Wikipedia:Reference desk/Archives/Mathematics/2009 April 14

= April 14 =

Puzzle
I am not a math professor, or expert but was posed with this question and cannot get it out of my mind. What multi-digit number is such that when the last digit is placed at the beginning, the number is doubled? [i.e. - abcd=1/2 (dabc)] There may be an obvious answer but I have tried it for a while and have come up with nothing yet.
 * I have come up with some parameters so far. If these are true or false could someone tell me either way and explain.
 * the second to last digit is 0 - since the only single digit that, when multiplied by 2 equals itself is 0, the second to last digit must be 0 due to the fact that it will end up being the last digit when the transfer is made.
 * The last digit is 5 - in order for the second to last digit to be 0 the last one must be 5 since it is the only number that, when multiplied by 2, will give you a number with a 0 in it.
 * In the original number, the first two digits must be 25 - if the last digit is 5 than the first two of the original number must be something that will give you 1/2 of the first digit in the final number(5), thus 25.

I understand that all of these premises are based off of the integrity of the first one. If the first one is wrong than I am really lost!!:):) Any help is appreciated.jondn (talk) 01:33, 14 April 2009 (UTC)


 * There are no 4-digit solutions to that problem other than 0000 (I used a computer to check). Also, your reasoning makes no sense. 207.241.239.70 (talk) 02:54, 14 April 2009 (UTC)
 * Too blunt a comment was posted above - ignore it. Note that the value of 'abcd' is 1000a + 100b + 10c + d, and the value of 'dabc' is 1000d + 100a + 10b + c. Therefore, 2a = d, 2b = a, 2c = b, and 2d = c, because twice 'abcd' is 'dabc'. Therefore, 8c = d (to see why, combine the first three equations) and 2d = c. In particular, we conclude that 8c is both equal to d and 16d. Therefore, we see that c = 0 and d = 0. From the initial equations, we conclude that a = 0 and b = 0. This demonstrates that the number is 0000, as desired. P.S. I must emphasize that this is not the type of problem that math professors solve, contrary to what you suggest. -- PS T  03:35, 14 April 2009 (UTC)


 * I don't think 2a=d, 2b=a, etc. necessarily hold, because there can be carries from one place to another. Note that if you don't mind ignoring a remainder, 1052 = 1/2 of 2105. 207.241.239.70 (talk) 04:48, 14 April 2009 (UTC)


 * A number that can be multiplied by moving the last digit to the front is called a parasitic number. The Wikipedia article on parasitic numbers gives 105263157894736842 as a solution to your puzzle and provides a general method for finding an n-parasitic number for any n.  The example I've given here shows that your first premise (that the next-to-last digit must be 0) is false, since the next-to-last digit here is 4.  In fact, all of the premises are false, but as you say, the truth of the second and third premises depends on that of the first.  By the way, I wouldn't call this solution obvious, but there is a way of thinking about the problem that leads straight to this solution. Michael Slone (talk) 04:03, 14 April 2009 (UTC)


 * There have been two recent New York Times articles on this puzzle:
 * Freeman Dyson’s 4th-Grade Math Puzzle
 * Prize for Dyson Puzzle
 * The second explains the solution. Jim (talk) 04:36, 14 April 2009 (UTC)
 * Btw, if you need to recall the nubmer 105263157894736842 (to show your ability in multiplications), note that the digit $$x_{n+2}$$ is $$\scriptstyle\left\lfloor\frac{10(x_n\mod 2)+\,x_{n+1}}{2}\right\rfloor$$. --pma (talk) 06:55, 14 April 2009 (UTC)

Thanks a lot for the help!! I was not necessarily implying that math professors do these problems rather was just saying that I am in no way an expert but the problem was still driving me crazy!! Thanks againjondn (talk) 14:02, 14 April 2009 (UTC)

What is $$\int \sqrt{1-\frac{dt^2}{dr^2}} dr$$?
What is $$\int \sqrt{1-\frac{dt^2}{dr^2}} dr$$? This is not homework. I just got stuck in integrating this by hand and Wolfram Mathematica's Integrator thought that the "d"s were constants (except for the last one, obviously!)! Please help me! The Successor of Physics  15:09, 14 April 2009 (UTC)


