Artificial intelligent assistant

evidently

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1
evidently
evidently, adv. (ˈɛvɪdəntlɪ) [f. evident a. + -ly2.] † 1. So as to be distinctly visible or perceptible; with perfect clearness, conspicuously. Hence in active sense, with vbs. of perceiving, knowing, explaining, etc.: Without possibility of mistake or misunderstanding; clearly, distinctly. Obs. or ... Oxford English Dictionary
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Evidently Chickentown
"Evidently Chickentown" is a poem by the English performance poet John Cooper Clarke. Clarke appears as himself reciting "Evidently Chickentown" in the 2007 British film Control, directed by Anton Corbijn. wikipedia.org
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evidently
evidentlyadv obviously; it appears that 明显地; 显然 Evidently he has decided to leave. 显然他已决定要离开. 牛津英汉双解词典
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Evidently... John Cooper Clarke
Evidently... John Cooper Clarke is a 2012 television documentary about the life of the Salford performance poet John Cooper Clarke. Radcliffe, Craig Charles, Plan B, Kate Nash, Alex Turner, Miranda Sawyer and Paul Morley; and poems by Clarke including "Things Are Gonna Get Worse", "Evidently wikipedia.org
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So Google's Gemini Doesn't Like Python Programming and ...
18 hours ago — So I asked Gemini Pro from Google. It gave me a few tips, but there was one thing I didn't understand: It said "Adding caching to functions".
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What does this inserted clause mean? 文法だけなくて > It is saying that: evidently, the way of thinking in English and Japanese is extremely different. What does the ? (Grammar only ...?) I always get confused when appea...
The sentence should be: > **** **** XXYYor XXYYXXYYXXYYetc.) means "Not only XX but also YY". So it literally means: "As for Japanese and English, it seems / they say that not only their grammars but also their ways of thinking are quite different." → "It seems / they say that Japanese and English a...
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$\sigma$ - compact and locally compact metric space Is the following sentence is true? Each complete, separable and $\sigma$ - compact metric space is locally compact. I suppose (but I'm not sure) it is a truth, bec...
For a counterexample let $e_i, i=1, 2, \ldots$ be the standard unit vectors in $\ell^2$, and $X$ the union of the line segments $L_i$ joining 0 to $e_i$ for all $i$.
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Evaluating $\sum_{k=0}^n \frac{1}{(2k+1)!(2(n-k))!}$ Evidently: $$\sum_{k=0}^n \frac{1}{(2k+1)!(2(n-k))!} = \frac{4^n}{(2n+1)!}$$ (says wolfram alpha) But what is a good way to come up with this?
Multiplying by $(2n+1)!\over (2n+1)!$ might help, as this is the result of adding $2k+1$ to $2(n-k)$. $$\frac 1{(2n+1)!}\sum_{k=0}^n\frac{(2n+1)!}{(2k+1)!(2(n-k))!}=\frac 1{(2n+1)!}\sum_{k=0}^{2n+1}{2n+1\choose k}-\frac 1{(2n+1)!}\sum_{k=0}^{n}{2n+1\choose 2k}\\\ =\frac 12\frac 1{(2n+1)!}\sum_{k=0}^...
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evidently - 搜索 词典 - Bing
1. 明显地；显然地 clearly; that can be seen or understood easily. She walked slowly down the road, evidently in pain. 她沿路慢慢地走着，显然很痛苦。. 'I'm afraid I couldn't finish the work last night.' 'Evidently not.'. "对不起，昨天晚上我没能完成工作。. ""显然完不成。. ". 2.
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Two reflections will evidently result in a (fractional) linear transformation I'm having hard time figuring out the following sentence in my textbook. _"Two reflections will evidently result in a (fractional) linear ...
Reflection with respect to the circle $|z-a|=\rho$ takes $a + r e^{i\theta}$ to $a + (\rho^2/r) e^{i\theta} = a + \dfrac{\rho^2}{\overline{r e^{i\theta}}}$, i.e. it is the mapping $R_{a,\rho}(z) = a + \dfrac{\rho^2}{\overline{z - a}}$. Thus with two reflections with two circles we get $$ \eqalign{R_...
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Vectors that are not the eigenvectors of any linear operator Let me preface this by saying that I am not a mathematician, so forgive me if my phrasing is not rigorous or not formally correct. An eigenvector is some v...
Yes every non sero vector is an eigen vector for some matrix. Note that the matrix $\lambda I$ has every nonzero vector as its eigenvector with the eigenvalue $\lambda$
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如何利用闲暇时间:给门板刷油漆 Az Oghat-e Faraghat-e Khod Chegouneh Estefadeh Konim?
　　Evidently the first installment in a series that didn’t continue, this instructional film shows idle twelve- and sixteen-year-old brothers learning (展开全部) 　　Evidently the first installment in a series that didn’t continue 豆瓣
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Holomorphic function approaches infinity on boundary Let $f:U \to \mathbb{C}$ be a holomorphic function on an open, relatively compact domain $U$ of $\mathbb{C}$. If $f$ is unbounded on U, then evidently _there exists...
No. Take$$\begin{array}{rccc}f\colon&D(0,1)&\longrightarrow&\mathbb C\\\&z&\mapsto&\frac1{1-z}\end{array}$$and $a_n=-1+\frac1n$ for each $n\in\mathbb N$.
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Non zero-divisor that divides a unit is a unit > Let $R$ be a ring with unity and $a, b \in R$. Assume $a$ and $b$ are not zero divisors. Show that $a$ and $b$ are units, if $ab$ is a unit. Clearly, $a$ has a right i...
Since $ab$ is a unit, there is an element $c\in R$ such that $abc=1=cab$. To show that $a$ is a unit, it is enough to show that $bca=1$, for then $bc$ is both a left and right inverse of $a$. Starting with $abc=1$ and multiplying both sides on the right by $a$, we obtain $abca=a$, or $a(bca-1)=0$. S...
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Using the Weierstrass M-test to show that $x^n$ converges uniformly on$ [-1/2,1/2]$ I need to use the Weierstrass M-test to show that $x^n$ converges uniformly on $[-1/2,1/2]$. Evidently one needs a sequence that conv...
We have $|x^n|\leqslant\left(\frac12\right)^n$ and $$\sum_{n=1}^\infty \left(\frac12\right)^n=1, $$ and so we conclude.
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