I think this involves an extra symmetry of the triangular region that can not be reflected in the symmetries of intervals, and so a proof by index shifts is not possible
I just don't want to fiddle with showing that the map $\langle j,k\rangle\mapsto\langle j-k,k\rangle$ is bijective on the triangular region, cause that's a pain
but I might be able to set up the sums with index shifts so that it ends up as $\sum_R=\sum_{R'}$ where $R$ and $R'$ are seen to be equal
Suppose I threw in a rule that says $$\sum_{j = 0}^n\sum_{k = 0}^{n - j} a_{j,k}=\sum_{k= 0}^n\sum_{j = 0}^{n - k} a_{j,k}.$$ Would that, plus index shifts, get us to the end?
That's more or less what arguing from a 2D sum will get you
If both $m$ and $n$ are odd, then $m+n$ is even and $mn$ is odd.
First we need to prove that $m+n$ is even if $m$ and $n$ are odd. By definition 2.3.1, there exists some $k \in Z$ and $l \in Z$ such that $n=2k+1$ and $ m = 2l+1$
can't wait to master this proof stuff thing one day... at least I'm reading and trying so I could get better at it.. I could choose to give up but that ain't happening home boy ^^
yay... I was trying to aim for the 2kl+1 or at least some manipulation that could get me to 2kl+1
because by def if $k \in Z$ then n is odd such that $n=2k+1$ but since I have another variable, l, then wouldn't my goal would be to use manipulation to achieve $2kl+1$?
@PedroTamaroff I guess the most concise way to prove that is to note that only one term from each $g$ and $h$ contributes to the minimal/maximal degree term of $gh$.
Ok... I need advice on close-voting... This question, math.stackexchange.com/questions/771930/…, should be on Physics.SE (not a math question at all). But, Physics.SE doesn't like hw questions, so the migration probably wouldn't go through. Does this fall under the "don't migrate junk" rule, or should I just vote to close as off topic anyway?
(or, should I just answer it, because I know how?)
how was I stuck ? hmm because i wasn't sure if that was it after I found the 2k+1 because we are manipulating to have it match the definition of $n = 2k+1$ but I figure that wasn't enough so I had to factor the 2 out
yay why is it that some proofs are easier to solve than others
?
because well I thought my question from last night was good, but apparently I flopped....bad....... http://math.stackexchange.com/questions/771119/prove-that-if-t-in-t-and-q-in-q-but-q-neq-0-then-qt-in-t-where-t
righhhhhhhttttttt since $q$ is rational... $t$ is transcendental but taking the product of $q$ and $t$ then it's going to be transcendental... like hmmm if we let q = 4 and t = $\pi$
I prefer office hours actually but sometimes it's not enough because there's a lot of students and sometimes you don't even get a chance to talk to the professor all week
@AlexanderGruber, pardon, will you please let me know how to use it here?I think ab + bc + ca is objective function, and $a^2 + b^2 + c^2 = 1$ is constraint, right?
I think Lagrangian is way better! But am stuck! I found out as @FernandoMartin told that max is 1 and min is -1/2. But How that relates to this answer?@Mike
