1,330 reputation
830
bio website
location
age 24
visits member for 3 years, 4 months
seen 1 hour ago

Sophomore Graduate Student in Mathematics


Aug
4
comment Additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity
I will say every $\mathbb{Q/Z}$ is isomorphic to $X = \left\{x : x \in [0,1] \text{ and }x \in \mathbb{Q} \right\}$
Aug
4
revised All possibilities of seven numbers in ascending order?
Made links pretty. and added LaTeX.
Aug
4
suggested approved edit on All possibilities of seven numbers in ascending order?
Aug
4
comment Boundedness in a topological space?
If you have found the answer helpful and correct, please accept it else let people know what more do you want from the answer. :-)
Aug
4
comment Graduate School/Career paths
@AdamSmith: Still, how much lower?
Aug
3
comment Prove that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers
@RajeshKSingh: Do you remember where you found it? There might be something more to it.
Aug
2
comment What is a good complex analysis textbook?
@ChangweiZhou: Seriously man? In high school? Swell!
Aug
2
comment Show $\lim_{h \to 0} \frac{f(h)+f(-h)}{h^2}=f''(0)$
@Ranabir: Yes, you can write $f''(0)=\lim_{h \to 0} \frac{f'(0)-f'(-h)}{h}$. This is because, $f''(0)$ is said to exist only when both the right and left hand limits exists and are equal to each other.
Aug
2
comment Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$
@LeoIzen: $R(t)$ is a constant?
Aug
2
revised Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$
Changed the hyperlinks to make it easy to read without taking the focus away.
Aug
2
comment Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$
I edited your answer to make the hyperlinks appear pretty so that it is easier to read and does not take focus away from your answer. :-)
Aug
2
suggested approved edit on Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$
Aug
2
comment What does “$f$ is a function on $S$” mean?
I am guessing the exact expression of $f$ will provide image.
Aug
2
comment What does “$f$ is a function on $S$” mean?
$S$ is a domain. The codomain is any superset of the image of $S$ under $f$.
Aug
2
comment Is this a function?
Thanks for that. :-) +1
Aug
2
comment Is this a function?
@HenningMakholm: Okay, thanks. I was not aware of this notation. So you are saying the notation is equivalent to the below? \begin{equation} \theta(x,y) = (3y,2x,x+y) \end{equation}
Aug
2
comment Is this a function?
@Mathh Pressland: I do not believe the co-domain is $\mathbb{R}^{3}$, rather I believe the co-domain is a vector space $W$ where an element of $W$ is $w$ of the form $(u,v)$ where $u \in \mathbb{R}^{2}$ and $v \in \mathbb{R}^{3}$. This space might be isomorphic to $\mathbb{R}^{3}$ though.
Aug
2
comment Is this a function?
Yes, it is a function. Domain is $\mathbb{R^{2}}$ and co-domain is $V\times W$ where $V$ is a vector space over $R$ of dimension $2$ and $W$ is a vector space over $R$ of dimension 3.
Aug
2
comment hints on solving DE
Yeah, can happen. ;-)
Aug
2
comment hints on solving DE
Sorry for the $\frac{a}{2}$, it should be only a