Wikipedia:Reference desk/Archives/Mathematics/2008 December 30

= December 30 =

Factorization
Can someone factor this number?

--Melab±1 &#9742; 01:02, 30 December 2008 (UTC)


 * 73369 × 7493033. You can use http://www.alpertron.com.ar/ECM.HTM another time. PrimeHunter (talk) 01:12, 30 December 2008 (UTC)

Hypersudoku question
I read that with ordinary Sudoku at least 17 numbers had to be given to get a viable puzzle, although that conjecture has not been formally proved. Now I am interested in Hyper Sudoku, and wondering what the minimum number of givens would be. I imagine it would be less than 17. Any ideas? Myles325a (talk) 01:09, 30 December 2008 (UTC)


 * Correct me if I'm wrong, but I believe the conjecture was that EXACTLY 17 is the minimum necessary to define a unique solution. And is has been proven that AT LEAST 17 are necessary, i.e. it's a lower bound but we don't know if it's the upper bound for the minimum necessary as well. A math-wiki (talk) 23:25, 2 January 2009 (UTC)
 * No, for regular sudoku it's an upper bound. That is, it is known that there is an arrangement of 17 numbers in a sudoku grid which determines a unique solution, but it is not known if there is an arrangement of 16 that does so. Algebraist 23:28, 2 January 2009 (UTC)

Polynomial
Can a polynomial have modulus operator, floor function, or ceiling function in it? --Melab±1 &#9742; 01:31, 30 December 2008 (UTC)

I am unsure if I understand your question, but I will try to answer it as best I can. In a polynomial we allow the coefficents and exponents to be fixed, that is they will not be a function of your input x. If your function is f(x), and a modulus operator or ceiling/floor function operates on x, that is to say you take x modulo a number, or similar action, then your function is not a polynomial. Please see polynomial for more information. -Damelch —Preceding unsigned comment added by Damelch (talk • contribs) 02:21, 30 December 2008 (UTC)


 * Haven't you asked this before? The answer is that although they are not commonly used if you wish such operators can be defined and they will correspond intutively to our understanding of the floor, ceiling functions. Cheers--Shahab (talk) 05:08, 30 December 2008 (UTC)
 * Meaning? --Melab±1 &#9742; 18:26, 30 December 2008 (UTC)
 * Was there anything unclear in the preceding answers? Or maybe it was something else, what you wanted to know? But what do you mean by "a modulus operator in it"?--PMajer (talk) 22:54, 30 December 2008 (UTC)
 * Remainder division represented by "%" --Melab±1 &#9742; 21:14, 1 January 2009 (UTC)

Question concerning two integrals
A textbook I was reading said that
 * $$\int_a^b \frac{\sin t}{t} dt$$

is uniformly bounded for positive numbers a,b. This was presented as an elementary fact, and I'm not seeing it. Am I missing something obvious? A similar statement was made that
 * $$\int_a^b \frac{\cos (\theta t) - \cos(t)}{t} dt$$

is uniformly bounded as well in terms of $$\theta$$, and converges to an expression involving $$\log (\theta)$$, which is also mysterious to me. Can anybody shed light on this? Thanks, Ray (talk) 12:55, 30 December 2008 (UTC)
 * The first equation is usually called the sinc integral or the sine integral. This webpage should be able to help you. Mathworld Link As you can see the integration is somewhat complex, which is probably why they did not give you the proof of this.  You can also look at Trigonometric integral for assistance.  From the trigonometric integral page we can see that the cosine part of the integrals will also be bounded, however I am unsure why they converge to $$\log (\theta)$$.  I hope this helps. --Damelch (talk) 17:56, 30 December 2008 (UTC)
 * I'd say it is just a fact of a continuous function F(b)on $$[0,\infty]$$ being bounded (w.l.o.g. a=0). It's therefore a consequence of the fact that the integral is convergent for b going to infinity. For the second it is the same, but first write (for$$ \theta>0$$):

$$\int_0^b \frac{\cos (\theta t) - \cos(t)}{t} dt = \int_0^b \frac{\cos (\theta t) - 1}{t} dt -\int_0^b \frac{\cos (t) - 1}{t} dt = \int_0^{b\theta} \frac{\cos (t) - 1}{t} dt- \int_0^b \frac{\cos (t) - 1}{t} dt $$, so you can bound your second integral with 4 times the sup norm of $$\scriptstyle F(b):=\int^b_0\frac{\cos (t) - 1}{t} dt$$, again a bounded function for the integral is convergent.--PMajer (talk) 18:49, 30 December 2008 (UTC)

