let $I_{p,q}=\int_{\gamma}p(z)\overline{q(z)}dz$ where $\gamma$ is the closed contour $\gamma(t)=e^{it},0\leq t \leq 2\pi.$ Then

Consider the polynomial $ p(z),q(z)$ in the complex variable $ z$ and let $ I_{p,q}=\int_{\gamma}p(z)\overline{q(z)}dz$ where $ \gamma$ is the closed contour $ \gamma(t)=e^{it},0\leq t \leq 2\pi.$ Then

  1. $ I_{z^m,z^n}=0,\forall m,n\in \mathbb Z^+,m\neq n$

  2. $ I_{z^n,z^n}=2\pi i,\forall n\in \mathbb Z^+$

  3. $ I_{p(z),1}=0,\forall $ polynomials $ p$

  4. $ I_{p,q}=p(0)\overline{q(0)},\forall $ polynomials $ p, q$

My Attempt

When I flashed through the options, I got (3) as the answer. Since Polynomial function is analytic. By Cauchy’s theorem, $ \int_{\gamma}p(z)dz=0$ . option (1) is wrong. Since, If $ m-n+1=0$ , I got $ I_{z^m,z^n}=2\pi i$ . For option (2), $ I_{z^n,z^n}=\int_{\gamma}z^n\overline{z^n}dz=\int_0^2\pi 1.e^{it}i dt=2\pi i.$ Using option (2), We can deduce that (4) is a wrong answer. But the answer given in the answer key is only (3). Can you please help me?

Why does Apple make it hard to run laptops with lid closed?

No, this is not a duplicate to Is there any way to set a MacBook Pro to not sleep when you close the lid? . (The answer to that question is that there are 3rd party applications that may or may not work, or you can make it work if you have an external display/keyboard attached.)

But what I’m asking here is does anyone know WHY there isn’t a simple system setting to control what the machine does with the lid closed. Has Apple ever offered any rationale for (what seems to me) unnecessarily crippling their machines this way?

Some people have commented that it’s dangerous to run with the lid close due to heat buid-up. That seems far-fetched. There are cooling slots on the bottom of my MacBook Pro, and it just seems silly that the hardware would be designed that way. Plus, you can run with the lid closed if you have an external monitor and keyboard connected. So what is it? I mean, it can’t be that Apple engineers couldn’t figure this out, it seems like they’ve gone to extra trouble to prevent it. Are they trying to stop people from running servers on Apple laptops?

Closed manifold with non-vanishing homotopy groups and vanishing homology groups

Is there a closed connected $ n$ -dimensional topological manifold $ M$ ($ n\geq 2$ ) such that $ \pi_i(M)\neq 0$ for $ i>0$ and $ H_i(M, \mathbb{Z})=0$ for $ i\neq 0$ , $ n$ ? The manifold $ S^1\times S^2$ satisfies the first requirement but not the second (generally, the direct product of two positive-dimensional manifolds does not seem to satisfy the second requirement because of Kunneth).

How to prove regular languages are closed by some operations?

I don’t how to prove these.

Show that the regular languages are closed under the following operations:

(a) $ $ \mathbb{DROPOUT}(L) = \{ xz \mid xyz \in L \text{ where }x,z \in \Sigma^*, y \in \Sigma \}. $ $ Namely, $ \mathbb{DROPOUT}(L)$ is the language containing all strings that can be obtained by removing one symbol from a string in $ L$ . For example, if $ L = \{012\}$ , then $ \mathbb{DROPOUT}(L) = \{12, 02, 01\}$ .

(b) $ $ \mathbb{INIT}(L) = \{ w\in \Sigma^+ \mid \text{ for some } x\in\Sigma^*,\ wx \in L\}. $ $ For example, if $ L = \{01, 110\}$ , then $ \mathbb{INIT}(L) = \{0, 01, 1, 11, 110\}$ . (HINT: Start with a DFA $ A$ for $ L$ and describe how to construct an FA for $ \mathbb{INIT}(L)$ using $ A$ . We assume that $ A$ has no sink states.)