You get a differentiable function by integrating a continuous one, so the extreme points would have to be at zeroes of that integrand, and it has none, unless all the $a_i$ are the same.
ok, @Hippa, got it. Thanks. Maybe I'll steal some later.
@TedShifrin When someone has a surface $S$ in $\bf R^3$ and a vector field there, "the flux of $F$ through $S$" always means $$\int_S F\cdot \vec ndS$$ right?
My students who don't do homework are failing my course. And I have told some 5 times that they need to come get help from me; they refuse. They're going to fail the course. Sigh.
So @Pedro @Studentmath: An interesting thing happened with one of the probability homeworks that two of my best students were working on. They were given a problem to find a probability that something happens in $n$ turns. They were supposed to conjecture the formula, then give a recursion formula and prove it.
They cheated and came up with the recursion formula $$f(n) = f(n-1)+\frac1{4n^2-1}.$$ The honest way gives the recursion formula $$f(n) = \frac{(2n-1)f(n-1)+1}{2n+1}.$$ Would you buy this?
No, @Studentmath, assuming their conjecture was right led them to the first. Probability theory leads one to the second. However, both turn out to be correct. Looking at the difference equations, one would say that can't happen. @Balarka, too
They cheated and came up with the recursion formula $$f(n) = f(n-1)+\frac1{4n^2-1}.$$ The honest way gives the recursion formula $$f(n) = \frac{(2n-1)f(n-1)+1}{2n+1}.$$ Would you buy this?
They're going well. At the last minute, I changed my mind about going to Virginia Tech and took my offer to go to University of Virginia instead to study CS.
@Ted the latter formula gives a statement about the probability, the first might give the right results - which is, if what asked, just fine - but I can't get its statement about the probability.
Since $G$ has no element of order $2$, the $C_2$ part is trivial, and since the elements of order $3$ are in $C_9=H$, the map is trivial, @DanielFischer
Then we're going to look at $\eta :G\to {\rm GL}(2,3)$.
@DanielFischer If $f(x) = \sum{c_n e^{2\pi i nx}}$, I can multiply through by $e^{2\pi i m x}$, integrate, interchange orders of summation and integration, and use orthogonality to kill off all but $c_m$, etc., the usual thing. But that's all formal. When is that actually justified? There are smoothness requirements on $f(x)$ or something, right?
Possibly. It's not hard to enumerate but I am too lazy to do it atm =P
@AbstractionOfMe On the other hand, if the two binary operators satisfy some fundamental relations while acting on the same underlying set X, they can be interesting.
$S=4x^5-4x+1=2(2x^5-2x)+1$ is odd, so if it is a perfect square its square root must be odd.
