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

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

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

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

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

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

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

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

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

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

terenzreignz
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1there must be some sort of trick here

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

terenzreignz
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1I don't think stirling would work out.

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

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

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

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

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

RolyPoly
 one year 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
 one year 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 :) )

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

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

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

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.0Whats the question exactly? Is there a summation involved?

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

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

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

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910 @saifoo

joemath314159
 one year 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.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1Roly, how did you get your font so big.....

joemath314159
 one year 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:\[a=\frac{m}{n}, b=1\frac{m}{n}\]

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.0Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]

joemath314159
 one year 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.

RolyPoly
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1does anyone want options btw?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1I think that would be helpful for the people solving the problem to crosscheck their end result

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

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

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

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

experimentX
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1some misplaced n with x sorry about that :/
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