talk) 05:19, 5 March 2010 (UTC)[reply]

And if you choose two positive integers at random, the probability (by natural density) that they are coprime is 6/π². In general, the probability that n random positive integers are all coprime is 1/ζ(n), where ζ is the Riemann zeta function. There's a bit about this at Greatest_common_divisor#Probabilities_and_expected_value. Staecker (talk) 12:36, 5 March 2010 (UTC)[reply]

Toilet paper calculus

I was just in the bathroom and this calculus problem is stumping me.

We have a cylinder, representing the toilet paper roll. Now, let's say the radius of the cylinder, prior to it being used at all, is 10cm. The thickness of the toilet paper is 1mm. As we use toilet paper, two things happen: The radius decreases, and the circumference also decreases. As more toilet paper is used, the rate of change of the length of the decrease of the radius starts to increase. That is, the more toilet paper you use, the faster the radius shrinks. The squares you are using are 10cm², and you are using them at a rate of one per second. What is the rate of change of the decrease in radius? --70.122.117.52 (talk) 05:03, 5 March 2010 (UTC)[reply]

well, it's probably easiest to model this as a spiral whose radius r is given by 1mm*ø/2π (ø in radians going from 0 to infinity). then you can calculate the change in ø and r for a point that's moving at a constant rate along the spiral. does that help for a start? --Ludwigs2 05:37, 5 March 2010 (UTC)[reply]

A spiral? You've lost me.--70.122.117.52 (talk) 05:40, 5 March 2010 (UTC)[reply]

lost you how? you don't see that toilet paper spirals (more or less) outward from the center of the roll? or is it something else that's not clear? --Ludwigs2 06:59, 5 March 2010 (UTC)[reply]

If you use the paper of thickness D at the rate V [paper tape length per time], then you decrease the side area of the roll at the rate dS/dt = V×D, so the area is S = S₀ – V×D×t where S₀ denotes the initial area. Area is S = π×r², so r = √(S/π) = √(S₀ – V×D×t) / √π.
Now differentiate r in respect to t, and you'll get the rate of change of the radius. --CiaPan (talk) 07:38, 5 March 2010 (UTC)[reply]

Perhaps a laxative would prevent you from spending so much time in the bathroom, contemplating TP rolls ? :-) StuRat (talk) 13:10, 5 March 2010 (UTC) [reply]

In case you want to find such exercises, they're usually told with video casettes instead of toilet paper rolls. As an example, see exercise 2.8. in Jurij B. Csernyak, Robert M. Rose, A minszki csirke (Akkord kiadó, 1999. original title: Yuri B. Chernyak – Robert M. Rose, Chicken from Minsk). (Do you young people still know what a video casette is, or is that ancient history now like typewriters? :) – b_jonas 11:01, 7 March 2010 (UTC)[reply]

Mobius Strip

Why is it that, when you cut a Mobius strip along its surface, one piece remains instead of two? Is there an simple way of seeing why this is true?

The strip has only one edge. When you cut the strip along its 'middle line', you do not cross the edge, so the whole edge remains in one piece. Additionally each point of the strip lies between your cutting line and the edge, so does not get separated from the edge. This way the whole strip remains with ist original edge, and the edge remains in one piece—consequently, the whole strip remains in one piece.
Note, however, that your cutting line becomes a new edge, so if you try the same trick on the strip already cut, it will fail: the strip will get into two pieces. --CiaPan (talk) 07:45, 5 March 2010 (UTC)[reply]

There is another, even simpler method to see this. You know how to make a real Möbius strip model: get a strip of the paper, twist one of its end, then bend the whole and glue the ends together, right? Then you cut it.
Now change the order of your manipulations: take a strip of paper, cut it along the central line (but before cutting mark carefully on both halves which edges get created from the central line, and which ends are adjacent). Then twist both halves, bend them and glue the ends to construct the resultant 'cut strip'. You will see that you join both half-strips into one twice-twisted loop, so you made them both to form a one-piece object. --CiaPan (talk) 08:05, 5 March 2010 (UTC)[reply]

See also orientable manifold (the article may be a bit advanced, but it formally pinpoints some of the above observations). PS T 05:30, 6 March 2010 (UTC)[reply]

A Simple Linear Functional Equation

Consider the linear functional equation

f(2t)=e^{-kt^{2}}f(t).