Equations for Πnundefinedf(x) and Σnundefinedf(x)
How do you change an equation of Πnundefinedf(x) into an equation without the Π and how do you change an equation of Σnundefinedf(x) into an equation without the Σ? The Successor of Physics  14:42, 30 December 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)
 * With difficulty, in general. If it's really important, one might invent a new notation, such as factorial. Did you have any particular f in mind? Algebraist 16:59, 30 December 2008 (UTC)


 * No, but I guessed if I could get the general equation for this, then it would make a question of mine to do with gravity and other scientific questions a lot easier. The Successor of Physics  14:45, 31 December 2008 (UTC)

Extension of Σnundefinedf(x) and Πnundefinedf(x) for n≠xINT1 or n≠an integer
How do you do Σnundefinedf(x) and Πnundefinedf(x) for n≠xINT1 or n≠an integer? The Successor of Physics  14:48, 30 December 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)
 * You simply do not do it this way, unless you are trying to give some mistic sense to it. There are nevertheless situations where $$\scriptstyle S(n):=\sum_{k=1}^n f(k)$$ has a natural extension to non integer $$n$$. For instance: if f(x) is defined for all real $$\scriptstyle x\geq0$$, $$\scriptstyle f(x)\geq0$$ and f(x)=o(1) as $$\scriptstyle x\to\infty$$ decreasing then a natural extension of S(n) is the unique increasing solution of the functional equation S(0)=0, S(x)=S(x-1)+f(x) . I suggest that you try and prove existence and uniqueness for it and your spirit will be placated! --PMajer (talk) 16:55, 30 December 2008 (UTC)(PS: it's just $$\scriptstyle S(x):=\sum_{k=1}^{\infty} (f(k)-f(k+x))$$, btw)

Collatz conjecture
In Collatz_conjecture, it states that: "It is also known that {4,2,1} is the only cycle with fewer than 35400 terms". I don't understand this. If you start with the number 2, you only have two terms {2,1}, right ? Do they mean something else by "terms" ? StuRat (talk) 14:52, 30 December 2008 (UTC)


 * If you start with the number 2, you get the cycle {2,1,4} – the same as {4,2,1}, just shifted a little. -- Jao (talk) 15:15, 30 December 2008 (UTC)


 * No, because, under the Collatz conjecture, you stop when you get to 1. StuRat (talk) 17:06, 30 December 2008 (UTC)


 * That doesn't seem to be stated in our article (which I have not read carefully). Certainly for the purposes of that sentence it is not true. Algebraist 17:09, 30 December 2008 (UTC)


 * If you don't stop at 1, then you always get an infinite number of terms, as once you get to 1, you then get {1,4,2,1,4,2,1,4,2,1,...}. StuRat (talk) 17:45, 30 December 2008 (UTC)


 * Sure, but you're stuck in a cycle, which has a length of 3. If the Collatz conjecture is false, then either there is some starting value for which the resulting sequence grows without bound, or else there is some cycle other than 1, 2, 4, 1, 2, 4, …. In the second case it is known that such a cycle cannot have a length less than 35400. —Bkell (talk) 17:51, 30 December 2008 (UTC)


 * Oh, so that's what they meant. I'll have to fix it up so it's a bit more obvious to non-mathematicians.  Thanks. StuRat (talk) 21:42, 30 December 2008 (UTC)

OK, I changed it to:


 * "All initial values tested so far eventually end in the repeating cycle {4,2,1}, which has only three terms. It is also known that {4,2,1} is the only repeating cycle possible with fewer than 35400 terms."

