Wikipedia:Reference desk/Archives/Mathematics/2008 October 1

= October 1 =

Recurrence
Can someone please explain this to me: I have the Ramsey function $$R(m,n)\le R(m-1,n)+R(m,n-1)$$. I want to show that $$R(m,n)\le \binom{m+n-2}{m-1}$$. My book says that use a certain binomial coefficient. Can someone help? Thanks--Shahab (talk) 07:44, 1 October 2008 (UTC)
 * Oh never mind. Pascal's identity was what I was looking for. The proof was by double induction.--Shahab (talk) 08:11, 1 October 2008 (UTC)

algebra
They gave me this equation to use and I was ookay with it for a while. x = a1 b1 + a2 b2 / a1 + a2 but now they expect me to find b2!

I know I should bring it around so that the one side is x/a1 b1 + a2

But what do I do with a1 + a2? Do I put it so that x + (a1 + a2) or should I do it like x(a1 + a2)? --Jeevies (talk) 15:24, 1 October 2008 (UTC)


 * Step by step, apply the same operation to both sides. Adding parentheses and operators as my assumption of what you mean with your equation we have x = ( ( ( a1 * b1 ) + ( a2 * b2 ) ) / a1 ) + a2. Thus your first step would transform to the equation ( x - a2 ) = ( ( ( a1 * b1 ) + ( a2 * b2 ) ) / a1 ) + a2 - a2 = ( ( ( a1 * b1 ) + ( a2 * b2 ) ) / a1 ). Carry on from there. -- SGBailey (talk) 15:39, 1 October 2008 (UTC)


 * From the rest of the question I'm guessing they actually mean
 * x = (a1 b1 + a2 b2) / (a1 + a2)
 * That's the only way I can interpret the first steps they took. If I'm right the first step would be to multiply each side by (a1 + a2). --  Mad031683  (talk) 20:33, 2 October 2008 (UTC)