Aryabhata
Reputation
60,384
412/400 score
 Feb 22 awarded Nice Question Feb 17 awarded Good Answer Feb 16 awarded Enlightened Feb 14 awarded Nice Answer Feb 11 revised Maths question from an IQ test edited tags Feb 10 comment Comparing $\pi^{e}$ and $e^{\pi}$ @NikolajK: Thank you for the suggestion! Done. Feb 10 revised Comparing $\pi^{e}$ and $e^{\pi}$ added 48 characters in body Feb 9 awarded Enlightened Feb 9 awarded Nice Answer Feb 8 awarded Nice Answer Feb 1 comment Irrational numbers in reality @AsafKaragila: Thanks! Did not know the 60 day thing. The flag was declined with a comment that this is not about physics (which I disagree with) and it is too late to migrate (they didn't mention the 60 day thing). Jan 30 revised Convergent or Divergent? $\sum_{n=1}^\infty\bigl(2^{\frac1{n}}-1\bigr)$ added 64 characters in body Jan 30 comment Convergent or Divergent? $\sum_{n=1}^\infty\bigl(2^{\frac1{n}}-1\bigr)$ @DepeHb: Uh. You are right! Thanks for pointing that out. That is easily fixed, fortunately (take $\frac{1}{\sqrt{2}}$ instead)... Jan 29 comment Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$ You are welcome. Jan 29 revised Proving that $\sin1$(radian) is irrational without using Taylor Series Expansion. edited tags Jan 29 comment Proving that $\sin1$(radian) is irrational without using Taylor Series Expansion. btw, if it was $\sin 1^{\circ}$, the proof would be easy: $\sin 45^{\circ}$ is a polynomial in $\sin 1^{\circ}$ with rational (in fact, integer) coefficients! Since $\sin 45^{\circ}$ is irrational... Jan 29 answered Proving that $\sin1$(radian) is irrational without using Taylor Series Expansion. Jan 29 revised Proving that $\sin1$(radian) is irrational without using Taylor Series Expansion. edited tags Jan 29 comment Irrational numbers in reality This is not a mathematical question. Seems like you found a bug around getting close votes by offering a bounty :-) I have flagged for migration to physics.se, so let's see. Jan 27 comment Can't figure out $O(n \log n)$ divide-and-conquer algorithm You could implement so it is $\Omega(n \log n)$ instead of $O(n)$ though...