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Freshman Graduate Student in Mathematics


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 suggested 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
Aug
2
answered hints on solving DE
Aug
2
comment hints on solving DE
You must have made different substitutions. I am getting the following. \begin{equation} \frac{dw}{dz} = \frac{w-\frac{a}{2}z}{z-aw} \end{equation} I will write a partial answer for you.
Aug
2
comment hints on solving DE
No, basically you then eliminate $x$ and $y$ and solve for $w$ in terms of $z$. If you want more hint, I can give you so.
Aug
2
comment hints on solving DE
Try putting $w^{2} = x^{2} + y^{2} + z^{2}$ and then use the identity \begin{equation} \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{k_{1} a + k_{2} c + k_{3} e}{ k_{1} b+ k_{2} d+ k_{3} f}. \end{equation}
Aug
1
awarded  Yearling
Jul
31
comment Limit of binomial coefficient
@J.M., Thanks for the info.