Artificial intelligent assistant

ourself

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1
ourself
ourself, pron. (aʊəˈsɛlf) Forms: see our and self. [A parallel formation to next, with self instead of selves, appearing first in 14th c. It may have arisen out of our selven, through our selve, our selfe, finally with e mute, as in the infinitive of vbs. (e.g. holden, holde, hold); but, on the othe... Oxford English Dictionary
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Ourself Lip Filler Review 2023 | The Strategist - New York Magazine
Jan 10, 2023I do worry that everyone is getting prettier around me while I turn to tar, but still, I'm firm about no needles before the age of 47. Ourself's Lip Filler requires zero injections, though. It ...
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Maybe Mars
Lonely Leary Alpine Decline Hiperson The Bedstars Future Orients Backspace Deadly Cradle Death Run Run Run Alumni Artists Joyside Low Wormwood Ourself wikipedia.org
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us self
▪ I. us self etc.: see self A. 3–4.▪ II. us self, pron. In later use Eng. regional (north. and north midl.). Emphatic and reflexive pronoun, corresponding to we, us. See ourself pron., ourselves pron. Brit. /ʌˈsɛlf/, /əˈsɛlf/, U.S. /əˈsɛlf/ Forms: α. OE us selfe[accusative], OE us sylfe[accusative],... Oxford English Dictionary
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Letters patent (United Kingdom)
same degree title honour and dignity belonging or appertaining thereto In Witness whereof We have caused these Our Letters to be made Patent Witness Ourself oath of allegiance and the judicial oath in his presence or before him; In Witness whereof We have caused these Our Letters to be made Patent Witness Ourself wikipedia.org
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Condition on the positivity of a quadratic form We place ourself in $\mathbb{R}^{n}$. Let's consider a positive definite matrix $M \in \mathcal{M}_{n} (\mathbb{R})$, $V$ and $E$ $\in \mathbb{R}^{n}$, and $\alpha > 0$...
Hint: put $X = -V - \frac{1}{2\alpha}(E-\alpha V)$. Then $f(X)=-\frac{1}{4\alpha}(E-\alpha V)^TM(E-\alpha V)$.
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Victory
Religion Buddha emphasized the immortality that exists within ourself, to act to conquer ourself is his victory and ours; "It is better to conquer yourself wikipedia.org
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Set from powerset which gives nonempty intersection Have such question: Let us have $S=\\{1,2,\dots,n\\}$ and let there be powerset $P(S)$. Prove that there exists such subset $F\subseteq P(S)$ that $A\cap B\neq\empt...
You're looking for a large family of subsets of $S$ whose members overlap. Notice that $2^{n-1}$ is a very nice number: it is the number of subsets of any set with $n-1$ elements. So, fix some $s$ in $S$, have a look at the collection $F_0 = \mathcal{P}(S\setminus\\{s\\})$, and throw $s$ back in to ...
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Awareness Sunday
"Awareness Sunday is an annual opportunity to reflect on the value of loving our neighbour as ourself, even if our neighbour is of a different religious wikipedia.org
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Equivariant homomorphism in a non-abelian group of order $pq$, where $p$ and $q$ are distinct primes. Let $G$ be a group and $H$ be a normal subgroup of $G$. A homomorphism $\phi$ from $H$ to $G$ is said to be $G$-equ...
Consider $G = \langle a\rangle:\langle b\rangle\cong \mathbb{Z}_p:\mathbb{Z}_q$, a semidirect product, by $bab^{-1} = a^{m}$ for some $m$ such that $m^q\equiv 1\pmod p$. Let $\phi:\langle a\rangle\to G$ be a homomorphism, then $\phi(a) = a^n$ for some $n$, because $\phi(a)$ has order $1$ or $p$. For...
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Pascal Le Boeuf
Discography Leader, featured artist Collaborator Tasha Warren & Dave Eggar, “Snapshots”, Ourself Behind Ourself, Concealed (Bright Shiny Things, 2022 wikipedia.org
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Upper bound of $\frac{\|AQx\|_2}{\|Ax\|_2}$ with $Ax\neq 0,\forall x$? I am studying matrix norm right now, based on the definition of norm on matrix, we know that $\|A\|_2 = \sup_{x} \frac{\|AQx\|_2}{\|Qx\|_2}$ for s...
$$\sup_{Ax \neq 0}\frac{\|AQx\|_2}{\|Ax\|_2}=\sup_{y\neq 0} \frac{\|AQA^{-1}y\|_2}{\|y\|_2}=\|AQA^{-1}\|_2$$
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Long Shot Party
They released their first mini-album titled Making Ourself Understood in our Sounds in January 2000. wikipedia.org
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Existence of a set with given Hausdorff dimension Today we had funny discussion whether one can extend the notion of homotopy groups for noninteger indices. Long story short, we asked ourself the following question: ...
**First step: answer the question for $0 0$. If $d$ is an integer, just take $A = [0,1]^d$. Otherwise, write $d = \lfloor d \rfloor + \\{d\\}$. Thanks to the first step, as $0<\\{d\\}<1$, we have a Cantor set $C \subset [0,1]$ _s.t._ $\mbox{dim}_H(C) = \\{d\\}$. The previous result yields $\mbox{dim...
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为什么业界P盛Q衰 而美国的金融工程 金融数学项目大量培养Q-quant？
So we try to make ourself different... But I still don't like the words 'P' and 'Q'. In fact, most works in financial industry are between them now.” zhihu
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