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 Post subject: Calculus (1)
PostPosted: Mon, 5 Feb 2007 05:12:38 UTC 
Evaluate $ \lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}.


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 Post subject: Re: Calculus (1)
PostPosted: Mon, 5 Feb 2007 20:50:14 UTC 
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commutative wrote:
Evaluate $ \lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}.


I may be falling into a trap, but isn't it just 1?


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 Post subject: Re: Calculus (1)
PostPosted: Mon, 5 Feb 2007 22:58:12 UTC 
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Zathras wrote:
I may be falling into a trap, but isn't it just 1?


It might depend on how you try to do it. There are plenty of ways to get correct results with the wrong method. So I suggest you post why you think it is one.

Personally, I'm not sure what this limit is (or even if it exists). It does look like a tricky little booger. Thanks for the nice problem commutative.

-Q


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 Post subject: Re: Calculus (1)
PostPosted: Mon, 5 Feb 2007 23:26:54 UTC 
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$ \lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=\lim_{n\to\infty}\frac{\sum_{k=0}^n\frac{n^k}{k!}}{e^{n}}.

Since the top approaches e^{n}, the limit is 1.

(Using that \sum_{k=0}^\infty\frac{n^k}{k!}=e^n)


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 Post subject:
PostPosted: Tue, 6 Feb 2007 00:11:50 UTC 
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Edit: Damn, I had some wonderful maths posted, but there was an mixup with some less thans and greater thans...

Anyway, the answer cannot be 1, as the sequence is decreasing after about n=4 (and is trivially bounded above by 1). I'll have to see if I can repair my proof...


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 Post subject:
PostPosted: Tue, 6 Feb 2007 06:53:38 UTC 
yeah Zathras, it is a trap! hehehe ... you are right qleak, it
is a nice problem. the limit exists and, as Beardy mentioned,
is not 1. it is a "nice" number between 0 and 1. :wink:


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 Post subject:
PostPosted: Tue, 6 Feb 2007 16:53:07 UTC 
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commutative wrote:
it is a "nice" number between 0 and 1. :wink:


Indeed it is a very nice number.

Spoiler:
The sum
$ \sum_{k=0}^n\frac{e^{-n}n^k}{k!}
is the probability of a sum of n independent Poisson(1) random variables attaining value at most n (since independent Poisson random variables add by Poisson(a)+Poisson(b)=Poisson(a+b)). So it is the probability of the mean of n Poisson(1) r.v.s is at most 1, which central limit theorem tells us the limit is \frac{1}{2}.


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 Post subject:
PostPosted: Wed, 7 Feb 2007 00:33:55 UTC 
kommodorekerz, $ \frac12 is the answer indeed, and the probabilistic solution
you gave is very nice! it would also be nice to see a calculus approach to
this problem, which, by the way, is not that complicated!


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 Post subject:
PostPosted: Thu, 8 Feb 2007 20:39:23 UTC 
Not a solution but a question.

Spoiler:
Does this non-complicated solution deal with cauchy products? :oops:


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 Post subject:
PostPosted: Mon, 5 Mar 2007 23:24:01 UTC 
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Perhaps n's presence in both the summation and the index has something to do with this?

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 Post subject:
PostPosted: Tue, 5 Jun 2007 07:01:23 UTC 
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My question is a request of the original poster for the calculus solution in some form (hide/show or PM I'm good either way). I have repeatedly assaulted this, but to no avail and I am extremely interested in how the calculus proof is done. This is greatly appreciated, thanks!

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 Post subject:
PostPosted: Wed, 6 Jun 2007 22:35:03 UTC 
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I have also just noticed that commutative is no longer a member here, does anyone perchance know a way I might find him to see if he is willing to assist me in my search? Perhaps another website he is still a part of? Any and all help is greatly appreciated. Thanks!

Shadow

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 Post subject:
PostPosted: Wed, 6 Jun 2007 23:14:36 UTC 
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Well shadow, I searched for the post, but apparently I mix what helmut tells me and what he posts, I do recall him saying that commutative asked to have his account deleted, because he was spending too much time on sosmath, and was becoming too busy :/

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 Post subject: Re: Calculus (1)
PostPosted: Tue, 20 Dec 2011 23:20:34 UTC 
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I'm bumping this in hopes that someone knows the calculus proof since it recently resurfaced in my memory due to another topic asking for the probabilistic solution. This has bothered me for years.

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 Post subject: Re: Calculus (1)
PostPosted: Wed, 21 Dec 2011 00:19:05 UTC 
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Shadow wrote:
I'm bumping this in hopes that someone knows the calculus proof since it recently resurfaced in my memory due to another topic asking for the probabilistic solution. This has bothered me for years.


I can't think of an easy calculus argument. Obviously we want the limit of normalised incomplete gamma function Q(n+1,n)=\dfrac{\Gamma(n+1,n)}{\Gamma(n+1)}. We have the well-known asymptotic expansion of Temme
\begin{aligned}
Q(a,z)&=\mathop{\mathrm{erfc}}(\eta\sqrt{a/2})+R_a(\eta)\\
\frac{1}{2}\eta^2&=\lambda-1-\log\lambda\\
\lambda&=z/a\\
R_a(\eta)&=\frac{e^{-\frac{1}{2}a\eta^2}}{\sqrt{2\pi a}} S_a(\eta)\\
S_a(\eta)&\sim\sum_{n=0}^\infty\dfrac{C_n(\eta)}{a^n}
\end{aligned}
(the sign of \eta is chosen so that \eta>0 for \lambda>1) obtained by saddle point method. The first term of the asymptotic series is \mathop{\mathrm{erfc}}(O(n^{-1/2}))=\dfrac{1}{2}+O(n^{-1/2}).

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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