In this question sos440 has mentioned about an integral that he computed:
$$\int_0^{\pi/2}\arctan(1-\sin^2 (x) \cos^2 (x))dx = \pi \left( \frac{\pi}{4}- \arctan \sqrt{\frac{\sqrt{2}-1}{2}}\right)$$
I would really like to know how this can be proved. I tried using differentiation under the integ...
@Ted Unless I'm misunderstanding, the question is whether you can find solutions e.g. to $\{f \in C^\infty(\Bbb R^2) : f_x=0,f_y=0\}$ by solving $f_x+f_y=0$, which isn't true...
Then it's like asking a question like this: If you add the equations $x+y=3$ and $x-y=1$, you get $x=2$. Is a solution of that a solution of the original?
I don't think the solution has nontrivial solutions, @UserX. One has solutions that are linear combinations of $e^{\omega t}$ where $\omega$ is a cube root of unity, and the other has linear combinations of $e^{\eta t}$ where $\eta^4+4 = 0$. They don't overlap.
Funny... there are a lot of questions on MSE about the topology needed for differential geometry, but nobody well tell me about the differential geometry needed for topplogy :)
Sharpe's book does look nice, and luckily, our library has it. I'll go check it out before class.
@DanielF: Since gauge theory in the 70s and 80s, that is no longer the case. We're not talking point-set here :P
We needed you last night, @DanielF. I've forgotten who, but someone was wrestling with compact operators on Banach spaces. He seemed confused about whether the implication compact $\implies$ continuous was reversible.
LOL, to give him intuition on compact operators? He didn't hang around long, and I was in the middle of three other things.
Like right now I need to be working on recommendation letters before I go to be carved by the dermatologist ... then more Thanksgiving shopping ... and the cooking commences.
I'm pretty sure differential geometry techniques are very powerful in low dimensions for the reasons Ted mentions, but I guess in high dimensions algebraic techniques reign supreme... I think all of the progress on exotic spheres in dimensions above 4 is due to homotopy theory.
@DanielFischer If a Maclaurin series converges for all values on the unit circle, does it necessarily converge absolutely for all those values? I'm specifically talking about the case when the radius of convergence is exactly 1.
@RandomVariable That is a difficult question. I don't know the answer to it. I think there was a question asking that a couple of months ago, iirc, it got no answers.
Most of his acting is mediocre, because he doesn't need to do better than mediocre. But he's a great actor when he wants to be; see Truman Show or Eternal Sunshine
@BalarkaSen to kill your question about $\Bbb C$ minus countably many points, here's a problem for you: show that $\pi_1(\Bbb C \setminus \Bbb Z)$ is countable, but $\pi_1(\Bbb C \setminus \Bbb Q(i))$ is not.
@Hippalectryon When one tries continuously, one ends up succeeding. Thus, the more one fails, the greater the chance that it will work. Good advice. A hell to consciously implement though.
@Hippalectryon Do you consume chocolate? I eat like 100g every day and still I'm in shape, no overweight. I couldn't live without chocolate, peanuts, milk, honey.
@Hippalectryon Why? What happened? Besides that, you only eat a bit of chocolate! If I were a movie character, then I'd definitely be Willy Wonka (... chocolate factory). :D:D:D
In this question sos440 has mentioned about an integral that he computed:
$$\int_0^{\pi/2}\arctan(1-\sin^2 (x) \cos^2 (x))dx = \pi \left( \frac{\pi}{4}- \arctan \sqrt{\frac{\sqrt{2}-1}{2}}\right)$$
I would really like to know how this can be proved. I tried using differentiation under the integ...
