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

Can someone please explain this to me:

I have the Ramsey function $R(m,n)\backslash le\; R(m-1,n)+R(m,n-1)$. I want to show that $R(m,n)\backslash le\; \backslash binom\{m+n-2\}\{m-1\}$. 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)

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)