Wikipedia:Reference desk/Archives/Mathematics/2009 April 5

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April 5

A question

I saw this question in an old newspaper but I can't figure it out, I know how to solve it for two variables (i.e. by finding the derivative) but I don't get how to solve this, do I need to use a partial derivative? The question is; three integers (let's just call them x, y and z) where x!=z!=y and x+y+z=30. What is the highest product you can get from x*y*z? --BiT (talk) 00:02, 5 April 2009 (UTC)

This is always symmetric, so the answer would be 10 * 10 * 10, except you require they are different. So, it's going to be 9 * 10 * 11 = 990. StatisticsMan (talk) 00:23, 5 April 2009 (UTC)

Yes, like I tried to explain with "x!=y!=z" or better yet; x≠y≠z. But I'm not talking about how to solve it (I'm looking for a word here) using your head, but rather arithmetically (like differentiating). You could always try solving every possible permutation as well, but I'm not interested in that kind of a solution. --BiT (talk) 00:45, 5 April 2009 (UTC)

StatisticsMan's answer is correct, but there are lots of similar optimisation problems whose answer is not symmetric (maximize a^2+b^2 for positive integers a,b, with sum 20; answer = (1,19) or (19,1)). So let's prove it (assuming that x,y,z are meant to be distinct positive integers). If two of the integers differ by 3 or more, increasing the smaller and decreasing the larger makes the product bigger (calculate the difference it makes). So the optimum occurs at a place where this operation cannot be done without making two of the values equal. The only such pattern with sum 30 is (9,10,11). McKay (talk) 00:42, 5 April 2009 (UTC)

Like I said above, I figured out the answer but I was rather looking for a way of solving it using calculus or arithmetic. I'm sorry, I should have been more clear on this in my question. --BiT (talk) 01:23, 5 April 2009 (UTC)

Usually when the variables are integers the complete solution cannot not found with calculus. Calculus can show that the variables must be equal for maximizing over the reals, see Lagrange multiplier, but from that point to apply the conditions of integrality and distinctness needs a more combinatorial argument. My method of moving variables a little at a time is a discrete form of differentiation that often works for integer problems. Incidentally, x≠y≠z still allows that x=z; you need to write x≠y≠z≠x but it is even clearer to write "x,y,z are distinct". McKay (talk) 02:29, 5 April 2009 (UTC)

I misread the first condition as (x!) = (z!) = y. —Tamfang (talk) 03:20, 5 April 2009 (UTC)

As did I. However, there is no solution, as x=4 gives us a total of 32, while x=3 only gives us a total of 12. StuRat (talk) 17:41, 6 April 2009 (UTC)

Reminds me of the C proof that 0! = 1: if (0!=1) printf("true\n"); else printf("no\n"); McKay (talk) 04:53, 5 April 2009 (UTC)

For the specific problem given, StatisticsMan's approach is the most direct: the product cannot be greater than 1000 because of the arithmetic-geometric mean inequality. Looking at the next largest numbers 999, 998, etc., we find all of them above 990 have a prime factor larger than 30, so we can rule them out immediately (assuming x,y,z are supposed to be positive integers). 990 factors as 2*3*3*5*11 and inspection shows this can be partitioned as (x,y,z)=(2*5,3*3,11) with x+y+z=30. 75.62.6.87 (talk) 08:35, 5 April 2009 (UTC)

DNE

Hello. Is DNE, which stands for Does Not Exist, a widely recognized abbreviation? Thanks in advance. --Mayfare (talk) 01:33, 5 April 2009 (UTC)

Yes, almost everybody will understand what DNE stands for without explanation. However, I find it a pretty ugly abbreviation, and it's not a phrase used so often for the abbreviation to be worth anything. And writing things like $\lim _{n\to \infty }(-1)^{n}={\text{DNE}}$ really bothers most people (well, myself anyhow). So, don't use it. Eric. 131.215.159.99 (talk) 03:01, 5 April 2009 (UTC)

I've been a mathematician for 33 years and I don't remember meeting DNE before except in computational logic where it has a technical meaning. So my answer to the question is "no, it is not a widely recognized abbreviation". McKay (talk) 04:50, 5 April 2009 (UTC)

For what it's worth, I don't recognize it at all, and I can't think of a circumstance in mathematics in which it wouldn't be horrible. Algebraist 10:38, 5 April 2009 (UTC)

I'd use ⊥ (bottom) for the computational logic case or NaN or an error for computing. Dmcq (talk) 11:54, 5 April 2009 (UTC)

Hmmm, maybe to some extent it is a generational / cultural thing. Ask current American college students about DNE in the context of calculus and I'd guess they'd get the reference. In any case, we're all agreed that it should never be used. Eric. 131.215.159.99 (talk) 12:13, 5 April 2009 (UTC)

I teach college mathematics in the US and I use DNE all the time. What's wrong with it? (I agree with Eric that obviously something like x=DNE is bogus, but something like x DNE sounds fine to me.) Staecker (talk) 13:31, 5 April 2009 (UTC)

Never heard of it before!The Successor of Physics 14:30, 5 April 2009 (UTC)

Actually I'm quite comfortable with the x = does not exist. Extending types with a does not exist in the sense of bottom type seems a fairly natural thing to me in many circumstances. Of course if you do that then even the results of a test like x<y is normally bottom if either of the operands is. Dmcq (talk) 20:23, 5 April 2009 (UTC)

You should not say "this limit equals `does not exist'"; rather you should say "this limit does not exist". If the "equals" sign is there, then that is the verb in that sentence. But "does not exist" should be the verb. So if you must use the "DNE" abbreviation, then write

$\lim _{n\to \infty }(-1)^{n}{\text{ DNE}},\,$

without the "=".

