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

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.1
summon @ParthKohli
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Stirling's
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
No idea , sorry!
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@mathstudent55
 one year ago

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

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

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

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

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

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

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

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

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

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

Jhannybean Group TitleBest ResponseYou've already chosen the best response.1
@saifoo.khan *
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.0
If x is an integer, then the answer is 0. If x is an integer, then that is one term out of the binomial theorem.
 one year ago

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

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

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

joemath314159 Group TitleBest ResponseYou've already chosen the best response.0
So 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.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
@bookylovingmeh Maybe you'd love to explain why you think it is two?
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
Answer is neither 0 nor 2
 one year ago

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

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
my 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
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
does anyone want options btw?
 one year ago

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

DLS Group TitleBest ResponseYou've already chosen the best response.1
\[\Large \frac{m^x}{x!}.e^m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]
 one year ago

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

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

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
so 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
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
its A as well as D
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
oh yeah same thing
 one year ago

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
using \( n x \approx n\) 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
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Woops!! 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 \( e^{x} \)
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
Here 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
 one year ago

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