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| bio | website | |
|---|---|---|
| location | Canada | |
| age | 36 | |
| visits | member for | 2 years, 3 months |
| seen | May 13 at 20:30 | |
| stats | profile views | 140 |
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Mar 19 |
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Experiences with Kumon the question was about exclusively about Kumon, not other program |
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Mar 17 |
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Show: grad f (x + y) = grad f (x) + grad f (y) saying x,y are in $R^2$ already implies x,y (hence x+y) are in the domain of $f$ and $g$. |
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Mar 15 |
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Question about limit! the name is l'Hospital , see en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule |
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Mar 14 |
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Determining the dimensions of the null space and column space of a matrix "use this to find rank A" can it get more simpler than that? |
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Mar 3 |
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Closed under equality this is known as the identity of indiscernibles |
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Mar 2 |
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Are all integers fractions? @Carl Mummert . In my comment to an answer I ended up with "It is equal because there is a mathematical convention that makes it equal.", which I think it is incorrect in light of your answer. So they are not equal as mathematical entities. Are we left to say they are equal as a language convention? That is when we say Z is a subset of Q or z=z/1 for z integer, we make a language convention (shortcut, compromise)? Thanks |
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Mar 2 |
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Are all integers fractions? I think it's more than that. One can ask why c is the same as c/1? One might answer because we see them as embedded in the quotient ring Q, that is c is the same as equivalence class of (c,1) . Or one can maybe say because "ordinary" division is preserved, that is c divided by 1 is c. In other words, the answer because "it is a subset" or because "c=c/1" is not sufficient. It is equal because there is a mathematical convention that makes it equal. |
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Mar 2 |
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Limit of a function Yes, what Berci says |
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Mar 2 |
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Limit of a function well why not choose epsilon such that l>epsilon>0 . Such epsilon exists |
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Mar 2 |
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The number of positive integers @Brian M. Scott not quite ok at the end, more justification is needed |
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Feb 11 |
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Questions on atoms of a measure Does this work? Suppose the intersection has probability 1, then its complement has probability 0. If we choose only the countable set, the probability of complement of intersection , which is union of compelments has probability 1. Then there exist one element with probability 1. Then its complement has probability 0. Contradiction with the choise of elements in intersection |
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Feb 11 |
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Questions on atoms of a measure in lemma's proof, why cant the intersection of all elements in sigma with probability 1 have probability 0? |
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Feb 8 |
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How should I understand the $\sigma$-algebra in Kolmogorov's zero-one law? I refer to "atomic" as used in Lemma. |
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Feb 8 |
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How should I understand the $\sigma$-algebra in Kolmogorov's zero-one law? May I know how you define "atomic" here? Thanks |
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Feb 5 |
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the expectation of a random variable of a random variable @Tim Unless you meant the collection of p.d.f's Z|x is a random variable, which is an interesting question and I think it is true. But perhaps a sidetrack. |
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Feb 5 |
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the expectation of a random variable of a random variable @Tim $E(Z|X)$ is a random variable and a mapping of X, not the p.m.f. of Z|X=x . Note that h(z) above is the p.m.f. of Z|X=x . Did you write h(z) explicitly? I don't want to spoil the hint for Shuai |
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Feb 5 |
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the expectation of a random variable of a random variable @Tim I think it's like this $\delta(a)(z)=1$ if $z=a$ and $\delta(a)(z)=0$ if $z\neq a$ and then define $h(z)=x\delta(1)(z)+(1-x)\delta(0)(z)$ So $\delta(a)$ is a function that depends on $a$-you can call it alternatively and more clearly $\delta_a$ |
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Feb 5 |
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the expectation of a random variable of a random variable you can start to write h(z)=... explicitly as a function. Also, $E[X] = \int xf_xdx$. |
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Jun 29 |
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Angle equations +1 for "The "co" in cofunction is related to the "co" in complement" I didn't know that |
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Jun 17 |
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Calculus question www.artofproblemsolving.com website has this feature (hidden hints or answers), but here I see it for the first time +1 |
