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Jan
26
comment Show that if $v\in (V_c)^{\perp}$ then $(Av)\in (V_c)^{\perp}$ for a normal matrix $A$ with an eigenvalue $c$
$\langle x, Av \rangle = \langle x, v c \rangle = ....$
Jan
25
reviewed Approve Bounding summations
Jan
23
awarded  Good Question
Jan
21
reviewed Approve Evaluating the limit $\displaystyle \lim _{x\to \infty }\frac{(x^2+1)}{(x-1)}\sin(\frac{1}{x})$
Jan
17
reviewed Reject What is the intuition behind the exponential distribution?
Jan
15
comment Proving compactness of an operator
You can use ideas from this question math.stackexchange.com/questions/320751/….
Jan
15
reviewed Approve on the definition of graded Betti numbers
Jan
15
comment Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 + 2|w|^2$
This is Parallelogram law.
Jan
12
comment Bounding $\frac{2}{\pi}\int_0^{\pi}\frac{|\sin(n+\frac{1}{2})t|}{2\sin\frac{1}{2}t}$ below by $\frac{4}{\pi^2}\log n$
From which book is this?
Jan
9
awarded  Yearling
Dec
28
reviewed Approve Likelihood Ratio and Neyman-Pearson factorization theorem
Dec
28
reviewed Reject Prove that $\frac{\sin n}{n}$ is a Cauchy sequence from the definition.
Dec
28
answered $f(x):S^{2n} \rightarrow S^{2n}$ continuous so that there is $x \in S^{2n}$ with $f(x)=x$ or $f(x)=-x$
Dec
26
reviewed Approve What is the explicit obstruction to almost sure convergence in stochastic integrals?
Dec
23
comment Is the function $f(x) = \begin{cases} 1 & \text{$x\in\Bbb Q$} \\[2ex] 0 & \text{$x\notin\Bbb Q$} \end{cases}$ Riemann integrable?
It's not Riemann integrable. Btw. this is known as Dirichlet function as far I know.
Dec
21
reviewed Approve Proving that if the semigroup (A, *) is a group, then the relation is an equivalence relation.
Dec
19
awarded  Constituent
Dec
18
comment Given $|f'(x)|\leq r<1$ show that $f(x)=x$ is unique solution
I think Lagrange's Mean Value Theorem is useful here.
Dec
17
comment Evaluate $\lim_{n\to1^+}\left({\zeta(n)-\dfrac{1}{n-1}}\right)$
For real number $x>0$ and $s\neq 1$ such that $\Re{(s)}>0$ we have $$\zeta(s)=\sum_{1 \le n \le x} \frac{1}{n^s} + \frac{x^{1-s}}{s-1}+\frac{\{x\}}{x^s}-s\int_x^{\infty}\frac{\{u\}}{u^{s+1}}\,du.‌​$$ From that you can (for $x=1$) find that $$\zeta(s)=\frac{s}{s-1}-s\int_1^{\infty} \frac{\{u\}}{u^{s+1}}\,du,$$ so zeta has pole in $s=1$ and...That might help.
Dec
15
reviewed Approve Why does this nonlinear ODE solution not work?