Wikipedia:Reference desk/Archives/Mathematics/2010 June 23

= June 23 =

hint needed for apparently difficult problem
Hello all... I have a problem which I have been grappling with for some time. Let b be a positive integer and consider the equation z = x + y + b where x,y,z are variables. Suppose the integers {1,2,...4b+5} are partitioned in two classes arbitrarily. I wish to show that at least one of the classes always contains a solution to the equation.

I have tried using induction on b. The case b = 1 has been solved entirely by me. But I cannot understand how to use the induction hypothesis to prove the result. The more I think of it, the more I feel that a different approach to the problem is needed, but I cant figure out what. It is sort of a special case of a research problem, which has been solved in a more general way. I have little experience of doing research on my own, and so will be glad if anyone can offer me any advice or hints. Thanks - Shahab (talk) 07:06, 23 June 2010 (UTC)
 * Do x, y and z need to be different? -- Meni Rosenfeld (talk) 07:26, 23 June 2010 (UTC)
 * No.-Shahab (talk) 07:34, 23 June 2010 (UTC)


 * It looks like a pigeonhole principle problem to me but I haven't figured it out yet. Dmcq (talk) 08:33, 23 June 2010 (UTC)
 * The other thing I can see is that if you consider x+x+b you tend to need to have the 0 mod 3 and 1 mod 3 ones into separate sets and then the 2 mod 3 ones have to be stuck in and cause problems. Dmcq (talk) 09:48, 23 June 2010 (UTC)


 * Here is an outline of a solution. Let's call the two partitions A and B and let's try to split the integers from 1 to 4b+5 between A and B so that neither A nor B contains a solution to z=x+y+b.
 * Let's put 1 in A. Then if b+2 is also in A we have x=1, y=1 and z=x+y+b all in A. So we must put b+2 in B.
 * Now, if 3b+4 is also in B, we have x=b+2, y=b+2 and z=x+y+b all in B. So we must put 3b+4 in A.
 * Now we have 1 and 3b+4 both in A. If 4b+5 is in A then we have ... what ? Or if 2b+3 is in A then we have ... what ? And if we put 2b+3 and 4b+5 both in B, along with b+2 which is already in B, then we have ... what ?
 * (There may be a more elegant solution that uses induction on b and/or the pigeonhole principle, but I can't see it.) Gandalf61 (talk) 14:45, 24 June 2010 (UTC)
 * That seems plenty elegant to me, thanks, I wish I'd spent a bit more time on it and got something like that. Dmcq (talk) 21:22, 24 June 2010 (UTC)

dy
In one of my textbooks it says $$\int dy=[function]dx$$ What is meant by the $$\int dy$$ —Preceding unsigned comment added by 76.230.251.114 (talk • contribs) 17:01, 23 June 2010
 * Generally the notation $$\int dy$$ means the same thing as $$\int 1 \cdot dy$$. --Trovatore (talk) 17:50, 23 June 2010 (UTC)
 * I bet your textbook says $$\int dy=\int f(x)dx$$ meaning $$y=\int f(x)dx$$. Bo Jacoby (talk) 19:33, 23 June 2010 (UTC).

Solving equations like this
How do you solve equations like this for y:

$$\frac{dy}{dx}=xy$$

Where the derivative of y with respect to x is equated to some function of y and x? —Preceding unsigned comment added by 203.22.23.9 (talk) 22:55, 23 June 2010 (UTC)
 * See separation of variables. --Tango (talk) 23:00, 23 June 2010 (UTC)


 * I'm not sure how rigorous this is, but we can multiply both sides by dx and then divide both sides by y. This gives a dx on one side and a dy on the other side, so we integrate to give
 * $$ \int\frac{1}{y}\,\mbox{d}y = \int\! x\,\mbox{d}x $$
 * Evaluating these two indefinite integrals gives ln(y) = ½x2 + c where c is the constant of integration. We can solve for y by taking the exponentiation of both sides to the base e. Writing k for ec to simplify things we have
 * $$ y = e^{\frac{1}{2}x^2+c} = ke^{\frac{1}{2}x^2} . $$
 * We can check this solution, and show that it is right:
 * $$ \frac{\mbox{d}y}{\mbox{d}x} = kxe^{\frac{1}{2}x^2}, \ \ xy = x\left(ke^{\frac{1}{2}x^2}\right) = kxe^{\frac{1}{2}x^2} . $$ •• Fly by Night (talk) 13:48, 24 June 2010 (UTC)


 * From what I remember, the method that is normally taught is the integrating factor method. Rewrite the equation as
 * $$\frac{\mbox{d}y}{\mbox{d}x}-xy=0$$
 * Now, our integrating factor is $$e^{\int-x\,dx} = e^{-\frac{1}{2}x^2}$$. Multiply through and we get
 * $$e^{-\frac{1}{2}x^2}\frac{\mbox{d}y}{\mbox{d}x}-e^{-\frac{1}{2}x^2}xy=0$$
 * This is actually the derivative of a product. Simplifying, we get
 * $$\frac{\mbox{d}}{\mbox{d}x}\left(ye^{-\frac{1}{2}x^2}\right) = 0$$
 * This can be easily integrated and rearranged to get
 * $$y = ce^{\frac{1}{2}x^2},\,c\in\mathbb{R}$$
 * Readro (talk) 14:16, 24 June 2010 (UTC)

See also separation of variables. 75.57.243.88 (talk) 21:45, 24 June 2010 (UTC)
 * It's not "also", because Tango already mentioned it. -- Meni Rosenfeld (talk) 15:02, 25 June 2010 (UTC)