## Is the language $L = \{(M,m,n)|\exists x \in \{0, 1\}^n:M$ uses $m$ space on input $x$$\}$ decidable?

I have stumbled upon this language: $$L = \{(M,m,n)|\exists x \in \{0, 1\}^n:M$$ uses $$m$$ space on input $$x \}$$. At first, it looked like an undecidable problem, but I have failed to prove it, and now I am beginning to wonder whether it is actually decidable.

I have designed the following algorithm. Let $$G_{M,x}$$ be the configuration graph of $$M$$ on input $$x$$ (each node represents a snapshot of $$M$$, starting from $$M(x)$$). To decide whether on $$x$$ we use $$m$$ space, we visit, DFS-style, each node of $$G_{M,x}$$ starting from the first node. Each node can be computed from the prior node and we can memorize every node we have encountered so far into a data-structure. Now:

• If we encounter a node which takes up at least $$m$$ space, we halt and say yes.
• If $$M$$ halts before reaching size $$m$$ or if we encounter a cycle (i.e. find a node we already visited), we stop and say no.

We apply the upper algorithm for each $$x \in {0, 1}^n$$, looking for at least an $$x$$ on which we say yes.

Does this algorithm work? Why or why not? To me it sounds like it works, but I don’t know how to prove it. I guess we need to prove that this algorithm actually decides the problem and that it always halts.

Informally, I believe a way to prove this would be to say that we only have finitely many snapshots which represent less-than-$$m$$-space configurations: in a finite time, either we encounter some of them more than once (so we enter in a cycle) or we exceed the $$m$$-space limit. Either way, we halt and answer the question.

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## В чём разница между {$x} и${x}?

В php есть минимум 3 способа вставки переменной в строку:

https://ideone.com/sN22RT

$x = 'test'; echo("$  x {$x}$  {x}"); 