I don't recall ever seeing the "DNE" abbreviation used by a mathematician, but I see it used by undergraduate students all the time, and have for many years. It's one of many memes that strangely persist without ever being taught. Michael Hardy (talk) 21:39, 5 April 2009 (UTC)

Never met this ugly notation. If I did, I was able to forget it soon and to be happy again. Hope to do this time too! --pma (talk) 22:59, 5 April 2009 (UTC)

I use this intialism on the blackboard, but never in print of any sort. Even on the blackboard I would never write "$\lim _{x\to 1}f(x)={\mbox{DNE}}$". Instead I would write "$\lim _{x\to 1}f(x)\,{\mbox{DNE}}$" without the equals sign. I also use abbreviations like "b/c" ("because") on the blackboard; it's just shorthand, not much different than writing the three dots ∴ for "therefore". In general, these sorts of abbreviation are all shunned in mathematical prose. — Carl (CBM · talk) 03:56, 6 April 2009 (UTC)

I was taught it in high school, with the equals sign. Haven't seen it since. Black Carrot (talk) 13:50, 6 April 2009 (UTC)

Largest Ordinal

I'm a little confused by the alleged paradox of the largest ordinal. Say you construct the ordinals as sets of previous ordinals, with zero being the empty set, and try to form the set O of all ordinals. O is also an ordinal. Clearly, the order relation on ordinals breaks down, since the successor of O is O itself, but how is that a contradiction? It just means the order doesn't work. Where's the actual paradox? Black Carrot (talk) 03:13, 5 April 2009 (UTC)

In ZF or ZFC, if O is its own successor, it would have to contain itself as a member, contradicting the axiom of foundation. In some other set theory (New Foundations?), perhaps something like O is possible. —Preceding unsigned comment added by 75.62.6.87 (talk) 08:51, 5 April 2009 (UTC)

You don't need the axiom of foundation. When you do have the axiom, an ordinal is sometimes defined to be a transitive set linearly ordered by elementhood, but this is a little bit of a trickological definition, frankly.

The idea of an ordinal is that it abstracts the order of a wellordering. So we see the paradox already from this informal concept of ordinal: If there were a set of all ordinals, then it would be wellordered by the relation of "shorter than" (which is coded as "elementhood" in the case of the von Neumann ordinals). So abstracting the order of the set of all ordinals, we come up with an ordinal that is strictly greater than all ordinals. But then it is strictly greater than itself, contradiction.

(I won't try to comment on the NF issue because I'd probably wind up saying something stupid, except to say that, as is common in NF, it's likely to depend on choices you make that, if you're not familiar with NF, would not appear to matter. However, if a coding of the ordinals in NF respects the informal idea outlined above, then the paradox must apply to it too.) --Trovatore (talk) 09:46, 5 April 2009 (UTC)

An ordinal can't be its own successor. Each ordinal is the order type ("order-isomorphism class"?) of the set of all lesser ordinals. The proposed biggest ordinal O would then be the order type of the set of all ordinals, including itself, and that set would have a largest member, namely O; but on the other hand, O is the order type of the set of all strictly smaller ordinals, which has no largest member. Michael Hardy (talk) 21:34, 5 April 2009 (UTC)

Here's a different way of saying the same thing. Suppose that there was a largest ordinal, call it α. The usual definition of von Neumann ordinals in set theory says that α is the set of all ordinals smaller than α. Now define a set $\beta =\alpha \cup \{\alpha \}$. Then β is also a von Neumann ordinal, but is larger than α. Thus there is no largest ordinal. The "paradox" is that the class Ord of all ordinals appears to be an ordinal itself; it contains only ordinals and is well ordered by ∈. This is paradoxical because there is no ordinal larger than the class of all ordinals (and thus larger than itself). The resolution of the paradox is that Ord is not an ordinal because it is not a set. — Carl (CBM · talk) 14:06, 6 April 2009 (UTC)

