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comment How to compare dispersion of data?
There are different measured of dispersion. The standard deviation is one of them, and it can be used sensibly for a large class of non-normal distributions. But, as you have observed, a few very large outliers can greatly influence it. If this makes it unsuitable for your application, then you'll have to explain what exactly you are trying to achieve better so that people can suggest alternatives.
Mar
27
answered Why is a function space considered to be a “vector” space when its elements are not vectors?
Mar
11
awarded  Enlightened
Mar
11
awarded  Nice Answer
Feb
21
comment How to Prove the divisibility rule for $3$
The induction methods is nice because it provides an insight into why this divisibility rule works. However, AFAICS, it only shows that the digit-sum being divisible by 3 is a necessary condition for the number being divisible by 3. I don't see how you'd get the sufficiency, though.
Feb
16
reviewed Close Show that $P_n=p(1-P_{n-1})+(1-p)P_{n-1}$ $n \ge 1$
Feb
16
comment Show that $P_n=p(1-P_{n-1})+(1-p)P_{n-1}$ $n \ge 1$
The answers to the question linked by @DilipSarwate seem to explain all the steps of the proof, therefore I'm closing this question.
Feb
16
awarded  Custodian
Feb
16
reviewed Leave Open Comparison test, with one negative term?
Feb
16
comment NFA from regular grammars
In this case, I suggest to the OP to consider the input $ab$, which his NFA should also accept.
Feb
16
comment NFA from regular grammars
The state $1$ seems redundant. And think about what happens if you feed $abaa$ (which it should accept) to your NFA. Finally, why is state $2$ colored blue? If that means that it's an acceptance state, you need to rethinkt this - your NFA is also supposed to accept $a$ and $aba$, neither of which makes it end up in state $2$.
Feb
16
answered Operator on continuous functions, its norm is not attained
Feb
16
revised differentiability at a point (0,0) based on partial derivatives
Latexified the question
Feb
15
comment Using Zorns Lemma, does my arguement hold?
You need to take the behaviour of the function into account. A good choice for the PO would be to interpret $f$ as a subset of $A \times B$, i.e. as the set of pairs $(a,f(a))$. $f_1 \leq f_2$ then means that $f_2$ extends $f_1$, i.e. that whenever $a$ lies in the domain of $f_1$, then $f_2(a) = f_1(a)$.
Feb
15
comment Using Zorns Lemma, does my arguement hold?
Your "partial order" is not really a partial order, because it's not antisymmetric. It is possible for two functions $f_1,f_2$ to have the same domain, yet be different functions. But your $\leq$-relation will then yield $f_1 \leq f_2$ and $f_2 \leq f_1$, which would need to imply $f_1 = f_2$ if $\leq$ truely were a PO.
Feb
15
revised What is the significance of the Uniform Cauchy Criterion vs just being uniformly convergent?
added 115 characters in body
Feb
15
answered What is the significance of the Uniform Cauchy Criterion vs just being uniformly convergent?
Feb
15
comment What is the significance of the Uniform Cauchy Criterion vs just being uniformly convergent?
Those are two orthogonal concepts. There's convergence vs. the cauchy criterion, and there's uniformity of convergence/cauchy criterion vs. non-uniformity. Which distincting is causing you difficulties?
Feb
15
comment Spinner probability when not pointing at C
Does "Conditional Probability" ring any bell?
Feb
15
comment What is wrong with the following function $f:ON \rightarrow ON$?
What's the problem? If $f$ were the identity, then $\{\alpha \,|\, f(\alpha) \leq \gamma \} = \{\alpha \,|\, \alpha \leq \gamma \}$ were bounded too, no?