Very simple and elegant solution, though going thru determinents was unnecessary.

@angelmendez-rivera351

2 жыл бұрын

Not at all. The audience may be unfamiliar with the concept.

I'm sure you could do this with the calculus of finite differences, since P(x)Q(X+1)+Q(X)P(X) is just delta PQ, so deltaPQ-2PQ =P(X)Q(X+1)-Q(X)P(X) =1. DeltaPQ-2PQ=1 should be easy to solve. 1+2PQ=DeltaPQ SUM(1+2P(x)Q(x))=Sum(Delta(PQ)) PQ=SUM(1+2PQ) -> PQ= X + 2SUM(PQ) 1/2(PQ-X)=SUM(PQ) You can then use umbral calculus to solve SUM(PQ) by converting the polynomial into a falling polynomial and back.

I have to admit I have confusion about many aspects of the solution. Maybe it’s just me.

From LTC🙂

@mohmohanplaylist9476

3 жыл бұрын

Same

That’s rad

I didn't really get it. Why p_m != q_n? What's wrong with this case? They cancel each other but when we take (x+1) into this power it's not that clear what happens. And I don't see where do we use the fact that they aren't equal. If they're equal it just means that t = 1.

@MetaMaths

3 жыл бұрын

We want the degrees of p and q to be m and n. If p_m = q_n then the highest order terms will cancel out (even with (x+1)^n, think about it!)

@Jop_pop

3 жыл бұрын

@@MetaMaths but don't we want those terms to cancel out, leaving only the degree one terms behind?

@jgilferez

3 жыл бұрын

I don't see either how this is used in the argument, but I think it's actually irrelevant. It seems to me that the leading coefficients of p(x) and p(x+1) are the same, p_n. So, the (m+n)-th coefficient of r(x) = p(x)q(x+1) - p(x+1)q(x) is r_{n+m} = p_n q_m - q_m p_n = 0. But, as the video mentions, the (m+n-1)-th coefficient of r(x) is more interesting. If my calculations are correct, this is (assuming that the degrees of p and q are at least 1): r_{n+m-1} = (p_{n-1} q_m + m p_n q_m + p_n q_{m-1}) - (q_{m-1} p_n + n q_m p_n + q_m p_{n-1}) = (m-n) p_n q_m. We are assuming that both m and n are at least 1, hence n+m-1 > 0, and therefore r_{n+m-1} = 0. Since both p_n and q_m are not 0, we conclude that m-n = 0, that is, m = n.

Am so confused. So many details seem to be glossed over or should be so obvious. Am just utterly lost and confused about what is going on after the observation of the determinant and matrix observation.

@MetaMaths

Жыл бұрын

This is Putnam which is not the easiest thing to grasp. Try following the video while writing down some notes

p_m and q_n must be != 0, so that p(x) and q(x) actually have degrees m and n. The coefficient of x^(m+n-1) in R(x) is: p_m(q_(n-1) + n q_n) + p_(m-1) q_n - q_n(p_(m-1) + m p_m) - p_m q_(n-1) which simplifies to p_m q_n(n - m). Since this must be 0, n = m, even for m = n = 1. The subsequent argument is that q'(x) = q(x) - (q_n/p_m) p(x) satisfies p(x)q'(x+1) - p(x+1)q'(x) = 1 and deg q'(x) = 1, this contradicts the previous proof that p(x) and q'(x) must have the same degree. Note that p(x) = ax + b and q(x) = cx + d, with bc - ad = 1, includes solutions where p(x) or q(x) is constant, ie, p(x) = ax + b, q(x) = -1/a; and p(x) = 1/c, q(x) = cx + d.

Leaving a comment for the algorithm.

@MetaMaths

3 жыл бұрын

Algorithm approves !