I think that's more clear. StuRat (talk) 21:50, 30 December 2008 (UTC)


 * Has it been proven that {4, 2, 1} is the only repeating cycle with fewer than 35400 terms? Or is that just an experimental result?  You might want to clear that up.  --Tigerthink (talk) 15:50, 3 January 2009 (UTC)


 * How are they different ? For a finite range it seems to me that experimental evidence is proof.  StuRat (talk) 21:24, 3 January 2009 (UTC)


 * But having period less than 35400 doesn't fix a finite range of cycles, a priori...--PMajer (talk) 14:46, 5 January 2009 (UTC)


 * I don't know what type of evidence was offered. The footnote lists the source as "Lynn E. Garner (1981) Proceedings of the American Mathematical Society Vol. 82, No. 1, May, 1981; ISSN:00029939, pp. 19-22".  If someone with access to that document can check, then we can update the article. StuRat (talk) 17:44, 5 January 2009 (UTC)

Triangle wave
Can someone generalize this function for a triangle wave of amplitude 0.5 so that I can plug in a 2nd number to change the amplitude? f(x)=((1/2)+2*floor(x/2)-floor(x))*(x-floor(x)-(1/2))

--Melab±1 &#9742; 18:39, 30 December 2008 (UTC)


 * If the function above has an amplitude of 1/2, then just multiply the entire thing by n to get a function of amplitude n/2. —Bkell (talk) 19:39, 30 December 2008 (UTC)


 * But I also want to keep the wavelength the same as the amplitude. --Melab±1 &#9742; 21:41, 30 December 2008 (UTC)


 * Then replace x with x/n throughout. —Bkell (talk) 00:20, 31 December 2008 (UTC)


 * Use a.f(x/a), you can do this trick for any function, you might notice this has the form g( f( g-1( x ) ) ) which occurs quite often in maths. Dmcq (talk) 00:25, 31 December 2008 (UTC)

Trigonomic approximation revisited
Hi. You may remember over a year ago, I asked a question about estimating the distance of something by using its size and angular size, or estimating its size using its distance and angular size. Let's consider an object in the distance has an angular diameter of 15 degrees. You know that the object is, say, 50 m in diameter. You want to figure out its distance. You divide 60 degrees by 15 degrees, which gives you 4. You then multiply this number by 50 m, and the result is 200m. However, this would assume that the altitude of an equilateral triangle is equal to its base length, which it's not. So, you multiply the calculated distance, 200 m, by the square root of 0.75, which ≈0.866. The result then ≈173 m. However, not all triangles are equilateral. Last time, I devised a number, ~1.1. I figured if you multiplied your initial estimate by sqrt0.75, you'd then multiply the result by ~1.1, and vice versa. Then, you have your distance, which =~190 m. Or, is it far more complicated than that? Or, is this step not nessecary, or does it involve a different or a variable number? Let's use an example where the unknown parameter is size. You see an object you know is 10,000 metres away. It has an angular diameter of 3 degrees. 60 degrees divided by 3 is 20. Then, you divide 10,000 by 20, then you divide the result by sqrt0.75. Then, you divide the result by ~1.1? This would be a useful way of estimating if you don't have a calculator or graph handy, and you have no way of making an exact estimate anyway. Is it a good approximation? Is there an article on this? How can it be improved? Thanks. ~ A H  1 (TCU) 22:47, 30 December 2008 (UTC)
 * What you ment to do was to compute the angular size in radians. 60 degrees equal &pi;/3~1.04720 radian. You compute 60 degrees equal to 1/(1.1&times;sqrt0.75)~1.04973 radian. Basicly you have computed the number Pi, &pi;, to be approximately 3/(1.1&times;sqrt0.75)~3.14918. An improved approximation is &pi;~3.14159. Bo Jacoby (talk) 07:22, 31 December 2008 (UTC).


 * A better approximation than yours is to take pi as 22/7≈3.143, when the three quantities are related by the equation
 * angle in degrees = (630 X diameter)/(11 X distance). Knowing any two, the third can be estimated. No calculator is needed if you can multiply or divide by 7, 9 or 11.→86.132.234.49 (talk) 13:34, 31 December 2008 (UTC)


 * You also don't need a calculator if you have a suitable scale on your tool to measure angles. The book Öveges József, A fegyverek fizikája (Zrínyi katonai kiadó, Budapest, 1972, p. 152) talks about such a device: a scale on hairlines in a military binoculars.  If the visible size of an object is the distance between two adjacent hairlines, then the object is 1000 times as far as its size, so for example if a 1 meter high object spans exactly from one hairline to the next one, it's 1 kilometer away.  &#x2013; b_jonas 10:19, 4 January 2009 (UTC)