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Apr
13
comment Motivating convex sets.
Probability distributions form a convex space, where convex combination means randomly choosing one of several random processes.
Apr
13
comment how to prove $P=M$?
Hint: What is the closure of $P$?
Apr
12
comment math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1$?
No, in that step, he just rewrites the $3$ as $6/2$, which is absolutely unproblematic. It's the next step that's the problem, namely equating $(-1)^{6/2}$ with $((-1)^6)^{1/2}$.
Apr
11
reviewed Approve suggested edit on radical expression foil which confuses me
Apr
9
comment is there an efficient algorithm for comparing collections of points?
@HagenvonEitzen: Interesting idea (that matrix is, of course, just the adjacency matrix of the graph). Since the eigenvalues are invariant under orthogonal transformations of the matrix (while orthogonal transformations of the space of course don't change the matrix at all), the question boils down to whether the eigenvalues plus the special form of the matrix (zero diagonal, non-diagonal elements fulfil triangle inequalities) already fully determine the matrix up to permutations.
Apr
8
comment is there an efficient algorithm for comparing collections of points?
I just notice that I forgot: You also have to include $0$ in both point sets because orthogonal transformations cannot move that (they don't contain translations). But other than that, the distance set should be sufficient for testing equivalence. I can't at the moment think of a formal proof, but for me it seems obvious that it is the case. After all, we are speaking about an Euclidean space and the full group of orthogonal transformations (if it were restricted to only special orthogonal transformations, your doubt would be justified).
Apr
6
reviewed Approve suggested edit on Prove a language is Context Free
Apr
6
reviewed Approve suggested edit on Proof polynomial has only one real root.
Apr
5
reviewed Approve suggested edit on How can I show whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $ converges or diverges?
Apr
5
comment A proportionality puzzle
Actually you can only derive $10/3 = 10/(5/2) = 2\cdot 10/5$, since we neither know that in that foreign country $10/5=2$, nor that $2\cdot2=4$.
Apr
5
comment A proportionality puzzle
But then, if in that country, $5/2=3$, then it seems that at least one of the symbols $5$, $2$ and $3$ has not the meaning we assign to it. We cannot know what meaning those people assign to $10$, therefore the question cannot be answered. Of course, it could also be that the numbers are the same, but they consistently round fractional results to the nearest odd integer. In that case, $10/3 = 3 + 1/3 = 3$.
Apr
5
comment What axiom makes it possible to take the union or intersection of an infinite number of sets $A_1, A_2, \ldots,$ and get a resulting set $B$.
@AsafKaragila: I see, thank you.
Apr
5
comment What axiom makes it possible to take the union or intersection of an infinite number of sets $A_1, A_2, \ldots,$ and get a resulting set $B$.
But doesn't that derivation mean the axiom of union is redundant because it can be derived from specification using the formula $\phi(a)=\exists B(B\in C \land a\in B)$?
Apr
5
comment What axiom makes it possible to take the union or intersection of an infinite number of sets $A_1, A_2, \ldots,$ and get a resulting set $B$.
I just noted that I should explicitly note that the union, as defined in that axiom, is not a sequential application of two-set unions, but a single operation that takes a set of arbitrary many (possibly even uncountably many!) sets to take the union of, and gives the union of those sety. Your unions are just the countably infinite special case. Note also that I didn't see a corresponding direct axiom for the intersection.
Apr
5
comment What axiom makes it possible to take the union or intersection of an infinite number of sets $A_1, A_2, \ldots,$ and get a resulting set $B$.
At least in the Wikipedia version the existence of the union (without any restriction to finite sets) is one of the axioms.
Apr
5
reviewed Reject suggested edit on Please can you check my proof of this about twice differentiable function
Apr
5
comment if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$
With that addition, the proof is correct (and actually quite nice). Note that I had already upvoted you before I thought about the negative number problem, and forgot to unupvote when I noticed it, so don't be surprised that I don't upvote after this comment.
Apr
5
comment Changing order of partial derivatives
However, are you guaranteed that those integrals exist under the conditions in which the commutativity of partial derivatives hold? Also, can you prove both Green's theorem and the gradient theorem without making use of the commutativity of partial derivatives (otherwise this would be a circular proof)?
Apr
5
comment What's the negation of “E is uncountable” ??
Anyway, "countable or finite" is correct no matter whether one interprets "countable" as "having the cardinality of $\mathbb{N}$", or as "having the cardinality of a subset of $\mathbb{N}$". At worst, "or finite" is redundant.
Apr
5
comment if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$
But what if $a<0$?