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 Apr14 comment Improper Integral with trigonometric functions Well this integral diverges so my integral will diverge by the comparison test. Correct? Mar17 comment Residue Theorem and Homologous to zero In your definition of being homologous to zero, if there is a point outside of $G$ that isn't so that the winding number is zero, then wouldn't that mean the the curve does not meet the hypothesis of the RT? Mar17 comment Residue Theorem and Homologous to zero This is the heart of my question I think. If $C_2$ is NOT homologous to 0, then it does not satisfy the hypothesis because in order to use the Residue Theorem, we need a curve that is homologous to 0. I apologize if I'm being stupid here... Mar17 comment Residue Theorem and Homologous to zero So this was my question; we don't consider such curves as $C_2$ for the residue theorem because it does not satisfy the hypothesis. Correct? Mar17 comment Residue Theorem and Homologous to zero Well in this picture the answer is no, you can't deform $C_2$ to a single point and still stay in $K$. Mar17 comment Residue Theorem and Homologous to zero So in this picture, $C_2$ is not homologous to 0? Then for the Residue Theorem we couldn't use a curve such as $C_2$ but can only consider curves that are either $C_1$ or curves that wrap around $H$ in such a way so that the winding number is 0, correct? Mar17 comment Residue Theorem and Homologous to zero I think I'm a little confused here. In this example $f=\frac{1}{z^2}$ and so is analytic on $\mathbb{C}-\{0\}$. Then if our contour is just the unit circle traversed once around this point, shouldn't the winding number be 1? Feb19 comment Consider the function I would try a sequential argument and use the denseness of $\mathbb{R}$. Feb19 comment A Step in the Proof of Green's Theorem Which limits are you talking about? Could you be more specific? Feb1 comment If $x_n \rightarrow 0$ and $\{y_n\}$ is a bounded sequence, then $x_ny_n \rightarrow 0$. Just be careful of your $\alpha$. It must be a point so that $\alpha \in \mathbb{R}^+$. And yes, the same $N$ works because $y_n$ is bounded independent of whether $x_n$ converges or not. Jan12 comment Part of Fubini's Theorem with almost everywhere This question was already asked math.stackexchange.com/questions/1092685/… Jan6 comment Tonelli and Fubini and almost everywhere I had forgotten about this theorem. Thank you. Jan5 comment Folland 2.36 portion of proof This is the conclusion I arrived at as well. Thank you! Dec22 comment Question about simple functions as described in Folland's Real Analysis I'm a fool. Thank you so much. Dec22 comment Question about simple functions as described in Folland's Real Analysis Yes, I understand that part of the proof. My question is why does this proof work for the more general case. I'm not seeing why $\phi$ being simple and 0 almost everywhere imply that it's integral is 0 since this is something we are trying to prove in the first place. Nov19 comment representing intervals as infinite intersectiom or union Yes, but I think my issue is I think of these as limits which I'm sure I wrong. Oct19 comment Different ternary representations Oh, no I have not. I will look this up. Oct15 comment Kernel of cononical ring homomorphism This helps clear things up. Thank you! Oct6 comment Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring Sorry, this isn't quite right what I meant was that I get $ax+ay\rho+bx\rho+by\rho^2=ax+by(\rho+\frac{D-1}{4})+(ay+bx)\rho$, which give $(ax+by(\frac{D-1}{4}))+(ay+bx+by)\rho$ Oct6 comment Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring So, $(a+b\rho)(x+y\rho)=(ax+\frac{1}{4}by(\rho+\frac{D-1}{4})+(ay+bx)\rho^2$. Then since $D\equiv 1$ $\frac{D-1}{4}$ must be an integer. And so the $by$ goes with the $\rho$ (I for some reason assumed it was already there?). Thank you!