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seen Jul 25 at 20:18

Sep
3
comment Method of Least Squares-Why is it preferred?
Please define "negative error", and in any case, clarify your question.
Sep
3
comment Method of Least Squares-Why is it preferred?
This answer does not explain why the method of least squares is as popular as it is.
Sep
3
comment Method of Least Squares-Why is it preferred?
SN77: this is not true. If the errors are $1,1,1$ then you would have a total error of 3, but an equal total error for $0,0,3$. Least squares would say that individual errors of $1,1,1$ (total 3) is a better fit than $0,0,3$ (total 9).
Sep
3
answered Method of Least Squares-Why is it preferred?
Aug
31
comment Intuitive significance open sets (and software for learning topology?)
@MichaelGreinecker: limit points are not unique in arbitrary topologies.
Aug
31
comment Intuitive significance open sets (and software for learning topology?)
@rafiki: yes, many $\varepsilon$-$\delta$ definitions can be rewritten in terms of open and closed sets.
Aug
31
comment Intuitive significance open sets (and software for learning topology?)
@MichaelGreinecker: limit points are rather shaky in arbitrary topological spaces.
Aug
31
comment Intuitive significance open sets (and software for learning topology?)
@BabyDragon: also, metrically you usually define open sets first and then closed sets, since it is not quite as trivial to define closed sets in terms of "for any $x$ there is a ball of radius $\varepsilon$"-type sentences.
Aug
31
comment Intuitive significance open sets (and software for learning topology?)
@BabyDragon: you are right; it's just that some things take 2 fewer characters to define in terms of open sets.
Aug
31
answered Intuitive significance open sets (and software for learning topology?)
Aug
29
comment Proving $(x^s)^t=x^{st}$
In fact, there is a mistake in your induction. To be proven in the second step should be that $(x^s)^{t+1}=x^{s(t+1)}$ for all $s$.
Aug
29
answered Show that $z^{-1} = \frac{\bar z}{|z|^2}$
Aug
29
comment Use a particular method to prove if $a^n \mid b^n$ then $a\mid b$
what do you mean by $(a,b)$? a representation of $a/b\in\mathbb{Q}$? or the gcd of $a$ and $b$?
Aug
29
comment If $A+A^T$ is negative definite, then the eigenvalues of $A$ have negative real parts?
Fabian: if I'd have a proof I'd give it. These facts can be concluded straight away. Of course the complicating factor here is that $A$'s eigenspaces are different from $A^T$'s.
Aug
29
comment If $A+A^T$ is negative definite, then the eigenvalues of $A$ have negative real parts?
$A+A^T$ is symmetric, thus diagonalizable, and since it is positive definite it is thus similar to a diagonal matrix with strictly positive eigenvalues. $A$'s eigenvalues are equal to $A^T$'s.
Aug
27
comment Products of groups with no non-abelian quotients
Nicky, that's a more interesting question but one I suppose is difficult to answer. Do all groups have nontrivial normal subgroups?
Aug
27
comment Decomposition of Permutation Group
Agreed, I'll leave this in place for reference.
Aug
27
revised Decomposition of Permutation Group
added 154 characters in body
Aug
27
comment Existence of an Infinite Length Path
You are right, I am a Mathematica noob and entered the integral incorrectly.
Aug
27
comment Existence of an Infinite Length Path
TonyK, your current arc length is bounded on $[0,1]$, according to mathematica.