Here's the question you clicked on:
DLS
Limit question()2
\[\LARGE ^{n}C_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\] where n->infinity
@terenzreignz @yrelhan4 @shubhamsrg
Is that a combination?
I'm thinking an approximation is due here?
I'm just tossing ideas :D
yeah I've heard about that :P
but it sort of resembles a binomial approximation thingy
So I'm thinking 0.
TBH,no freaking idea about this one. we can try stirlings
I have options,but 0 isn't one of them. D:
That's a shame :D okay, let's see...
\[\Large _nC_x = \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]
\[\Large = \frac{\left(\frac{n}e\right)^n}{\left(\frac{n-x}e\right)^{n-x}\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]
there must be some sort of trick here
Not that I've heard of :D
The e's probably cancel out... \[\Large = \frac{\left(n\right)^n}{\left(n-x\right)^{n-x}\left(x\right)^x\sqrt{2\pi n}}\]
I don't think stirling would work out.
Then we need a new plan :D
We need to summon people :O
Proceed, summoner :)
I don't want to further torment people :P
Well, we'll just stare mindlessly at this limit :D
Why is \[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]?
Approximations. It led to a dead end, pay no heed to it (unless you can make use of it, of course :) )
Why is it \[\sqrt{2 \pi n}\] for the denominator?
Stirling's approximation for factorials?
Shouldn't it be\[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi (n-x)}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi x}}\]?
Whats the question exactly? Is there a summation involved?
oh right.. typo sorry about that.
But even then, I'm not sure this was the correct way to do it :)
@joemath314159 The question is \[\LARGE \lim_{n\rightarrow \infty} \ ^nC_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\]
@jim_thompson5910 @saifoo
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.
Roly, how did you get your font so big.....
\[(a+b)^n=\sum_{x=0}^{n}\left(\begin{matrix}n \\ x\end{matrix}\right)a^xb^{n-x}\]let:\[a=\frac{m}{n}, b=1-\frac{m}{n}\]
Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]
So if x is an integer, it follows that:\[\left(\frac{m}{n}+\left(1-\frac{m}{n}\right)\right)^n=\sum_{x=0}^n \left(\begin{matrix}n \\ x\end{matrix}\right)\left(\frac{m}{n}\right)^n\left(1-\frac{m}{n}\right)^{n-x}\]\[1=\sum_{x=0}^n \left(\begin{matrix}n \\ x\end{matrix}\right)\left(\frac{m}{n}\right)^n\left(1-\frac{m}{n}\right)^{n-x}\]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.
@bookylovingmeh Maybe you'd love to explain why you think it is two?
idk if this will help, but it might http://www.cut-the-knot.org/arithmetic/algebra/HarlanBrothers.shtml
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
does anyone want options btw?
I think that would be helpful for the people solving the problem to cross-check their end result
\[\Large \frac{m^x}{x!}.e^-m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]
its e raise to power -m in the first option sory about that^^
I think I found it, but I hardly understand what half of it means lol http://mathworld.wolfram.com/PoissonDistribution.html
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
hmm strangely enough, yeah, A and D are equivalent
so maybe it's not A afterall
If you are talking about binomial to poisson, then http://openstudy.com/users/rolypoly#/updates/518bc047e4b062a8d1d94aa1
lol, indeed it is... Where m/n = mu / n = p in that post.
using \( n- x \approx n\) sould give you \[ \frac{n!}{(n-x)!x!} \left( \frac m n \right )^x \approx \frac{ m^x}{x! n^x} \cdot \frac{ \sqrt{2 \pi n}}{\sqrt{2 \pi (n-x)}} \cdot \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{n-x}{e} \right )^{n-x}} \\ \approx \frac{ m^x}{x!n^x} \cdot \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{n}{e} \right )^{n-x}} \approx \frac{ m^x}{x!} \cdot \left(\frac{1}{e} \right )^{x} \] the other part use the definiton of 'e '\[\left( 1 - \frac m n \right )^{n-x} \approx \left( 1 - \frac m {n-x} \right )^{n-x} =\left( 1 - \frac m {n-x} \right )^{\frac{n-x}{m} \cdot m } = e^{-(m-x)} \] multiply those and get A or D
Woops!! there is a slight error \[ \left( 1 - \frac m n \right )^{n-x} \approx \left( 1 - \frac m {n-x} \right )^{n-x} =\left( 1 - \frac m {n-x} \right )^{\frac{n-x}{m} \cdot m } = e^{-(m)} \] And \[ \frac{\left(\frac{n}{e} \right )^n}{\left(\frac{n-x}{e} \right )^{n-x}} = e^{-x} \cdot \frac{n^n}{(n-x)^{n-x}} = e^{-x} \cdot \frac{n^{x}}{ \left(1 - \frac x n \right )^{n-x}} \\ \approx e^{-x} \cdot \frac{n^x}{\left(1 - \frac x {n-x} \right )^{n-x}} = e^{-x} \cdot \frac{n^x}{e^{-x}} = n^x \] The first part gives \( \frac{m^x }{x!}\) and second part gives \( e^{-x} \)
Here is the simplified original solution \[\Large \frac{n(n-1)(n-2)....x~factors}{x!}. \frac{m^x}{n^x}\] \[\Large \frac{\frac{n}{n} \frac{(n-1)}{n} \frac{(n-2)}{n}....x~factors}{x!}.{m^x}\] The other part..which is 1^infinity \[\Large e^{(n-x)(\frac{-m}{n})}=>e^{(\frac{m}{n}-1})\] Combine them.. \[\Large 1.1.1.1------x ~factors . \frac{m^n}{x!}.e^{(-m)}\] @terenzreignz and everyone :D
some misplaced n with x sorry about that :/