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  • 0 posts edited
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  • 31 votes cast
May
27
comment $\lim_{n\to \infty} n^{1/n^2}$
Ahh, this is much better. Thanks!
May
26
comment Bounded sequence of positive numbers
Do you mean $r $ instead of $K $?
Apr
14
comment Improper Integral with trigonometric functions
Well this integral diverges so my integral will diverge by the comparison test. Correct?
Mar
17
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?
Mar
17
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...
Mar
17
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?
Mar
17
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 $K$.
Mar
17
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?
Mar
17
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?
Feb
19
comment Consider the function
I would try a sequential argument and use the denseness of $\mathbb{R}$.
Feb
19
comment A Step in the Proof of Green's Theorem
Which limits are you talking about? Could you be more specific?
Feb
1
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.
Jan
12
comment Part of Fubini's Theorem with almost everywhere
This question was already asked math.stackexchange.com/questions/1092685/…
Jan
6
comment Tonelli and Fubini and almost everywhere
I had forgotten about this theorem. Thank you.
Jan
5
comment Folland 2.36 portion of proof
This is the conclusion I arrived at as well. Thank you!
Dec
22
comment Question about simple functions as described in Folland's Real Analysis
I'm a fool. Thank you so much.
Dec
22
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.
Oct
19
comment Different ternary representations
Oh, no I have not. I will look this up.
Oct
15
comment Kernel of cononical ring homomorphism
This helps clear things up. Thank you!
Oct
6
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 $(ax+by(\frac{D-1}{4}))+(ay+bx+by)\rho$