3,347 reputation
21034
bio website
location Buenos Aires, Argentina
age 21
visits member for 4 years, 2 months
seen 58 mins ago

(my about me is currently blank)


Mar
21
comment Prove that the set of positive real numbers is not bounded from above
@Arturo: I didn't say it, but I can use the basic axioms of the real numbers.
Mar
21
asked Prove that the set of positive real numbers is not bounded from above
Mar
19
answered Using polar form to prove $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$
Mar
16
comment See the sign of the double derivative from just looking at the graph?
In this case, it looks like it, but I don't think you can be completely sure without knowing the function's formula.
Mar
13
accepted Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$?
Mar
13
comment Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$?
Well, that was simple :). Thanks.
Mar
13
asked Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$?
Mar
8
comment directional derivative unit vector
If you have an angle $\theta$, the unit vector in that direction is $(\cos \theta, \sin \theta)$.
Feb
29
accepted How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
Feb
29
comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
It wasn't my physics test, it was one some friends in my same year did. But never mind that, thank you for all your help.
Feb
29
comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
I end up with $\int \frac{\mathrm{d}y}{\sqrt{\frac{2k}{y}+C}} = \int \mathrm{d}x$. According to WolframAlpha, the integral on the left hand side is quite complicated, and it seems impossible to solve for $y=y(x)$. Am I doing something wrong?
Feb
29
comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
Well, I get $y' = \pm \sqrt{2k(\frac1{y}-\frac1{y(0)})}$ (I made it clearer in the title that the right hand side of the equation should be negative), but I don't know how to go from there.
Feb
29
revised How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
edited title
Feb
28
comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
@Henry: I know, but mathematically does it make any difference?
Feb
28
asked How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$?
Feb
19
comment A basic question about integration
@Gingerjin: the limit when $x$ goes to $0$ is $\infty$, not $0$, so it doesn't make sense to define it as such. It would not be continuous.
Feb
19
comment A basic question about integration
$x^{-\frac1{2}}$ is not integrable in $[0, 1]$, because it's not defined at $x = 0$.
Feb
16
accepted Confused about characteristic equation of a linear ODE
Feb
16
comment Confused about characteristic equation of a linear ODE
I guess that makes sense. If I understand correctly the idea is that you want a real space of solutions so you try to find a base made out of real functions, right?
Feb
16
asked Confused about characteristic equation of a linear ODE