Wikipedia:Reference desk/Archives/Mathematics/2009 December 10

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December 10

sudoku

though i m fond of solving sudokus i m unable 2 complete them. i feel very irritated when i get struck in the last step. so pls provide me the details of how to solve a sudoku, the tricks behind it. thanx —Preceding unsigned comment added by Srividhyaathreya (talk • contribs) 06:46, 10 December 2009 (UTC)

See Sudoku. One good techinique is to complete the puzzle square by square; in each empty square, neatly write the possible numbers that could be within the square at the particular stage you are at in solving the puzzle. Once you progress furthur, you will get more information which will allow you to eliminate numbers you had thought could be in a square at an earlier stage, eventually finding only one possibility for that particuar square. Continue in this manner to solve the puzzle. (Wikipedia:Reference Desk/Miscellaneous would be more appropriate; puzzles like this have nothing to do with proper mathematics, since mathematics is not about arranging numbers in a 9x9 grid). --PST 07:18, 10 December 2009 (UTC)

Wow, PST, that's a bit of a sweeping statement. I would say that there is a lot of very nice combinatorial mathematics to be found by "arranging numbers in a 9×9 grid." I think that the Mathematics of Sudoku article shows exactly that. In fact this subsection of the main Sudoku article says that "A completed Sudoku grid is a special type of Latin square". From the Latin Square article: "The name Latin square originates from Leonhard Euler, who used Latin characters as symbols." ~~ Dr Dec (Talk) ~~ 19:04, 10 December 2009 (UTC)

I am aware of all that but I would not call someone who can solve sudoku puzzles a mathematician. One could argue that there is a lot of mathematics in chess, but a chess player alone does not constitute a mathematician. Both these traits (chess and sudoku) require (admirable) intelligence but I doubt that they constitute a mathematical intelligence. --PST 02:25, 11 December 2009 (UTC)

But nobody has ever claimed that someone who can solve sudoku puzzles is automatically a "mathematician", so what is it you're so vehemently denying? Of course sudoku puzzles involve mathematical concepts, whether the solver realises it or not. Sheesh, making a cake involves mathematical concepts. -- Jack of Oz ... speak! ... 11:00, 11 December 2009 (UTC)

I never denied anything; I just suggested that perhaps the miscellaneous reference desk would be more appropriate for this sort of question (...since there may be people there who are used to solving sudoku puzzles; that is what "miscellaneous" is for! On the other hand, people here are used to responding to mathematics questions and may not have experience in sudoku (and arranging numbers in a 9x9 grid...)). When Dr. Dec commented on my post (as he usually does to promote his superiority (only joking!)), I responded to justify the reasoning behind my post. Thus, I never really denied anything said by anyone. On a different note, I would not say that making a cake involves mathematics; rather it involves following the instructions given on the packet. ;) --PST 12:50, 11 December 2009 (UTC)

I wasn't pr0moting anything; just pointing out that what you said was nonsense (only joking!) ;) ~~ Dr Dec (Talk) ~~ 16:08, 11 December 2009 (UTC)

Start with the 3x3 block that has the fewest blank spaces, and whose squares' lines have the fewest blank squares (if that makes sense). Comet Tuttle (talk) 22:21, 10 December 2009 (UTC)

What if the board comprises a single completed 3 × 3 block? ~~ Dr Dec (Talk) ~~ 22:26, 10 December 2009 (UTC)

That would not be a valid Sudoku puzzle, because it would have more than one solution. —David Eppstein (talk) 22:30, 10 December 2009 (UTC)

Who says that a Sudoku board need have a unique solution? ~~ Dr Dec (Talk) ~~ 23:00, 10 December 2009 (UTC)

This is a near-universal convention, which, oddly enough, doesn't seem to be mentioned at Sudoku. It's discussed at Mathematics of Sudoku. A puzzle with a non-unique solution would require guessing at some step (or at least some arbitrary choice), which most people regard as lame. (Most people who care at all, that is.) Staecker (talk) 02:33, 11 December 2009 (UTC)

Technically sudoku is not math, it is logic, since one could substitute something else for the numbers, letters, simple animal drawings, other symbols etc... 65.121.141.34 (talk) 16:18, 15 December 2009 (UTC)

Integral estimate

Hello, I'm having a lot of problems while trying to prove that

$\left|\int _{\epsilon }^{1}\left({\frac {1}{\left(\left(z+1\right)^{2}z^{2}+\epsilon ^{2}\right)^{3/2}}}-{\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}\right)dz\right|\leq {\frac {C}{\epsilon }}$

for $\epsilon$ small enought (it seems to be true from my numerical approximations).

