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Apr
5
comment Would like some pointers on this geometry problem
Sorry, it's $AB \cdot BC = BD^2$, the blockquote messed it up. But the idea of assigning some random value to a side might be helpful.
Apr
5
revised Would like some pointers on this geometry problem
deleted 4 characters in body
Apr
5
asked Would like some pointers on this geometry problem
Mar
26
revised A problem with a trigonometric equation
latex and translated the trig functions' names
Mar
26
suggested approved edit on A problem with a trigonometric equation
Mar
22
comment Prove that the set of positive real numbers is not bounded from above
@PeterT.off: Well, wouldn't you have to show that $\mathbb{N}$ is unbounded as well, using a similar proof?
Mar
21
revised Prove that the set of positive real numbers is not bounded from above
added 213 characters in body
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