oscillate

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oscillate
oscillate, v. (ˈɒsɪleɪt) [f. L. oscillāt-, ppl. stem of oscillāre to swing: see -ate3.] 1. a. intr. To swing backwards and forwards, like a pendulum; to vibrate; to move to and fro between two points.1726 Stone Math. Dict. s.v. Oscillation, If a single Pendulum be suspended between two Semi-Cycloids... Oxford English Dictionary
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oscillate
oscillate/ˈɔsɪleɪt; `ɑslˌet/ v1 [I, Tn] (cause sth to) move repeatedly and regularly from one position to another and back again (使某物)摆动 A pendulum oscillates. 摆锤能摆动.2 [I, Ipr] ~ (between sth and sth) (fml fig 文, 比喻) keep moving backwards and forwards between extremes of feeling, behaviour, opinion,... 牛津英汉双解词典
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Discontinuous derivative *not* by oscillation All the differentiable functions that I have ever seen whose derivative is discontinuous, achieve this discontinuity by oscillating: See, e.g., this question. Is it poss...
For a function of one variable (from $\mathbf{R}$ to $\mathbf{R}$), the derivative (if defined everywhere) can't have a jump discontinuity. This is a consequence of Darboux's theorem, which says that derivatives have the intermediate value property, so that the image of an interval is an interval.
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Proof that the sequence $a_{n} = n(-1)^n, n \in \mathbb{N} $ diverges The sequence is $-1,2,-3,4,-5,...$ I understand that it will oscillate between arbitrarily large positive and negative values as $n \rightarrow \in...
Using @Sujaan Kunalan suggestion, There are subsequences $a_{2j} = 2j(-1)^{2j} = 2j \rightarrow +\infty$ when $n=2j \rightarrow \infty$ and $a_{2j+1} = (2j+1) \cdot (-1)^{2j+1} = -(2j+1) \rightarrow -\infty$ when $n=2j+1 \rightarrow \infty$ The original sequence cannot converge and this could be con...
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Compound Interest and Oscillation If we calculate compound interest, it will approach continuous interest as we compound more and more frequently. However, in the region of $3.0e10^7$, compound interest exceeds contin...
The problem has nothing to do with compound interest calculations, but about how machines work. Although numbers can become infinitely large and require more and more precision, computer memory is _not_ infinite. (Floating-point) numbers are not represented perfectly in computers or calculators, hen...
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What property of a matrix causes $\|e^{tA}\|_2$ to oscillate as $t\rightarrow\infty$? What property of a matrix causes $\|e^{tA}\|_2$ to oscillate as $t\rightarrow\infty$? The best I can come up with is that $A=bi\cd...
Most non-normal matrices whose eigenvalues are pure imaginary will have this property. (Normal means $AA^* = A^*A$.) E.g. $A = P D P^{-1}$ where $P$ is an invertible $2\times 2$ matrix that is not a multiple of a unitary matrix, and $D = \begin{bmatrix} 0 & -1 \\\ 1 & 0 \end{bmatrix}$: then $$ e^{tA...
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Bounded Harmonic Functions on the Disk Denote by $\mathbb{D}$ the open unit disk in $\mathbb{R}^2$. Is it possible to find a _bounded_ harmonic function $u : \mathbb{D} \to \mathbb{R}$ that is not uniformly continuous...
For any bounded measurable function $f\in L^\infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous. Here you can find what the Poisson kernel is and how it's used ...
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Feature scaling's effect on gradient descent In Andrew Ng's machine learning class, he mentioned feature scaling will make gradient descent goes faster. < Specifically: > We can speed up gradient descent by having ...
The gradient descend uses one fixed learning rate for all $\theta$'s, so we need to choose the value based on the input value having the smallest range. Otherwise the gradient descent might not converge for that small range. Now with that small learning rate it takes ages for the large range to conv...
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A differentiable function that is $\mathcal{O}(1)$ which has a derivative that is not Let $E$ be a compact subset of a bounded open set $\mathcal{D},$ both of which are subsets of the real numbers. Let a differentiabl...
Take $E = [-1,1]$, and $\mathcal D = (-2, 2)$. Let $$ f(x) = \begin{cases} 0 &\text{if $x = 0$},\\\ x^2\sin(x^{-3}) &\text{else.}\end{cases}$$ Then $f$ is trivially bounded on $\mathcal D$. Also, the derivative of $f$ exists on all of $\mathcal D$: we have $$ f'(x) = \begin{cases} 0 &\text{if $x = 0...
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What is the intuitive reason for a function to be continous but not differentiable I was just wondering if the following is the right idea for the function to be continuous, but is not differentiable. So, intuitively ...
No. The function is not differentiable at $a$ when there is _no_ tangent line to the graph at the point $\bigl(a,f(a)\bigr)$. The most typical reason for non-existence of a tangent there is when there is a corner at that point. This has nothing to do with oscillations. However, the oscillation pheno...
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Show that $f$ is not Riemann integrable on $[0,1]$ If $x$ is any rational number, $f(x)=0$. If $x$ is any irrational number, $f(x)=1$. I know that $f(x)$ oscillate between $0$ and $1$ on $[0.1]$. But I have not idea...
I assume that you mean that this function is not Riemann Integrable, as this function is actually Lebesgue Integrable. A function is defined as Riemann integrable if the upper sums and the lower sums of arbitrary partitions get arbitrarily close. Let $0=x_0<x_1<...<x_{n-1}<x_n=1$ be any partition of...
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How do we say the behavior of the limit of cosine square? > $\lim_{n\rightarrow \infty } (cosx)^{2n}$ Is it correct to say that this sequence diverges? > $\lim_{n\rightarrow \infty } ((cosx)^{2})^{n}$ > $ 0\le (...
The sequence of functions converges to a discontinuous function. For any $x$ such that $|\cos{x}| < 1$, the sequence tends to $0$, and otherwise (because of the $2$ in the power), it tends to $1$. Hence, the limit function is simply $$\lim_{n \to \infty} (\cos{x})^{2n} = \left\\{\begin{array}{lr} 1 ...
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