Выводится

test test test 

То, что ${x} и {$ x} (в отличие от $x) можно использовать перед буквенноцифровым символом понятно, а чем сами эти два способа отличаются между собой? ## Measure of compact sets$C_{1}$and$C_{2}$given that$m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$for any real number$x$. I’m trying to solve the following question in the context of another problem. Suppose that $$E$$ is a Lebesgue measurable subset of $$\mathbb{R}$$ and let $$C_{1}, C_{2}$$ be compact subsets of $$\mathbb{R}$$ such that $$C_{1} \subseteq E$$, $$C_{2} \subseteq E^{c}$$ and $$m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$$ for any $$x \in \mathbb{R}$$ (where $$m$$ denotes the Lebesgue measure). Then, is it true that either $$m(C_{1})=0$$ or $$m(C_{2})=0$$? I have been thinking about this for a while and I attempted to prove it by assuming that $$m(C_{1}) >0$$ and then trying to conclude that $$m(C_{2})=0$$ via the continuity of the real-valued function that maps every $$x \in \mathbb{R}$$ into $$m(C_{1} \cap (C_{1}+x))$$ and the fact that $$m(C_{1} \cup (C_{2}+x)) = m(C_{2} \cup (C_{1}+x)) = m(C_{1})+m(C_{2})$$ for every $$x \in \mathbb{R}$$ (which is implied by the condition $$m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$$ for any $$x \in \mathbb{R}$$). My intuition tells me that since we can guarantee that $$C_{2}+x$$ will intersect $$C_{1}$$ for an adequate real number $$x$$ and $$m(C_{1}) >0$$, then the fact the equalities $$m(C_{1} \cap (C_{2}+x)) = m(C_{2} \cap (C_{1} + x))=0$$ hold for any $$x \in \mathbb{R}$$ is a very strong condition that will imply that $$m(C_{2})=0$$ but I haven’t really gotten anywhere besides those intuitive thoughts. Maybe the answer to the question is negative and there is a pathological counterxample with the fat Cantor set or something like that? Any hints or ideas would be greatly appreciated. Thanks in advance. Posted on Categories proxies ## Is there any equivalent of Taylor/Maclaurin series of$ln(1+x)$for$|x| > 1$I happened to come across Taylor series and Maclaurin series recently, but everywhere I read about the expansion for $$ln(1+x)$$, it was stated that the approximation is valid for $$-1 < x < 1$$. I understand that the bounds for $$x$$ are because the series doesn’t converge for $$|x|>1$$, but is there any equivalent of this series for the value of $$|x|$$ as greater than $$1$$? Please note that I am not asking if we can compute for $$|x|>1$$ or not, as that can be done by computing for $$\frac 1x$$, which will then lie between $$-1$$ and $$1$$. Also, I’m quite new to all this, so new that today was the day I read the name ‘Maclaurin’ for the first time. So any answers understandable with high school mathematics are highly appreciated. Thanks! Posted on Categories proxies ##$M \leq X$. Is it true that$M^*$is a subspace of$X^*$where$*$denotes the dual space? Let $$X$$ be a Banach space and $$M$$ be a closed subspace i.e. $$M \leq X$$. Is it true that $$M^*$$ is a subspace of $$X^*$$ where $$*$$ denotes the dual space? What I have tried: I know by Hahn Banach theorem that every $$f \in M^*$$ has an extension $$\hat f \in X^*$$ such that $$\hat f |_M =f$$ and the operator norm doesn’t change i.e. $$\|\hat f\| =\|f\|$$. I found that such an extension can be chosen to be linear if $$X$$ is a Hilbert space here, hence my claim might not hold in the general case. Any help is appreciated. ## Let$X$be a complete metric space. Suppose$\{x_n\}$is a sequence in$X$… Let $$X$$ be a complete metric space. Suppose $$\{x_n\}$$ is a sequence in $$X$$ with the property that $$|x_n – x_m| < \frac{1}{n}+\frac{1}{m}$$ for all $$n, m \in \mathbb{N}$$. Show that $$\{x_n\}$$ is convergent. I’m struggling to find where to start for this problem and would appreciate some help. ## Find harmonic of order$3$for$2\pi$periodic funciton$f(x) = \pi -|x|$Let $$f$$ be a $$2\pi$$ periodic function such that $$f(x) = \pi -|x|$$ in $$[-\pi,\pi]$$. Find its harmonic of order $$3$$ I’m learning about orthogonal families of polynmials and discrete polynomial approximations by fourier functions. I’ve came upon this exercise, but I don’t know what an harmonic of order $$3$$ means. Posted on Categories proxies ## What will be$\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$Evaluate $$\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$$ I tried by taking log but it wouldn’t work because there are infinitely many points in any neighbourhood of $$0$$ where $$\ln \left ( \sin {\frac {1} {x}} \right )$$ doesn’t exist. How to overcome this situation? Any help will be highly appreciated.Thank you very much for your valuable time. Posted on Categories proxies ## Given an algebra structure$(X,*)$s.t.$(x*y)*y = y*(y*x) = x$, prove$x*y=y*x\$.

Suppose $$(X,*)$$ is arbitrary algebraic structure such that $$\forall x,y\in X$$, we have $$(x*y)*y = y*(y*x) = x$$, prove that $$x*y=y*x$$.

This question seems pretty simple but I tried and I failed.

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## How to update the apt-check count of “X packages can be updated” message

Is there a command to update the apt-check results so I can use bash script to check them?

I run upgrade and it shows 0 but apt-check still shows 12 packages can be updated. How often is this suppose to update on its own? I have numerous servers. Some seem to update instantly. Others go days still showing updates are available when they aren’t.

I have read that rebooting can reset it but thats silly. Why would you need to reboot to refresh a simple message especially when most updates do NOT require a reboot.

# apt-get update && apt-get upgrade 0 upgraded, 0 newly installed, 0 to remove # /usr/lib/update-notifier/apt-check --human-readable 12 packages can be updated. 9 updates are security updates. 
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