I completely agree with Trovatore that this is more directly seen in terms of well-ordered sets (the explanation via von Neumann ordinals is very clear and elegant, but has more theory behind). In terms of well-ordering, the claim "for any ordinal α, α is not its own successor" reads: "if X is a well-ordered set with greatest element $\scriptstyle m$, then $\scriptstyle X$ and $\scriptstyle X\setminus \{m\}$ are not order-isomorphic". The reason of the latter is, that if there were an order isomorphism $\scriptstyle \varphi :X\to X\setminus \{m\},$ the $\scriptstyle \varphi$-orbit of $\scriptstyle m$, $\scriptstyle \{m>\varphi (m)>\varphi (\varphi (m))>...\},$ would be a non-empty subset of X with no least element, a contradiction. (That this is a strictly decreasing sequence follows immediately from $\scriptstyle m>\varphi (m)$, applying inductively $\scriptstyle \varphi$). --pma (talk) 19:23, 6 April 2009 (UTC)

I think I see. Thanks. Is there an intuitive way around this, or does it just come down to fiddling with the rules until it works? Black Carrot (talk) 02:12, 7 April 2009 (UTC)

I'm not certain what you're asking. — Carl (CBM · talk) 05:17, 7 April 2009 (UTC)

Cubic Residues

Just using some basic algebra and basic number theory, I have been introduced to quadratic residues. I just know about the definition and some basic properties of the Legendre symbol. One of the statements made by the book is that if a prime number p is of the form 3k+2 then all integers in a reduced residue system mod p are cubic residues. This fact is easy to prove and the cubic residue page on Wiki actually has an easy proof. The next line says, if p is of the form 3k+1, one one-third of the member of a reduced residue system are cubic residues. My question is how to prove it.

Now the only thing I get out of this is that p is prime of the form 3k+1 so minus the zero element, we have 3k elements in the reduced residue system (because everything else is coprime to p) so the RRS can be split up into three equal parts. The page here on wikipedia says that let e be a cubic nonresidue. The first set is the cubic residues, the second one is e times the numbers in the first set, and the third is $e^{2}$ times the numbers in the first set. How do we know this? If anyone can perhaps get me started down the correct path. Thanks!-Looking for Wisdom and Insight! (talk) 10:50, 5 April 2009 (UTC)

One way of making this easy is to remember that the multiplicative group of Z_p is cyclic. Algebraist 10:58, 5 April 2009 (UTC)

Integral with one of its bounds appearing in the integrand

A while ago I came across the integral $\lim _{n\to \infty }\int _{0}^{n}{\frac {\sqrt {n^{2}-x^{2}}}{2+x^{-x}}}dx$ in a friend's book and have been unable to make any meaningful progress in solving it. How should one go about solving this integral, or for that matter any integral with a similar form (i.e., 'n' approaching a limit as both a bound and a part of the integrand?) Is it allowable to bring the limit inside the integrand if a trig substitution such as $x=n\sin {u}$ is made? Any help would be greatly appreciated. Korokke (talk) 22:21, 5 April 2009 (UTC)

On the interval [1,2] the integrand is larger than $\scriptstyle {\frac {\sqrt {n^{2}-4}}{3}}$ so the limit is infinite.--pma (talk) 22:43, 5 April 2009 (UTC)

...and generally with things like this you should work with inequalities rather than with equalities. Trying to find something the integral is equal to, that doesn't involve integrals, may be futile, but trying to find something the integral is bigger than, and that diverges to infinity, may nonetheless be easy. Michael Hardy (talk) 01:48, 6 April 2009 (UTC)

ahh interesting...thank you! I guess I get caught up with trying to find the standard "textbook solution" and overlook the more elegant solutions. However, why was the interval [1,2] chosen? Korokke (talk) 02:31, 6 April 2009 (UTC)

Why [1,2]: just because, at a first glance, for all $\scriptstyle x\geq 1$ we have clearly $\scriptstyle x^{-x}\leq 1$, so the denominator is between 2 and 3; the numerator is decreasing and looks like big, so as a first thing one just checks if an easy lower bound on [1,2] is sufficient to conclude.

If you prefer a kind of more systematic approach to estimate the integral: compare it with a simpler one that you can use as bound, as Michael Hardy says. Here, the numerator would be easily integrated, were it alone. The denominator makes the integral very hard or impossible to compute in terms of elementary functions. But it has simple bounds: computing the maximum of $\scriptstyle x^{-x}=\exp(-x\log x)$ on $\scriptstyle ]0,\infty [$ you find:

$0<x^{-x}\leq e^{\frac {1}{e}},\quad \forall x>0.$

Therefore you have

$\int _{0}^{n}{\frac {\sqrt {n^{2}-x^{2}}}{2+e^{\frac {1}{e}}}}\,dx\leq \int _{0}^{n}{\frac {\sqrt {n^{2}-x^{2}}}{2+x^{-x}}}\,dx\leq \int _{0}^{n}{\frac {\sqrt {n^{2}-x^{2}}}{2}}\,dx,$

that is

${\frac {\pi n^{2}}{8+4e^{\frac {1}{e}}}}\leq \int _{0}^{n}{\frac {\sqrt {n^{2}-x^{2}}}{2+x^{-x}}}\,dx\leq {\frac {\pi n^{2}}{8}}$.

--pma (talk) 09:34, 6 April 2009 (UTC)

That makes so much sense now! Thank you! Korokke (talk) 23:00, 7 April 2009 (UTC)