# Tag Info

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It's called the canonical map into the double dual. It does not have another name that I know of. In a monoidal category with duals (I'm being vague), the uncurried version of it, $\operatorname{ev} : X \otimes X^* \to I$, is called evaluation.

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Let $\Phi$ the initial set of four clauses as above. Def 15. A pure literal is a literal $l$ that appears in at least one clause of $\Phi$ while $\lnot l$ doers not appear in any clause of $\Phi$. Thus, initially, there are no pure literals. Def 16. A unit clause is a clause with exactly one unknown literal in it. Thus, initially, there are no ...

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In linear algebra the double-dual comes to mind, but the general term that's most relevant is "evaluation map". One somewhat common notation is $\mathrm{ev}_{x}(f)=f(x)$. Either ev (your ~) is the evaluation map, or each $\mathrm{ev}_{x}$ is the evaluation map at $x$.

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In the context of functional analysis, the map $$\Phi: A \to (\Bbb F)^{\Bbb F^A}\\ \Phi:x \mapsto (f \mapsto f(x))$$ is called the "evaluation map".

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Maybe I have misunderstood, but it seems to me that you are mixing up two different questions. The number of steps required to compute a sum of $n$ terms is $O(n)$. The final result of adding the numbers $1$ to $n$ is $\frac12n(n+1)$, which is $O(n^2)$.

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Since N is a constant, we can factor the $\dfrac1{N!}$ out. Let $r=1-\dfrac1{N!}$, so we need to compute $S=1+2r+3r^2+\cdots$. Let $X=1+r+r^2+\cdots$. Note that $X=S+rS+r^2S+\cdots=S^2$. By the geometric series formula, $S=\dfrac1{1-r}$, so $X=\dfrac1{(1-r)^2}$. Plugging in the expression for r gives that $X=\dfrac{1}{1-\left(1-\dfrac1{N!}\right)^2}=N!^2$. ...

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To sum the series $$\sum_{n=1}^\infty n\cdot x^{n-1}$$ notice that it's the term-by-term derivative (with respect to $x$) of the geometric series $\sum_n x^n$, which sums to $\frac{1}{1-x}$. So the series above is the derivative of that. However, in your case, instead of working with series, an easier approach is to say that if you get a success in the ...

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Looking at $x \in X$ as a map $x \colon 1 \to X$, the operator $\tilde x$ is precisely the natural transformation given by co-Yoneda's embedding $$\mathsf{Set}(X,-) \stackrel{\mathsf{Set}(x,-)} \longrightarrow \mathsf{Set}(1,-) \simeq \mathrm{id}_{\mathsf{Set}}$$ So you could call it the restriction along $x$, or precomposition by $x$, and denote it $x^\... 1 Recall that all regular langauges satisfy the pumping lemma, which we capture by the following informal proposition: $$\text{regular}\implies\text{pumping lemma}.$$ Applying modus tollens (a.k.a. contrapositive), we get $$\neg\text{pumping lemma}\implies\neg\text{regular}.$$ Therefore, to prove a language is not regular, it suffices to show that it does not ... 1 Symmetric:$xRy \iff yRx$Is it fair to say that the statements "$x$lives in the same house as$y$" and "$y$lives in the same house as$x$" are equivalent? If so, then the relation is symmetric. Reflexive:$xRx$Is it always true that "$x$lives in the same house as$x$"? If so, then the relation is reflexive. Transitive:$xRy \wedge yRz \implies xRz$... 1 This would make sense if we interpret$A \cap B$as the bitwise "and" ($\land$) (or intersection, if you look at sets interpretation),$A \cup B$as the bitwise "or" ($\lor$) of the numbers (which is their union in the set interpretation). E.g.$7 \cap 11 =$0b0111$\land$0b1011$=$0b0011$= 3$,$7 \cup 11 =$0b0111$\lor$0b1011$=$0b1111$=15 $, so ... 1 Suppose$T$is some Turing machine, and construct this$U\$: read (y,n); simulate T on input y for n steps; if T halts after exactly n steps then loop forever else halt

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