 * As a first remark, your $$\frac{dt^2}{dr^2}$$ looks a bit like a weird thing. Either you mean $$\frac{d^2 t}{dr^2}$$, which is a notation for the second derivative of a function t(r) wrto the variable r; or you mean $$\Big(\frac{dt}{dr}\Big)^2$$, which is a notation for the square of the first derivative of t(r). In both cases there is no general identity that allows a reduction to a simpler formula: you have to keep the integral as it is. Moreover, even if there were such an identity, that integrator will not give it to you, for it only contains a data-base of antiderivatives of explicit functions. Using it to look for the integral of "f(x)", without specifying what's "f(x)", is like to send a letter addressed "to my grandfather": it will not be able to understand what you want. If you specify what is the unknown function t(r), for instance if you choose t(r)=cos(r) or t(r)=r2, then you may look for an antiderivative of your expression in terms of elementary functions, and in this case that integrator may help.  --pma (talk) 16:06, 14 April 2009 (UTC)


 * This integral gives you the spacetime interval along a path in 1+1 dimensional spacetime. As pma said, you can't integrate it unless you have a relation between t and r. If you have a function t(r) (which would be rather unusual) then the integral is $$\int \sqrt{1-t'(r)^2} \; dr$$. If you have a function r(t) (more likely) then it's $$\int \sqrt{1-\frac{1}{r'(t)^2}} \; r'(t) \, dt$$. If you have functions t(q) and r(q), where q is an arbitrary parameter, then the integral is $$\int \sqrt{1-\frac{t'(q)^2}{r'(q)^2}} \; r'(q) \, dq$$. If you don't care about the sign of r', which is the usual case, then you can bring it inside the square root and simplify the last two integrals to $$\int \sqrt{r'(t)^2-1} \; dt$$ and $$\int \sqrt{r'(q)^2-t'(q)^2} \; dq$$. -- BenRG (talk) 23:05, 14 April 2009 (UTC)


 * Well, I meant $$(\frac{dt}{dr})^2$$. Also, $$dr^2 = ds^2 + dt^2$$(I know the answer is s after the substitution but I don't wan't it to be done). Does that help? The Successor of Physics  14:12, 15 April 2009 (UTC)
 * The answer will depend on the limits of integration. I don't think any simple limits will give an answer of s (assuming s is a constant). Are you sure you don't mean $$dr^2 = ds^2 + dt^2$$? -- BenRG (talk) 18:38, 15 April 2009 (UTC)
 * Yeah, thats what I meant! I edited it! Thanks! The Successor of Physics  11:26, 18 April 2009 (UTC)
 * In that case try eliminating dt from the integral by substituting $$dt^2 = dr^2 - ds^2$$. -- BenRG (talk) 11:52, 19 April 2009 (UTC)

Repeated Root
Following on from an earlier question, how does one describe the behaviour of $$y= \sin x - 1$$ at $$x=\frac{\pi}{2}$$? Is this a repeated root? If not what is it? 92.0.38.75 (talk) 17:38, 14 April 2009 (UTC)


 * Yes, it is a repeated root. See Multiplicity_of_a_root. Both the function and its derivative are zero at that point, but its second derivative is nonzero, so that point is a root with multiplicity 2. Black Carrot (talk) 19:22, 14 April 2009 (UTC)


 * Also, we can define multiplicity of zeros via power series expansions. If $$\scriptstyle y(x)=\sum_k c_k(x-x_0)^k$$ and n is the index of the least non-zero term, we say that n is the multiplicity of $$x_0$$ as a zero of y(x) (in this case it's 2). So we say that the multiplicity of $$x_0$$ is zero if $$x_0$$ is not a zero ;-)
 * (Notice also that the order of a pole is defined analogously. Indeed, if you allow any integer value, the two concepts are unified: one can then say that the multiplicity as a zero is minus the order as a pole. Of course it should sound a bit strange to speak of negative multiplicities, but it is only a matter of language, while in the substance it's very natural). --pma (talk) 19:54, 14 April 2009 (UTC)