Wikipedia:Reference desk/Archives/Mathematics/2010 January 8

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January 8

Simple Mathematics, Requiring Data

Resolved

--See recent at Science Desk

I'm not exactly asking this because I can't do it, but because someone here has the data to do it more quickly, and s/he may as well have the opportunity to see the answer first.

What is the side-length of the regular octahedron composed of all anthropogenic Carbon currently in the atmosphere transformed into the diamond structure?

You can consider the question as illustrative, unless you think like me.Julzes (talk) 09:55, 8 January 2010 (UTC)

Differentiable?

Let f: ℝ→ℝ a function with |f(x)|≤x². How to prove that f′(0) is always 0? --84.62.205.233 (talk) 10:02, 8 January 2010 (UTC)

You should be made aware that not all functions have a derivative and that differentiability is a local characteristic (i.e., in this case $f(x)\$ may be differentiable only at zero if the problem has a solution). The problem's statement is not very good, but it does have a solution nevertheless. Use the basic definition of the derivative (at x=0) and see what you get.Julzes (talk) 10:22, 8 January 2010 (UTC)

You'll find that definition at Derivative Dmcq (talk) 10:29, 8 January 2010 (UTC)

You don't have anything to prove. Just by definition of derivative |f(x)|≤x² means in particular that f is differentiable at 0 and that f'(0)=0.--pma 13:16, 8 January 2010 (UTC)

Of course there is something to prove. Just because it takes about 2 lines, one of which is just stating a definition, doesn't mean it isn't a proof. --Tango (talk) 16:20, 8 January 2010 (UTC)

Well, there is always a proof, sure. To be precise you just have to recognize " |f(x)|≤x² " as a particular case (i.e. x₀=l=0) of the definition of derivative, that is f(x₀+x)=f(x₀)+lx+o(x) as x→0. To do this you don't have anything to write. Here the only thing to prove, if anything, is that x→0 as x→0. --pma 17:56, 8 January 2010 (UTC)

I'm not convinced. While what you say is obviously true, I think there is still something to prove in there - I would write down something about A+Bx+o(x)<x² (as x->0) implying A=B=0, I don't think that would be immediately obvious to someone taking a course at the level where this question would come up. I think an elementary proof directly from the definition (as the limit of a quotient) is preferable. --Tango (talk) 18:52, 8 January 2010 (UTC)

I'd say it depends on what is the definition of derivative you have given. I personally believe that the right definition of derivative (at every level, even at the level you mention) is not by means of a limit of a quotient, but it is via the first order expansion. There are several reasons of fundamental importance to do that, both from theory and from practice. Therefore, if you have introduced the derivative properly, and with a reasonable comment on the definition, then the OP's statement is really self-evident. --pma 23:55, 8 January 2010 (UTC)

Informally, you can see that f(x) is squeezed between x² and −x². If you want a pathological example that is differentiable only at x=0, consider f(x) = x² if x is rational; −x² if x is irrational. Gandalf61 (talk) 17:02, 8 January 2010 (UTC)

$f'(0)=\lim _{h\to 0}{\frac {f(h)-f(0)}{h-0}}.$

So what is ƒ(0) if |ƒ(x)| ≤ x² ? After you've answered that, you can simplify the numerator of the fraction, and then further simplify the fraction after that, and finally use the inequality |ƒ(x)| ≤ x² one more time. Michael Hardy (talk) 17:37, 8 January 2010 (UTC)

Number of partitions of a set in classes of given cardinalities.

Suppose there were 150 people in a building with six rooms, and only 25 people were allowed in each room. How many possible combinations can those people be divided into the rooms? jc iindyysgvxc (my contributions) 11:22, 8 January 2010 (UTC)

I just put a generic label on your section for you. I will let others deal with the question you've asked.Julzes (talk) 11:56, 8 January 2010 (UTC)

See combinations. How many ways are there of picking 25 people from the 150 to go into the first room? How many ways of picking 25 from the remaining 125 to go into the second room? What should be done with the two numbers obtained? Proceed likewise to get the answer.→86.160.55.155 (talk) 12:21, 8 January 2010 (UTC)

Alternatively, how many of the 150! permutations of the 150 people leave each person in the same room? Then you have to divide 150! by this number. The general problem is the same. I put a more specific header.--pma 13:06, 8 January 2010 (UTC)

