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DLS
 3 years ago
Limit question()2
DLS
 3 years ago
Limit question()2

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DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE ^{n}C_x (\frac{m}{n})^x(1\frac{m}{n})^{nx}\] where n>infinity

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1@terenzreignz @yrelhan4 @shubhamsrg

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Is that a combination?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I'm thinking an approximation is due here?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I'm just tossing ideas :D

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1yeah I've heard about that :P

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0but it sort of resembles a binomial approximation thingy

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0So I'm thinking 0.

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1TBH,no freaking idea about this one. we can try stirlings

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I have options,but 0 isn't one of them. D:

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0That's a shame :D okay, let's see...

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large _nC_x = \frac{n!}{(nx)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{nx}e\right)^{nx}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large = \frac{\left(\frac{n}e\right)^n}{\left(\frac{nx}e\right)^{nx}\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1there must be some sort of trick here

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Not that I've heard of :D

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0The e's probably cancel out... \[\Large = \frac{\left(n\right)^n}{\left(nx\right)^{nx}\left(x\right)^x\sqrt{2\pi n}}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I don't think stirling would work out.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Then we need a new plan :D

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1We need to summon people :O

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Proceed, summoner :)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I don't want to further torment people :P

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Well, we'll just stare mindlessly at this limit :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why is \[\Large \frac{n!}{(nx)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{nx}e\right)^{nx}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Approximations. It led to a dead end, pay no heed to it (unless you can make use of it, of course :) )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why is it \[\sqrt{2 \pi n}\] for the denominator?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Stirling's approximation for factorials?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Shouldn't it be\[\Large \frac{n!}{(nx)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{nx}e\right)^{nx}\sqrt{2\pi (nx)}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi x}}\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Whats the question exactly? Is there a summation involved?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0oh right.. typo sorry about that.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0But even then, I'm not sure this was the correct way to do it :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@joemath314159 The question is \[\LARGE \lim_{n\rightarrow \infty} \ ^nC_x (\frac{m}{n})^x(1\frac{m}{n})^{nx}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @saifoo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If x is an integer, then the answer is 0. If x is an integer, then that is one term out of the binomial theorem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Roly, how did you get your font so big.....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(a+b)^n=\sum_{x=0}^{n}\left(\begin{matrix}n \\ x\end{matrix}\right)a^xb^{nx}\]let:\[a=\frac{m}{n}, b=1\frac{m}{n}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So if x is an integer, it follows that:\[\left(\frac{m}{n}+\left(1\frac{m}{n}\right)\right)^n=\sum_{x=0}^n \left(\begin{matrix}n \\ x\end{matrix}\right)\left(\frac{m}{n}\right)^n\left(1\frac{m}{n}\right)^{nx}\]\[1=\sum_{x=0}^n \left(\begin{matrix}n \\ x\end{matrix}\right)\left(\frac{m}{n}\right)^n\left(1\frac{m}{n}\right)^{nx}\]Take the limit as n goes to infinity, the summation converges (since the left hand side is 1), and if a summation converges, then its terms must converges to 0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@bookylovingmeh Maybe you'd love to explain why you think it is two?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1idk if this will help, but it might http://www.cuttheknot.org/arithmetic/algebra/HarlanBrothers.shtml

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1my guess (and again this is a guess) is that your original problem is a binomial pdf (see link below) http://en.wikipedia.org/wiki/Binomial_distribution as n> infinity, it seems like the binomial distribution is becoming more and more like the normal distribution (based on the central limit theorem) so that makes me guess that as n > infinity, the binomial pdf is slowly approaching a normal pdf (see link below) http://en.wikipedia.org/wiki/Normal_distribution

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1does anyone want options btw?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think that would be helpful for the people solving the problem to crosscheck their end result

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{m^x}{x!}.e^m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1its e raise to power m in the first option sory about that^^

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1I think I found it, but I hardly understand what half of it means lol http://mathworld.wolfram.com/PoissonDistribution.html

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1so it looks like it matches with \[\Large \frac{m^x}{x!}e^{m}\] as for the "how", not 100% sure on that, but the page will hopefully clear that up

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1hmm strangely enough, yeah, A and D are equivalent

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1so maybe it's not A afterall

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0If you are talking about binomial to poisson, then http://openstudy.com/users/rolypoly#/updates/518bc047e4b062a8d1d94aa1

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0lol, indeed it is... Where m/n = mu / n = p in that post.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1using \( n x \approx n\) sould give you \[ \frac{n!}{(nx)!x!} \left( \frac m n \right )^x \approx \frac{ m^x}{x! n^x} \cdot \frac{ \sqrt{2 \pi n}}{\sqrt{2 \pi (nx)}} \cdot \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{nx}{e} \right )^{nx}} \\ \approx \frac{ m^x}{x!n^x} \cdot \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{n}{e} \right )^{nx}} \approx \frac{ m^x}{x!} \cdot \left(\frac{1}{e} \right )^{x} \] the other part use the definiton of 'e '\[\left( 1  \frac m n \right )^{nx} \approx \left( 1  \frac m {nx} \right )^{nx} =\left( 1  \frac m {nx} \right )^{\frac{nx}{m} \cdot m } = e^{(mx)} \] multiply those and get A or D

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Woops!! there is a slight error \[ \left( 1  \frac m n \right )^{nx} \approx \left( 1  \frac m {nx} \right )^{nx} =\left( 1  \frac m {nx} \right )^{\frac{nx}{m} \cdot m } = e^{(m)} \] And \[ \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{nx}{e} \right )^{nx}} = e^{x} \cdot \frac{n^n}{(nx)^{nx}} = e^{x} \cdot \frac{n^{x}}{ \left(1  \frac x n \right )^{nx}} \\ \approx e^{x} \cdot \frac{n^x}{\left(1  \frac x {nx} \right )^{nx}} = e^{x} \cdot \frac{n^x}{e^{x}} = n^x \] The first part gives \( \frac{m^x }{x!}\) and second part gives \( e^{x} \)

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1Here is the simplified original solution \[\Large \frac{n(n1)(n2)....x~factors}{x!}. \frac{m^x}{n^x}\] \[\Large \frac{\frac{n}{n} \frac{(n1)}{n} \frac{(n2)}{n}....x~factors}{x!}.{m^x}\] The other part..which is 1^infinity \[\Large e^{(nx)(\frac{m}{n})}=>e^{(\frac{m}{n}1})\] Combine them.. \[\Large 1.1.1.1x ~factors . \frac{m^n}{x!}.e^{(m)}\] @terenzreignz and everyone :D

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1some misplaced n with x sorry about that :/
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