robjohn

West Hills, CA

none

Age: 54

the mean square The Mean Square
(with one standard deviation and several unusual ones)

aka Rob Johnson

$\LaTeX$ support for Chat

44m
comment Calculate the following sequence $\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{n!},\; \alpha >0$
@achillehui: Nice catch! Adjusting this answer yields $$\frac2{\sqrt{\pi}}\int_x^\infty e^{-t^2}\,\mathrm{d}t \sim\frac1{\sqrt{\pi}}e^{-x^2} \sum_{k=0}^\infty\frac{(-1)^k(2k)!}{4^kk!x^{2k+1}}$$ which transforms into your expansion with $x=\sqrt\alpha$.
6h
revised Calculate the following sequence $\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{n!},\; \alpha >0$
incorporate Lucian's suggestion
7h
answered Calculate the following sequence $\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{n!},\; \alpha >0$
8h
comment How to handle the complex integration of this function around a branch point
With the last correction, this looks good (+1)
8h
comment How to handle the complex integration of this function around a branch point
@RonGordon: with the correction involving multiplying by $2$ instead of dividing by $2$, I think the only difference is the branch cut. Looking at $$-\oint\frac{\mathrm{d}z}{\sqrt{z}}=-2\sqrt{z}$$ I get $4\sqrt{r}$ on my contour and $-4i\sqrt{r}$ on yours, which matches each of our answers.
12h
answered How to handle the complex integration of this function around a branch point
21h
comment Showing $f(z_0) = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0+Re^{i\theta}) \ d\theta$
@Blaris: :-) Any time!
21h
comment Showing $f(z_0) = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0+Re^{i\theta}) \ d\theta$
The $f(z)$ is not the integrand...
21h
answered Showing $f(z_0) = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0+Re^{i\theta}) \ d\theta$
21h
answered Series convergence or divergence
22h
answered convergence of Taylor series on R
1d
comment Show that there is no integer n with $\phi(n)$ = 14
@ArthurCollé: if $n=\prod\limits_ip_i^{e_i}$, where $p_i$ is prime and $e_i\gt0$, then $\phi(n)=\prod\limits_i(p_i-1)p_i^{e_i-1}$.
1d
revised Show that there is no integer n with $\phi(n)$ = 14
fix a bit
1d
comment Show that there is no integer n with $\phi(n)$ = 14
@ArthurCollé: the set of divisors of $\phi(n)$ is $\{1,2,7,14\}$, therefore, the possible set of primes that divide $n$ is $\{2,3,8,15\}$, which can be reduced to $\{2,3\}$ since neither $8$ nor $15$ are primes.
1d
answered Show that there is no integer n with $\phi(n)$ = 14
1d
revised $ \lim\limits_{x \to +\infty}x\, e^{-x^2}\int_{0}^{x}e^{t^2}dt $
simplify the argument
2d
revised $ \lim\limits_{x \to +\infty}x\, e^{-x^2}\int_{0}^{x}e^{t^2}dt $
added 90 characters in body
2d
answered $ \lim\limits_{x \to +\infty}x\, e^{-x^2}\int_{0}^{x}e^{t^2}dt $
2d
comment Does $\sum _{n=1}^{\infty } \frac{\sin(\text{ln}(n))}{n}$ converge?
@i707107: I've recently posted a proof in chat that $$\sum_{n=1}^\infty\frac{\sin(H_n)}{nH_n}$$ diverges and the proof should be almost identical for your series.
2d
revised mapping from $2^A$ to $P(A)$
improve my first rough statement
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