Yes, sorry, I meant that there was no general solution. By closed form I meant a formula for f(n, k) that does not rely on recursion.
Neither statement though, as you rightly point out, is actually true.
It does seem that based on your generalized Pascal's triangle, one can find the...
I'm not sure what you mean by 'embedded', Stephen.
Your way of looking at the problem suggests that a two-parameter problem like that cannot be solved.
With linear recurrences you have a certain number of initial 'seed values' (like 0, 1 with Fibonacci). You can represent the seed value...
Hi,
I am trying to understand the binomial theorem, and would appreciate any insight or pointers.
To make notation simpler I'll call the binomial coefficient f(n,k).
I understand the combinatorial argument that f(n,k) = f(n-1, k-1) + f(n-1, k). This is, to my understanding, a two...
Given two nxn matrices A, B,
The eigenvalues of AB = the eigenvalues of BA.
Essentially I had proved that the trace and determinant of both matrices was the same, and if this lemma were true that's all I would need.
Oh wow, good catch Citan! (but in fact negative integers are allowable, so even Aleph's counterexample works).
Thanks guys. I guess it's back to the drawing board with the proof. :(
Hey,
I'm trying to prove a larger theorem; in order to complete my proof I need to use the following lemma (or, if it turns out to be false, try a completely different method of proof):
Consider any two sets of n nonzero integers, A and B. If their respective sums and products are equal...