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monotonically

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monotonically
monoˈtonically, adv. [f. prec. + -al1 + -ly2.] 1. In the manner of a monotone.1890 Lippincott's Mag. Jan. 100 Hear'st thou that rush of homeward-hurrying things, And word-calls monotonically harsh? 2. In the manner of a monotonic function, i.e. either without ever increasing or without ever decreasi... Oxford English Dictionary
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Monotonically normal space
The Sorgenfrey line is monotonically normal. The image of a monotonically normal space under a continuous closed map is monotonically normal. wikipedia.org
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Is [$x$] monotonically increasing?(where $[x]$ means greatest integer function) Is [$x$] monotonically increasing? (where $[x]$ means greatest integer function). In my book it is given as non monotonically increasing ...
Yes it is. Informally: The floor function never decreases (it is either constant or increases strictly). Formally: $[x] \leq [x]+1 = [x+1]$ so by definition the floor function is increasing.
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Understanding the relation between countably paracompact and monotonically normal Does monotonically normal imply countably paracompact? Thanks ahead:)
Yes, it does. This is essentially Theorem $2.3$ of Mary Ellen Rudin, _Dowker Spaces_ , in the _Handbook of Set-Theoretic Topology_ , K. Kunen & J.E. Vaughan, eds., North-Holland, 1984.
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Limit of a monotonically increasing sequence Task: $a_n$ is a monotonically increasing sequence, which don't converges. Proof, that $\lim_{n\to\infty}a_n =+\infty$. Idea: We know, if a monotonically increasing sequen...
By definition, $\lim_{n\to\infty} a_n=+\infty$ if for all $M>0$, there exists $N$ such that $n\geqslant N$ implies $a_n > M$. So let $M>0$. You've shown that $\\{a_n\\}$ is unbounded, so $M$ cannot be an upper bound for $\\{a_n\\}$. Therefore there exists an $N$ such that $a_N > M$. Then if $n\geqsl...
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Is $\frac{x}{e^x-1}$ a monotonically decreasing function? On the first sight, the function $f(x)=\frac{x}{e^x-1}$ is a monotonically decreasing function (< But when I zoom the graph to the neighborhood of the point $(...
Formally, your question makes no sense, because $f$ is undefined at $0$. But suppose that you extend the domain of $f$ to $\mathbb R$, defining$$f(0)=\lim_{x\to0}\frac x{e^x-1}=1.$$Then, yes, $f$ is decreasing near $0$. That's so because $f(x)=\frac1{g(x)}$, where\begin{align}g(x)&=\begin{cases}\fra...
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$\alpha$ monotonically increasing on [a, b] $\implies$ finite at $a$ and $b$, why? The following statement is part of the definition 6.2 in Baby Rudin: > Let $\alpha$ be a monotonically increasing function on $[a,b]$...
If $f$ is a real-valued function on $[a,b]$, then $f(a)$ and $f(b)$ are real numbers and hence finite. The function $\tan{x}$ isn't defined at $\pm\pi/2$, so its domain isn't on $[-\pi/2, +\pi/2]$.
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pyspark里加自增ID_pyspark monotonically_increasing_id-CSDN博客
这个需求好多时候是建立在想横向合并两个pyspark_dataframe，但是pyspark_dataframe与pandas_dataframe有所不同，无法用concat这类函数硬拼接，pyspark里的monotonically_increasing_id函数到一定长度之后两个df自增的中间会隔断，突然从一个比较大的数开始，合并之后就是空或缺行 ...
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Is it possible to define a monotonically increasing sequence on $\mathbb Z$ in such a way that the sequence is $\ldots,-3, -2, -1, 0, 1, 2, 3,\ldots$ Sorry if the question is a bit vague. We know that any monotonicall...
In a **monotonically increasing sequence** you can only find smaller number by "going backwards" on the indexes (since $a_{n+1} \geq a_n \; \forall n \ Note that on a not monotonically increasing sequence you don't have any restrictions on how small your sequence could get (take for example $a_n = -n$)
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Is such space monotonically monolithic or strongly monotonically monolithic? Let $R$ denote the set of all real numbers. $B$ is a subset of $R$ with $B=2^{\aleph_0}$. We topologize $R$ now: the set $B$ is discrete and...
point-countable base (like the Michael line; use a countable base for the usual topology on $R$ with all singletons from $B$), and all such spaces are monotonically
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Is there a difference in definition between an increasing sequence and a monotonically increasing sequence? Is there a difference in definition between an increasing sequence and a monotonically increasing sequence? ...
A monotone sequence is a sequence that is increasing or decreasing. $$\begin{align} \text{monotone increasing}&=\text{increasing}\\\ \text{monotone decreasing}&=\text{decreasing} \end{align}$$
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show $f_n = \frac{1}{n} \chi_{[n,\infty]}$ is monotonically decreasing How do I show $f_n = \frac{1}{n} \chi_{[n,\infty]}$ is monotonically decreasing? I know that $\frac{1}{n}$ is monotonically decreasing, but I am ...
It is sufficient to prove $f_{n+1}\leq f_n$ for each $n$. Let $x \in \Bbb{R}$, then either $x < n$ or $n \leq x < n+1$ or $n+1\leq x.$ If $x <n$, then $f_n(x)=0=f_{n+1}(x),$ If $n \leq x <n+1$, then $f_{n+1}(x)= 0 < \frac{1}{n}=f_n(x)$ and if $x \geq n+1$, then $f_{n+1}(x) = \frac{1}{n+1} < \frac{1}...
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What is an example of monotonically decelerating/accelerating function? I came across the term monotonically decelerating function. How can it be written as a mathematical function?
So in this case, $f''(t)$ would have to be a monotonically decreasing function.
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Prove that function $f(n) = \cos(1/n)$ is monotonically decreasing? Ho can we prove that this function $$f(n) = \cos(1/n)$$ is monotonically decreasing? I computed derivative of this function and got: $$f'(n) = \fr...
Is $n$ supposed to be a positive integer? If so you should probably call $f$ a _sequence_ instead of a function, to prevent confusion. Assuming that yes you are talking about the sequence $\cos(1/n)$: That sequence is _not_ decreasing, it's increasing. In detail: The sequence $1/n$ is decreasing. An...
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Ways to represent functions of bounded variation as the difference of two monotone functions I learnt (it is called Jordan's representation theorem) that every function $f : [a,b] \to \mathbb R$ with bounded variation...
The answers are 1) yes and 2) no. If $g$ is increasing and $h$ is decreasing (or vise-versa), then $g-h$ is monotone. A non-monotone BV function cannot be represented this way. On the other hand, if $f$ is $BV$ and is represented as $f = g - h$ where $f$ and $g$ are increasing, you can also write $f...
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