I think that sign should be = \sum_{k=0}^n a_n x^n \ $where$ \ a_k \in \mathbb{R} \ $and$ \ a_k > 0 \ $for$ \ k=0,1,2, \dots, n \\ \\ $Show that for any complex root of$ \ P \ $, that is$ \ \alpha \ $then$ \ |$arg$ \ \alpha| \leq \frac{\pi}{n} )
Yep my badI think that sign should be.
This is true because for fixedwith
, each
is nonzero and has the same sign (since
).
So
We have shown that P does not vanish in the sector, which is equivalent to the claim. (It is also trivial that P(0) is nonzero).
Nice proof by contradiction. and the conclusion can be a tiny bit stronger: the inequality sign can be strictly greater than. lolI think that sign should be
This is true because for fixed
So
We have shown that P does not vanish in the sector, which is equivalent to the claim. (It is also trivial that P(0) is nonzero).
I already posted a HSC method for it, and I know of one other HSC method for it too.
I'm pretty sure A has to be 1 so that we don't get a singularity at x=1.
 = \frac{\ln (Ax)}{x-1} \ $for some constant$ \ A > 0 )
Yes, that's what I got anyway.Oh, right, so it would be:
 = \displaystyle \begin{cases}\frac{\ln x}{x-1} , \ x > 0 | x \neq 1 \\ 1, x=1 \end{cases} )