Does anybody have any suggestion?--Pokipsy76 (talk) 12:04, 10 December 2009 (UTC)

I'd begin by noting that the integrand is negative, thus it can be negated and the absolute value dropped. Then I'd use the substitution $t={\frac {\epsilon }{z}}$ to make the integral more manageable. The Taylor polynomial of $x^{3/2}$ around $t^{2}+1$ may be useful for bounding the integrand. -- Meni Rosenfeld (talk) 13:53, 10 December 2009 (UTC)

These things usually involve the triangle inequality. It says that |x + y| ≤ |x| + |y|. Replacing y with −y we have |x − y| ≤ |x| + |y|. We know that | ∫ ƒ(x) dx | ≤ ∫ |ƒ(x)| dx. Thus:

$\left|\int _{\epsilon }^{1}{\frac {1}{\left(\left(z+1\right)^{2}z^{2}+\epsilon ^{2}\right)^{3/2}}}-{\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}\ {\text{d}}z\right|\leq \int _{\epsilon }^{1}{\frac {1}{\left|\left(z+1\right)^{2}z^{2}+\epsilon ^{2}\right|^{3/2}}}\ +\ {\frac {1}{\left|z^{2}+\epsilon ^{2}\right|^{3/2}}}\ {\text{d}}z$

Then if you play around with this second expression you might get somewhere. ~~ Dr Dec (Talk) ~~ 22:50, 10 December 2009 (UTC)

Be careful, you must exploit the fact that x is close to y. You'll loose everything writing |x-y|≤|x|+|y| (think to x=1,000,000 and y=1,000,001). Can you see that the RHS you wrote is O(ε^-2)? pma

I think you can argue that ${\frac {1}{(z^{2}+\epsilon ^{2})^{3/2}}}-{\frac {1}{((z+1)^{2}z^{2}+\epsilon ^{2})^{3/2}}}<{\frac {1}{(z^{2})^{3/2}}}-{\frac {1}{((z+1)^{2}z^{2})^{3/2}}}$ since the magnitude of the derivative of x^-3/2 is decreasing. Then you want to show that as z goes to zero, that expression grows no faster than 1/z² which is not so bad once the ε's are gone. Rckrone (talk) 23:46, 10 December 2009 (UTC)

Well, sounds good. Here I'm using more TeX than thought, so it could be possibly made shorter. Let's start with Meni's remark on the sign, and factor out the first term of the integrand

$0\leq \int _{\epsilon }^{1}\left({\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}-{\frac {1}{\left(\left(z+1\right)^{2}z^{2}+\epsilon ^{2}\right)^{3/2}}}\right)dz\leq \int _{\epsilon }^{1}{\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}\left[1-\left({\frac {z^{2}+\epsilon ^{2}}{(z+1)^{2}z^{2}+\epsilon ^{2}}}\right)^{3/2}\right]dz$

To bound the term into square brakets use the convexity of $x^{3/2}:$ for all $\scriptstyle 0\leq \theta \leq 1$ we have $\scriptstyle 1-\theta ^{3/2}\leq {\frac {3}{2}}(1-\theta )$ so we can go on with:

$\leq {\frac {3}{2}}\int _{\epsilon }^{1}{\frac {1}{\left(z^{2}+\epsilon ^{2}\right)^{3/2}}}\,{\frac {z^{3}(z+2)}{(z+1)^{2}z^{2}+\epsilon ^{2}}}dz$

Now change variable: $z=\epsilon x,$ and bound the resulting ugly integral:

$={\frac {3}{2\epsilon }}\int _{1}^{1/\epsilon }\left({\frac {x^{2}}{x^{2}+1}}\right)^{3/2}{\frac {2+\epsilon x}{(1+\epsilon x)^{2}x^{2}+1}}dx\leq {\frac {3}{2\epsilon }}\int _{1}^{\infty }{\frac {3}{x^{2}}}dx={\frac {9}{2\epsilon }}$

(PS: in fact there was no need of changing variable) pma--(talk) 01:02, 11 December 2009 (UTC)