One obvious solution is

f(t)=Ce^{-{kt^{2} \over 3}},

but is this the only solution? --Andreas Rejbrand (talk) 13:06, 5 March 2010 (UTC)[reply]

It can't be. Define an equivalence relation

a\sim b

iff b = a2ⁿ for some integer n, and take an arbitrary function

C\colon \mathbb {R} /{\sim }\to \mathbb {R}

. Then

f(t)=C(t/{\sim })e^{-{kt^{2} \over 3}}

is also a solution. A better question is whether yours are the only continuous solutions.—Emil J. 13:43, 5 March 2010 (UTC)[reply]

Thank you for your reply. I am only interested in continuous solutions. --Andreas Rejbrand (talk) 17:06, 5 March 2010 (UTC)[reply]

Your solutions are the only ones continuous at 0. Algebraist 17:34, 5 March 2010 (UTC)[reply]

That sounds great. May I ask how you know that? --Andreas Rejbrand (talk) 17:44, 5 March 2010 (UTC)[reply]

Write your solution in the form f(t):=C(t)exp(-kt²/3) and plug it into the equation: you'll get C(2t)=C(t): if f(t), hence C(t), is continuous at 0, C is a constant --pm a 18:01, 5 March 2010 (UTC)[reply]

Yes, because every function can be written on the form f(t):=C(t)exp(-kt²/3), for C(t) might contain the factor exp(kt²/3), and C is continuous iff f is. Thank you very much. --Andreas Rejbrand (talk) 19:01, 5 March 2010 (UTC)[reply]

Resolved

400 million French Francs in 1940

400 million French Francs in 1940. What was the value then in 1940 and also now in 2010? USD or GBP please. —Preceding unsigned comment added by Martin K Hanson (talk • contribs) 14:08, 5 March 2010 (UTC)[reply]

This is the Mathematics Reference Desk. Historical values of French Franc are the subject of history or economics, not mathematics (assuming your problem is not how to multiply a given number by 400 million, in which case you should use a calculator, not ask at RD).—Emil J. 14:20, 5 March 2010 (UTC)[reply]

Economics doesn't fit neatly into our desks, being partly math, partly science, and partly humanities, so let's have some patience with people who post such questions here, please. StuRat (talk) 19:53, 5 March 2010 (UTC)[reply]

The French don't use the Franc anymore; that have used the Euro since January 1999. •• Fly by Night (talk) 18:16, 5 March 2010 (UTC)[reply]

Can Francs still be exchanged for Euros ? If so, then we need to know the exchange rate, and that should be factored in. If not, we need to know the exchange rate back when they could be exchanged, which will then be factored in with the inflation of the Franc from 1940 until then, and the inflation on the Euro from then until now. StuRat (talk) 19:56, 5 March 2010 (UTC)[reply]

There is a difficulty in comparing prices over such a long period, as many of the things we buy are now quite different from then. How do you compare the price of a 1940's car with one now, for example, since they are so totally different ? One method to get around this is to compare something which hasn't changed, like a quart of milk, or, this being France, perhaps a loaf of French bread ? StuRat (talk) 20:00, 5 March 2010 (UTC)[reply]

There were some Vichy France Francs issued at that time, but hopefully you are talking about prewar Francs. According to French Franc there was a 100 to 1 devaluation when they moved from "old Francs" to "new Francs", around 1960, meaning you would have then had 4 million new Francs. The exchange rate with the Euro was 1 EUR=6.55957 FRF in 2002, meaning you would then have some 610,000 Euros. Obviously, there's been some inflation since 2002, but that should give you a rough idea of the value. StuRat (talk) 20:23, 5 March 2010 (UTC)[reply]

This page [1] gives the value of €1 (2009) as equal to 37 centimes (1940). So 400,000,000 FF in 1940 is €1.1 billion today. EamonnPKeane (talk) 17:33, 8 March 2010 (UTC)[reply]

I think you got it backwards. That chart says that one 1940 FF is now woth €0.37. This means 400,000,000 FF would convert to €148,000,000. This is still more than 200 times my answer, so I suspect I made an error in my logic relating to the 100 to 1 devaluation, with the remaining error due to different measures of inflation and the inflation between 2002 and 2009, which wasn't included in my calcs. StuRat (talk) 18:49, 8 March 2010 (UTC)[reply]