@BalarkaSen if I have a cubic $ax^3+bx^2+cx+d$ with positive solutions and I create a cuboid with sides the roots of the cubic. What is the Edge length, surface area and volume of my cuboid?
@TheGame Nope, these days I only come to chat. I come to chat for conversation, emotional support and advice. No more answering questions for me, for ever!
@TheGame Let $\epsilon_i$ the $i$th row of the identity matrix. Then $$(\epsilon_1,\ldots,\epsilon_{n-1},0) + (0,\ldots,0,\epsilon_n) = I$$ and $I$ is invertible so $I\not\in E$
One last thing, 'Let E be a subspace of Mn(ℝ) that contains no invertible matrix.' it's ambiguous, because there can be several spaces that contain non invertible matrices
@DanielFischer Can you help me with something :) .. I was trying to use perron summation formula to estimate $\sum\limits_{n \le x} |\mu(n)|$ and I was stuck at trying to estimate $\displaystyle\left|\frac{1}{2\pi i}\int\limits_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)x^s}{\zeta(2s)s}\,ds\right|$ as $\mathcal{O}(\textrm{something with $x^{\sigma}$ and $T$})$ (for $\sigma \in (\frac{1}{2},1)$) .. (the vertical integral after shifting the line integral to a $\sigma$)
@r9m Urk. I don't see how to estimate that integral. You presumably know that $$\sum_{n\leqslant x} \lvert \mu(n)\rvert \sim \frac{6}{\pi^2}x,$$ so do you want something better from the integral?
Hey, The Game. Do you remember the name of the dude who was in here yesterday helping us ou tw/ some stuff?
He suggested I preform some calculation because it'd be enlightening, and I neglected to write it down, but I have the time to do so now and don't remember what it was. :l
@r9m Maybe there is something known about the values of $\zeta$ that allows to get a good estimate, but I don't know much about $\zeta$. Perhaps Balarka knows something?
@r9m Oh, whoa, those. Complex zeta contour integral estimations never get old do they? ;)
OK, so wait let me write this up. You need an estimation for $$\int_{c-iT}^{c+iT} \frac{\zeta(s)}{\zeta(2s)} \frac{x^s}{s} ds$$
Let's think about this one instead $$\int_{c-iT}^{c+iT} \frac{\zeta(s+w)}{\zeta(2(s+w))} \frac{x^w}{w} dw$$
The integrand blows up at $w = 0$, so leaves out $\zeta(s)/\zeta(2s)$. Make a dent $[c-iT, -\delta -iT] \cup [-\delta-iT, -\delta+iT] \cup [-\delta+iT, c+iT]$ for some $\delta > 0$ such that $\zeta(2(s+w))$ has no zeros in $\Re[2w] > -\delta$ and the imaginary part bounded by $T + \text{blah}$ or some such.
@r9m OK, I think knowing that $\sigma > 1 - A\log^{-9}(t)$ is $\zeta(s)$ zero-free for some constant $A$ should do. Sub in $\delta = A\log^{-9}(t)$. The rest looks very tedious estimations.
@BalarkaSen I need estimation for the vertical integral emerging form shifting the line integral to the region $(1/2,1)$ .. (I have an idea for the estimates of horizontal integrals) ..
@BalarkaSen just $\displaystyle\left|\frac{1}{2\pi i}\int\limits_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)x^s}{\zeta(2s)s}\,ds\right|$ for $\sigma \in (0.5,1)$
I want it in terms of $\mathcal{O}$(something to do with $x^{\sigma}$ and $T$) ..
btw can you give me some link or book .. where I can find application of perron summation formula to estimate average of arithmetic functions ? @Balarka
@r9m Titchmarsh never explicitly proves PNT that way but does mention Perron and gives some examples. If you combine $\sum \mu(n)/n$ estimates with the number theoretic form of pnt ($M(x) = o(x)$) you'd have proved it anyway..
Well, I'm pleased, nevertheless, @Kaj. Your diff geo homeworks were pretty consistent. I just wished you'd had time to do the more interesting problems -- which were deserving of your attention.
@TheGame: I suspect that even in the rank 1 case, the dimension must be less. But surely we can show that if the dimension is too big, we'll have a matrix of rank $\ge 2$ in there.
@robjohn and everyone else: We all know that you can make a cone by gluing together a pacman figure (sector of a circle). If we slice said cone with a plane and get an ellipse, what does that look like in the pacman figure?
I have lots of students trying to do the actuarial certificate. The guy in Terry said he'd accept 4600 in lieu of STAT 4510, @Kaj. Some of them are actually quite good, but most aren't.
@robjohn we can say that $1<\zeta(k)<2$ for $k\ge 2$ therefore we have $n-1<S<2n-2$ where S is that sum. I can't get that upper bound tighter so that I can show it is inbetween 2 integers.
@TheGame For all but exceptional cases, you can determine the bottom right element so that the matrix is non-invertible. Thus, you have $n^2-1$ degrees of freedom in choosing the elements.
@robjohn I gtg to sleep now, it's 00:40 here. i've already tried thinking for a day or two at least on the question but haven't succeeded, if you find something let me know !
@PedroTamaroff I'm confus. Consider the Riemann surface of $w^3 = z^2 - 1$ I'm getting the branch cuts at $[-2, 2]$ and $[-i \infty, i \infty]$. This is apparently incorrect...
Hint:
Use the definition
$$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$ and then apply L'Hospital rule
