Reputation
Next privilege 75 Rep.
Set bounties
Badges
1 3
Newest
 Enthusiast
Impact
0 people reached

  • 0 posts edited
  • 0 helpful flags
  • 39 votes cast
Jun
23
comment T and T* have the same Eigenvectors then T is normal
@copper.hat - if $J$ is diagonal then $A$ is diagonalizable. But to prove it is normal we must show that $A$ is unitarily diagonalizable. Is it something we can conclude from $J$ being diagonal? If yes - how?
May
18
comment Proving that if $f$ is continuous then $f(a_n)$ converges
@N.S. - actually, I meant the same conditions as in the question of the OP (including $f(0)=f(1)$) excluding the sequence thing. I also thought that you deduced $f$ being constant on $[0,1]$ from the fact that $f(0)=f(1)$ (because you showed that in $[0,1)$, $f=f(0)$ and because $f(0)=f(1)$ it is also constant in $[0,1]$).
May
18
comment Proving that if $f$ is continuous then $f(a_n)$ converges
Great, exactly what I thought. Thank you.
May
18
comment Proving that if $f$ is continuous then $f(a_n)$ converges
A bit unrelated, but is true that if $f(x)=f(x^2)$ for all $x \in [0,1]$ then $f$ is constant on $[0,1]$?
May
14
awarded  Enthusiast
Apr
6
comment True or false statement about a simple limit of product
What if for all but a finite number of terms $b_n\geq c, c>0$? Can we say then that $\lim_{n\to\infty}(a_n)=0$? (see my reasoning in the comments above)
Apr
6
comment True or false statement about a simple limit of product
@Winther - I think it is true because $$\lim_{n\to\infty}(a_nb_n)=0\Rightarrow\forall\varepsilon>0,\exists N,\forall n>N:|a_nb_n|<\varepsilon\Rightarrow|a_n|<\frac{\varepsilon}{b_n}\Rightarrow|a_n|‌​<\frac{\varepsilon}{c}\Rightarrow\\\Rightarrow\lim_{n\to\infty}a_n=0$$ Am I correct?
Apr
6
comment True or false statement about a simple limit of product
@Winther - what if $c>0$, and for all but a finite number of terms $b_n\geq c$. Can we say then that $\lim_{n\to\infty}(a_n)=0$?
Sep
18
awarded  Altruist
Sep
15
awarded  Investor
Feb
5
awarded  Supporter