If you regard the rooms as distinguishable, then you'll get a bigger number than if you regard them as indistinguishable. It would be 6! = 720 times as big. More later.... Michael Hardy (talk) 17:28, 8 January 2010 (UTC)

${\frac {150!}{(25!)^{6}}}.$

You'll find in some very elementary books the statement that the number above is the number of distinguishable "words" that can be made from 25 indistinguishable "A"s, 25 indistinguishable "B"s, etc. through the sixth letter. Now let the letters A–F correspond to the six rooms. Michael Hardy (talk) 17:33, 8 January 2010 (UTC)

Letter grid problem

Here's a simple example of a more general problem which I'm considering without success - suppose that distinct letters are at the corners of a square, and that 4-letter words are formed by traversing each possible route linking all corners without repetition. There are two cases (a) only horizontal and vertical connections are possible, giving 8 words (b) diagonal connections too are possible, giving 24 words.

What I want is the number of possible complete words from the mn letters put on an mXn grid (m,n>1), for linking rules (a) and (b). Is this a standard result?→86.160.55.155 (talk) 12:12, 8 January 2010 (UTC)

Sorry, I'm not sure I got what you want. Will you clarify it better maybe with an example? Thanks --pma 13:12, 8 January 2010 (UTC)

I don't know the answer to your question, but I can phrase it in different terms that might lead you somewhere. In case (a), you're looking for the number of (directed) Hamiltonian paths in a square grid graph. Case (b) is similar, but the graph includes diagonal edges too. —Bkell (talk) 13:16, 8 January 2010 (UTC)

Case (b), if I understand the extension beyond 2×2 correctly, is the king's graph. --Tardis (talk) 15:22, 8 January 2010 (UTC)

As a further example, consider a grid of 2 rows and 4 columns, with the points in the top row A, B, C and D consecutively from L to R and in the bottom row E, F, G and H consecutively from L to R. Then AEFBCDHG would be a valid word in the case of either linking rule, but AEFBCHDG would be valid only with the permitted diagonal linking of case (b). If x(m,n) is the number of words with m rows and n columns without diagonal linking, it's easy to show, I think, that x(n,m)=x(m,n), x(1,n)=2, x(2,2)=8, x(2,3)=16, x(2,4)=28 and x(3,3)=56. I'm guessing that a recurrence relation might give general x(m,n), but I can't reliably count the number of words in larger grids so as to determine its form. Hence my question as to whether this is a standard result. If y(m,n) is the corresponding number of words with diagonal linking possible, y(n,m)=y(m,n), y(1,n)=2, y(m,n)>x(m,n) for m,n>1. Again, is there a known result?

The stimulus for this query was a common kind of letter puzzle, to find the 9-letter word in a 3X3 grid with diagonal linking possible, where I wondered how many possibilities there were. A particular one had the letters FUN, LET and GEC in consecutive rows (not all letters different), where you'll find GENUFLECT without too much trouble.→86.160.55.155 (talk) 16:10, 8 January 2010 (UTC)

I still don't know the answer, but I have a hunch that a general recurrence relation, if one exists, is not going to be pretty. Why do I say this? Well, if you start with a Hamiltonian path for a grid graph, and you slice off the last column (say), usually you will not get a Hamiltonian path for the smaller grid graph, but rather a collection of paths that together cover all the vertices. The "interaction" of such paths may be complicated, and whether such a collection of paths can be extended to a Hamiltonian path for the original grid graph may depend on what happens in the very first column of the grid.

Now, having said that, it is clear that the 1×n case is trivial, and you may be able to work out recurrence relations for the 2×n and 3×n cases, for example, but as the shorter dimension increases my intuition tells me that the number of special cases you will have to consider will grow quickly. Of course, I could very well be wrong. —Bkell (talk) 21:54, 8 January 2010 (UTC)

2xn seems easy (isn't that a Hamiltonian path is determined by its endpoints?). Anyway, it seems the general problem has been studied. If you state the question as you suggested, "number of Hamiltonian paths in a rectangular grid" and plug it into Google, you get a number of interesting results and articles; there is also something in the On Line Encyclopedia of Integes Sequences: the case nxn is [1] --pma 23:18, 8 January 2010 (UTC)

Limits

Is there a number $a$ such that

$\lim _{x\to -2}{\frac {3x^{2}+ax+a+3}{x^{2}+x-2}}$

exists? If so, find the value of a and the limit.

