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Aug
7
comment How to find eigenvalues of matrix $\begin{bmatrix} 3& a+1\\a+1&3 \end{bmatrix}$
Have you calculated the determinant of that matrix?
Jul
7
comment How to simplify this integrand,
At first you might want to note that $e^{-2t}+4e^{-2t}=5e^{-2t}$.
Jun
7
comment How do I show that the following is a basis for the weak topology on $X$?
Maybe you'll find this thread useful: math.stackexchange.com/questions/305808/…
Jun
7
comment How do I show that the following is a basis for the weak topology on $X$?
What is $p$ and what do you mean by that set being a semi-norm?
Jun
7
comment How do I show that the following is a basis for the weak topology on $X$?
Isn't this the definition of the weak topology? What is your definition?
Jun
7
comment When is the series converges?
Remember to include the absolute values inside the root.
Jun
7
comment Is this a metric on the shift space?
Did you check the axioms of a metric space, or what makes you "think" yes? Is there some particular part that puzzles you?
Jun
7
comment Error in the reasoning?
@dietervdf. but that's the thing. $d(x,a)=r$ does not translate to $x\in \partial B(a,r)$.
Jun
5
comment Error in the reasoning?
@dietervdf: That property does hold. Note that $\overline{B}(x,r)$ is not the closure of any given set of our interest. The upper bar is just a notation here, different from closure.
Jun
2
comment Compactness of a group with a bounded left-invariant metric
right! Because it's an additive group. Thanks.
Jun
2
comment What is the topology here with three elements in sets?
@Topology: The complement of $\{b\}$ is $\{a,c\}$, which is neither an element of $\tau$, nor $\tau$ without $\{b\}$ in it.
Jun
2
comment Compactness of a group with a bounded left-invariant metric
Why is this left invariant? Maybe it's obvious but I'm not seeing it.
Jun
1
comment Prove the dominated convergence for $f_n(x)=\frac{x^n}{x^2+3x+2}$
@Lucas: The conclusion is that $\lim_{n\to\infty}\int_{0}^{1}f_{n}(x)=\int_{0}^{1}f(x)$. And $f(x)=0$ if $x\in [0,1)$ and $f(1)=\frac{1}{6}$. This limit function is defined on the whole interval $[0,1]$ and is discontinuous. The reason why $\int_{0}^{1}f(x)=0$ requires measure theory: the set $\{1\}$ has measure zero and thus its contribution to the integral $\int_{0}^{1}f(x)$ is negligible. To fully understand this, a good place to start is to google "Lebesgue measure" and "Lebesgue integral".
Jun
1
comment Prove the dominated convergence for $f_n(x)=\frac{x^n}{x^2+3x+2}$
@lucas: Yes that's true. You can ignore the comment about measure zero: I thought you we're dealing with integration through measure theory, but you can also view the integral as a Riemann integral. That's fine too.
Jun
1
comment Prove the dominated convergence for $f_n(x)=\frac{x^n}{x^2+3x+2}$
@lucas: The point wise convergence to zero happens for all $x\in [0,1)$. At $x=1$ the limit is not zero, but the single point $x=1$ has measure zero and thus does not contribute to the value of the integral.
Jun
1
comment Prove the dominated convergence for $f_n(x)=\frac{x^n}{x^2+3x+2}$
@Lucas: I edited the answer and included that part too. For further reference for dominated convergence theorem, the wikipedia article has a lot of information: en.wikipedia.org/wiki/Dominated_convergence_theorem
Jun
1
comment Show that $\mathbb{R}^m$ is not homeomorphic to $\mathbb{R}^n$
@NoahOlander: True. I'll make that clear. Thanks.
May
29
comment What interpretation of the Lie braket is this?
What is the definition of $\frac{\delta}{\delta x}$?
May
29
comment Are these two definitions of basis equivalent?
@xhimi: The set $\beta$ belongs to $B$ which is by assumption a sub collection of $T$.
May
28
comment The terminology for particular subsets of the power set of R
The way you defined $X$ it follows that $X=\emptyset$, because for any $x\in\mathbb{R}$ there exists $a\in\mathbb{R}$ with $x\geq a$.