PeroK's latest activity
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PeroK replied to the thread B Average velocity as a weighted mean.The proportional of the total time that ##v_1## applies is ##\frac{v_2}{v_1+v_2}##. This is a relative weighing of ##v_2##. And vice versa. -
PeroK reacted to Chestermiller's post in the thread I Entropy and configurations of microstates with
Like.
In statistical thermodynamics, the following approximation is frequently used: $$\ln{(n!)}=\ln(1)+\ln(2)...\ln(n-1)+\ln(n)\approx... -
PeroK reacted to anuttarasammyak's post in the thread B Average velocity as a weighted mean with Like.
$$\bar{v}=\frac{2L}{L/v_1+L/v_2}=\frac{2v_1v_2}{v_1+v_2}$$ I am not sure what are harmonic mean and weighted average you say in this result.
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PeroK reacted to Demystifier's post in the thread I Entropy and configurations of microstates with Like.
Here it is tacitly assumed that ##x\gg n##. Under this approximation $$x(x-1)(x-2)....(x-(n-1)) \approx x^n$$
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PeroK reacted to jedishrfu's post in the thread B A Time Crystal Visible to the Human Eye with
Informative.
https://interestingengineering.com/science/worlds-first-time-crystal-visible-human-eye -
PeroK reacted to Nugatory's post in the thread I Quantum Entanglement and Synchronization with Like.
No. There is no way of distinguishing a so-called collapsed state from an uncollapsed state (the “so-called” is there because even...
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PeroK replied to the thread I Quantum Entanglement and Synchronization.I guess your idea is to measure A and simultaneously B "changes state", which is instantly observed by clock B. That misunderstands...
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PeroK replied to the thread I Quantum Entanglement and Synchronization.The mechanism described there is not what you propose in your original post.
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Part of the problem is that there are two common systems which many (most) conflate. Either imply the uncertainty from the number of...
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When I was a practicing EE, none of this ever made sense to me. These problems rarely mention the probability distribution of the data...
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My article cites the exact wording (Latin) and its English translation: “Cubum autem in duos cubos, aut quadratoquadratum in duos...
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"I have discovered a truly marvelous proof of this, which this margin is too small to contain."
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PeroK replied to the thread Errors and significant figures.As I said, my instinct is that the true area must lie within the given range. In this case, area could be 20.8, which is not ## 19\pm...
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PeroK replied to the thread Errors and significant figures.It's not a question of being precise. It's a question of the area definitely lying within the range given.
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PeroK replied to the thread Errors and significant figures.PS 20.5 is not an upper limit for the area. It would have to be ##19 \pm 2##, assuming the error must be symmetric.