No idea how to approach this problem; although I guess solving for $a$ would be a start, if I could set it up for that. Any ideas? (ಠ_ಠ) - 17:26, 8 January 2010 (UTC)

What is the value of the denominator at that point? If we assume the limit exists, what does that say about the numerator? --Tardis (talk) 17:54, 8 January 2010 (UTC)

You want to be able to factor x+2 out of the numerator, in order to cancel with x+2 in the denominator. That is, you want the numerator to evaluate to 0 when x=-2. --COVIZAPIBETEFOKY (talk) 17:55, 8 January 2010 (UTC)

Okay, that's fine then - $3(-2)^{2}+(-2)a+a+3=0$, which gives $a=15$. Although I don't quite understand the motivation. As by Michael Hardy's post below, why won't the limit exist if the numerator is not 0 when x = -2?(ಠ_ಠ) - 18:20, 8 January 2010 (UTC) —Preceding unsigned comment added by Damien Karras (talk • contribs)

Again, what is the value of the denominator there? What does any fraction with such a denominator do at that point if its numerator remains non-zero? --Tardis (talk) 18:35, 8 January 2010 (UTC)

Well, around the point, I assume it would shoot off to infinity. I'm not sure. There are many functions that are just discontinuous at a single point. (ಠ_ಠ) - 19:01, 8 January 2010 (UTC)

The denominator f(x) = x² + x - 2 is a continuous function, so not only is f(x) = 0 when x = -2, but as x gets near -2, f(x) gets near zero. When the denominator gets small that means the fraction gets large. When x approaches -2, f(x) gets arbitrarily small, which mean the fraction gets arbitrarily large (i.e. approaches infinity), unless the numerator does something to counter that growth. You're looking to find the value of a that let's the numerator do that. Rckrone (talk) 19:50, 8 January 2010 (UTC)

Solving WHAT for a??

The limit won't exist if the numerator is not 0 when x = −2. So plug −2 in for x in the numerator, set that equal to 0, and solve that for a. Whether the limit exists then is a question of what happens when the resulting polynomial is divided by (x + 2). Michael Hardy (talk) 18:04, 8 January 2010 (UTC)

No, that's not correct. In a more general case somewhat like this, whether a limit exists at x=-2 depends on whether x=-2 is a multiple root. If not, there is a limit regardless of the denominator. Generally, there is a limit if the multiplicity of the root in the numerator is not lower than the multiplicity in the denominator.Julzes (talk) 23:12, 8 January 2010 (UTC)

No, sorry, it is correct, and it is the efficient way to solve the given problem (nobody is speaking of more general or different cases: we are speaking of this case).--pma 23:28, 8 January 2010 (UTC)

That was a response to Michael Hardy's last sentence. In the specific case, once you divide out all of the factors of (x+2) from the denominator, you can be certain there is a limit.Julzes (talk) 00:10, 9 January 2010 (UTC)

Although, we should check that the denominator doesn't have x=-2 as a repeated root, since if it does the numerator would have to be equal to the denominator (up to multiplication by a constant) for there to be a limit. That doesn't take long, though, since the denominator is quadratic so the only way it could have a repeated root is if it is a perfect square, which it clearly isn't. --Tango (talk) 00:01, 9 January 2010 (UTC)

Note that ${\frac {d(x^{2}+x-2)}{dx}}=2x+1$ which has only one root at $x=-{\frac {1}{2}}$. Therefore, −2 cannot be a repeated root of $x^{2}+x-2$. --PST 03:54, 9 January 2010 (UTC)

In general, that's a good method, but in this case I would have just said x²+x-2 ≠ x²+4x+4, QED. --Tango (talk) 16:39, 9 January 2010 (UTC)

Anyway I'd follow MH's directions (find the proper value of a that makes the numerator vanish; consider the fraction only for that value of a --the fraction should be 3 + 12/(x-1); compute the limit); there's nothing to check. --pma 00:36, 9 January 2010 (UTC)