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.

В чём разница между {$x} и ${x}?

В php есть минимум 3 способа вставки переменной в строку:

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


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.

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.


$M \leq X$. 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.

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.

What will be $\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$


$ $ \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.

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.