Jayesh Badwaik
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 Aug4 comment $L_{2}$ norm of the gradient of a vector valued function. I know your answer is assuming things due to lack of information from the OP, but as I guess the the "gradient" is basically a jacobian matrix between the two sets of variables and hence its norm must be the Frobenius norm or a spectral norm of a matrix. I am not sure though. Aug4 comment Solving $x^4-y^4=z^2$ @Brhn Please accept the answer if the answer is correct and you are satisfied with it, else please let us know if you have some more doubts. Aug4 comment 'Linux' math program with interactive terminal? @Magpie Its QtOctave Aug4 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 $X = \left\{x : x \in [0,1] \text{ and }x \in \mathbb{Q} \right\}$ Aug4 revised All possibilities of seven numbers in ascending order? Made links pretty. and added LaTeX. Aug4 suggested approved edit on All possibilities of seven numbers in ascending order? Aug4 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. :-) Aug3 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. Aug2 comment What is a good complex analysis textbook? @ChangweiZhou: Seriously man? In high school? Swell! Aug2 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. Aug2 comment Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$ @LeoIzen: $R(t)$ is a constant? Aug2 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. Aug2 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. :-) Aug2 suggested approved edit on Solving for $v$ in $v'(t) + R(t)\cdot v^{2/3}=J(t)$ Aug2 comment What does “$f$ is a function on $S$” mean? I am guessing the exact expression of $f$ will provide image. Aug2 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 $f$. Aug2 comment Is this a function? Thanks for that. :-) +1 Aug2 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? $$\theta(x,y) = (3y,2x,x+y)$$ Aug2 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. Aug2 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.