Request for critique of proof
I have started working through the Berkeley Problems in Mathematics for
fun and it would help to have some input on solutions I come up with that
are different from those included in the book (especially since I've just
started working with this text for the first time). One problem requests
that I prove that $\cos^p(\theta)\le\cos(p\theta)$ for
$0\le\theta\le\pi/2$ and $0<p<1$.
For my solution, I let $\theta$ be fixed, thus for all $p$, if not
$\cos^p(\theta)\le\cos( p\theta)$ then we have
$\cos^p(\theta)>\cos(p\theta)$. If we let $\theta=0$, then we have
$1>1^{(1/p)}$, that is, $1>1$. This is clearly a contradiction and thus
the original inequality holds.
 
No comments:
Post a Comment