## Could we always find a curve on the manifold whose tangent vector always belongs to a linear subspace?

Suppose we have a smooth manifold $$M$$ and the tangent space of every point $$x \in M$$ has non-empty intersection with a given linear subspace. Could we find a curve on $$M$$ such that the tangent vector of point on thus curve always belongs to this linear subspace?

$$\textbf{My attempt:}$$ If the dimension for the linear subspace is one or two, I think I can find the curve. But I don’t know if the dimension is bigger than two?

I will appreciate for any useful answers and comments

## If $f$ belongs to $M^{+}$ and $c \ge 0$ then $cf$ belongs to $M^{+}$ and $\int cf = c\int f$

If $$f$$ belongs to $$M^{+}$$ and $$c \ge 0$$ then $$cf$$ belongs to $$M^{+}$$ and $$\int cf = c\int f$$.

I need to proove that, using the following observation:

if $$f\in M^{+}$$ and $$c>0$$, then the mapping $$\varphi \rightarrow \psi = c\varphi$$ is a one-toone mapping between simple function $$\varphi \in M^{+}$$ with $$\varphi \le f$$ and simple functions $$\varphi$$ in $$M^{+}$$ with $$\psi \le cf$$.

I know that this question is already answer here:One-to-one mapping of simple functions $\phi \to \psi = c\,\phi$ implies $\int cf\,d\mu = c \int f\,d\mu$ ?

But I can’t follow the verbal explanation.

My original idea was to proove $$c \int f \le \int cf \le c\int f$$ But I can’t… some idea?

## Is the language { | p and n are natural numbers and there’s no prime number in [p,p+n]} belongs to NP class?

I was wondering if the following language belongs to NP class and if its complimentary belongs to NP class:

\begin{align} C=\left\{\langle p,n\rangle\mid\right.&\ \left. p \text{ and n are natural numbers}\right.\ &\left.\text{ and there’s no prime number in the range}\left[p,p+n\right]\right\} \end{align}

I am not sure, but here’s what I think: for each word $$\langle p,n\rangle \in C$$ we know that the word belongs to C because there exists a primal certificate – an nontrivial divisor to any of the numbers between $$[p,p+n]$$, though I am not really sure it is in NP.

regarding the complement: I think it is in NP because the compliment compositeness can be decided by guessing a factor nondeterministically. But again I am not so sure about it and I don’t know how to correctly prove and show it.

Would really appreciate your input on that as I am quite unsure and also checked textbooks and internet (and this site) about it.

## Where f(n) = n! belongs to? P, co-P, NPComplete or NPHard? [duplicate]

This question already has an answer here:

• What is the definition of $P$ , $NP$ , $NP$ -complete and $NP$ -hard? 6 answers
• What is the difference between an algorithm, a language and a problem? 1 answer

Where f(n) = n! belongs to? P, co-P, NPComplete or NPHard?

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## How can I identify the cache if I see its name? One belongs to malware

If Device maintenance app is launched in Samsung S8, after pressing Clean Now button, storage cleanup briefly shows different apps, storage space of which is being cleaned:

System cache OneNote TopBuzz

I know that the last one is a malware and is not present in my list of apps, but how can I find out to which app it actually belongs? No antivirus which I tried so far reported this as a malware. But maybe there is some system analyzer which shows similar cache ownership like the above app, but with more details?

(Or maybe there is another method how to find origin of that TopBuzz symptom.)

The phone is not rooted.

## Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of $x$,$x^*$ belongs to $U$.

I have been stuck with this problem for a while now. I have a proof that letting $$U$$ be an ultrafilter, exactly one of $$x,x^*$$ belongs to $$U$$ for all $$x$$ in $$B$$, I did this by showing that both belong to $$U$$ implies that $$U=B$$ which cannot be the case, and if neither long to $$U$$ then a contradiction can be derived by de-Morgan’s laws. However, I’m stuck with the reverse implication, I need to prove that:

“If $$U$$ is a subset of a Boolean algebra $$B$$ such that $$\forall x \in B$$, exactly one of $$x \in U$$ and $$x^* \in U$$ is true.

Is this even the case? Any help would be greatly appreciated, thanks in advance.

## How can I check if an ID belongs to a user or a group?

I’ve got a SharePoint list with a field where authorized users and/or groups are listed. I know I can use CanCurrentUserViewMembership to check if the user belongs to a specific group, but if the ID I pass belongs to a user I get an Internal Server Error.

Is there a way to check whether the ID belongs to a person or group before making this call? Taking into consideration that I’m doing this on a loop, so a lot of calls are being made.

## MacOS Contacts App — How can I tell to which address book an entry belongs?

I have two entries that appear identical to each other. One is in the iCloud address book. The other is On My Mac.

I have opened up both entries, but I can’t tell which is which.

Is there a way to discern this?