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Mar 1	revised	Is the set of integers with respect to the p-adic metric compact? added 4 characters in body
Mar 1	asked	Is the set of integers with respect to the p-adic metric compact?
Feb 22	accepted	Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}\|q\|=1$
Feb 22	comment	Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}\|q\|=1$ I do! =] ok thanks everyone for the quick responses.
Feb 22	comment	Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}\|q\|=1$ Ohh ok! I get it, so it's for a fixed rational, and it's the absolute value that is running through all primes, is that correct?
Feb 22	revised	Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}\|q\|=1$ added 170 characters in body
Feb 22	comment	Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}\|q\|=1$ @git gud thanks good tip
Feb 22	asked	Prove that $\displaystyle\prod_{q\in \mathbb{Q}^{\times}}\|q\|=1$
Feb 19	accepted	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$
Feb 17	answered	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$
Feb 17	comment	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ @WillJagy That's probably the same contour I'm using, it is a parallelogram(rhombus), not a rectangle. It's definitely doable anyways, since the contour is given to me explicitly in my book.
Feb 17	revised	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ deleted 2 characters in body
Feb 17	comment	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ Ok that makes sense
Feb 17	revised	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ edited title
Feb 17	comment	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ Something I'm not completely confident about is are these poles in fact of order 1?
Feb 17	revised	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ added 1 characters in body
Feb 17	comment	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ Yes but $a=\sqrt{\pi}e^{\frac{\pi i}{4}}$, so bringing it up to the numerator and making the powers negative simplifies to $a$ up above, since $i=e^{\frac{\pi i}{2}}$. Or I guess simpler would be just to refer to my note that $\pi i=a^2$.
Feb 17	comment	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$ The only thing different between mine and yours is you have an extra factor of 2 in the denominator, and I think that might have been a mistake on your part.
Feb 17	asked	Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$
Feb 15	accepted	The Group of points on the Elliptic curve $y^2=x^3+1$ over $\mathbb{F}_5$