Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

DLSBest ResponseYou've already chosen the best response.1
@terenzreignz @yrelhan4 @shubhamsrg
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Is that a combination?
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
I'm thinking an approximation is due here?
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
I'm just tossing ideas :D
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
yeah I've heard about that :P
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
but it sort of resembles a binomial approximation thingy
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
So I'm thinking 0.
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
TBH,no freaking idea about this one. we can try stirlings
 11 months ago

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

terenzreignzBest ResponseYou've already chosen the best response.0
That's a shame :D okay, let's see...
 11 months ago

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

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

DLSBest ResponseYou've already chosen the best response.1
there must be some sort of trick here
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Not that I've heard of :D
 11 months ago

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

DLSBest ResponseYou've already chosen the best response.1
I don't think stirling would work out.
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Then we need a new plan :D
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
We need to summon people :O
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Proceed, summoner :)
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
I don't want to further torment people :P
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Well, we'll just stare mindlessly at this limit :D
 11 months ago

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

terenzreignzBest 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 :) )
 11 months ago

RolyPolyBest ResponseYou've already chosen the best response.0
Why is it \[\sqrt{2 \pi n}\] for the denominator?
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Stirling's approximation for factorials?
 11 months ago

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

joemath314159Best ResponseYou've already chosen the best response.0
Whats the question exactly? Is there a summation involved?
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
oh right.. typo sorry about that.
 11 months ago

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

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

JhannybeanBest ResponseYou've already chosen the best response.1
@jim_thompson5910 @saifoo
 11 months ago

joemath314159Best 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.
 11 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Roly, how did you get your font so big.....
 11 months ago

joemath314159Best 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:\[a=\frac{m}{n}, b=1\frac{m}{n}\]
 11 months ago

joemath314159Best ResponseYou've already chosen the best response.0
Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]
 11 months ago

joemath314159Best 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.
 11 months ago

RolyPolyBest ResponseYou've already chosen the best response.0
@bookylovingmeh Maybe you'd love to explain why you think it is two?
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
Answer is neither 0 nor 2
 11 months ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
idk if this will help, but it might http://www.cuttheknot.org/arithmetic/algebra/HarlanBrothers.shtml
 11 months ago

jim_thompson5910Best 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
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
does anyone want options btw?
 11 months ago

JhannybeanBest 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
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\Large \frac{m^x}{x!}.e^m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]
 11 months ago

DLSBest ResponseYou've already chosen the best response.1
its e raise to power m in the first option sory about that^^
 11 months ago

jim_thompson5910Best 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
 11 months ago

jim_thompson5910Best 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
 11 months ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
hmm strangely enough, yeah, A and D are equivalent
 11 months ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
so maybe it's not A afterall
 11 months ago

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

CallistoBest ResponseYou've already chosen the best response.0
lol, indeed it is... Where m/n = mu / n = p in that post.
 11 months ago

experimentXBest 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
 10 months ago

experimentXBest 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 \( e^{x} \)
 10 months ago

DLSBest 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
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
some misplaced n with x sorry about that :/
 10 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.