DLS
  • DLS
Limit question()2
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

DLS
  • DLS
\[\LARGE ^{n}C_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\] where n->infinity
DLS
  • DLS
@terenzreignz @yrelhan4 @shubhamsrg
terenzreignz
  • terenzreignz
Is that a combination?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DLS
  • DLS
yo!
DLS
  • DLS
summon @ParthKohli
terenzreignz
  • terenzreignz
I'm thinking an approximation is due here?
DLS
  • DLS
what kind of?
terenzreignz
  • terenzreignz
Stirling's
terenzreignz
  • terenzreignz
I'm just tossing ideas :D
DLS
  • DLS
yeah I've heard about that :P
terenzreignz
  • terenzreignz
but it sort of resembles a binomial approximation thingy
terenzreignz
  • terenzreignz
So I'm thinking 0.
DLS
  • DLS
TBH,no freaking idea about this one. we can try stirlings
DLS
  • DLS
I have options,but 0 isn't one of them. D:
terenzreignz
  • terenzreignz
That's a shame :D okay, let's see...
terenzreignz
  • terenzreignz
\[\Large _nC_x = \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]
terenzreignz
  • terenzreignz
\[\Large = \frac{\left(\frac{n}e\right)^n}{\left(\frac{n-x}e\right)^{n-x}\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]
DLS
  • DLS
there must be some sort of trick here
terenzreignz
  • terenzreignz
Not that I've heard of :D
terenzreignz
  • terenzreignz
The e's probably cancel out... \[\Large = \frac{\left(n\right)^n}{\left(n-x\right)^{n-x}\left(x\right)^x\sqrt{2\pi n}}\]
DLS
  • DLS
I don't think stirling would work out.
terenzreignz
  • terenzreignz
Then we need a new plan :D
DLS
  • DLS
We need to summon people :O
terenzreignz
  • terenzreignz
Proceed, summoner :)
DLS
  • DLS
I don't want to further torment people :P
terenzreignz
  • terenzreignz
Well, we'll just stare mindlessly at this limit :D
DLS
  • DLS
lol xD
DLS
  • DLS
@mathslover
mathslover
  • mathslover
No idea , sorry!
mathslover
  • mathslover
@mathstudent55
DLS
  • DLS
@ParthKohli
anonymous
  • anonymous
Why is \[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]?
terenzreignz
  • terenzreignz
Approximations. It led to a dead end, pay no heed to it (unless you can make use of it, of course :) )
anonymous
  • anonymous
Why is it \[\sqrt{2 \pi n}\] for the denominator?
terenzreignz
  • terenzreignz
Stirling's approximation for factorials?
anonymous
  • anonymous
Shouldn't it be\[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi (n-x)}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi x}}\]?
anonymous
  • anonymous
Whats the question exactly? Is there a summation involved?
terenzreignz
  • terenzreignz
oh right.. typo sorry about that.
terenzreignz
  • terenzreignz
But even then, I'm not sure this was the correct way to do it :)
anonymous
  • anonymous
@joemath314159 The question is \[\LARGE \lim_{n\rightarrow \infty} \ ^nC_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\]
Jhannybean
  • Jhannybean
@jim_thompson5910 @saifoo
Jhannybean
  • Jhannybean
@saifoo.khan *
anonymous
  • anonymous
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.
Jhannybean
  • Jhannybean
Roly, how did you get your font so big.....
anonymous
  • anonymous
\[(a+b)^n=\sum_{x=0}^{n}\left(\begin{matrix}n \\ x\end{matrix}\right)a^xb^{n-x}\]let:\[a=\frac{m}{n}, b=1-\frac{m}{n}\]
anonymous
  • anonymous
Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
@bookylovingmeh Maybe you'd love to explain why you think it is two?
DLS
  • DLS
Answer is neither 0 nor 2
jim_thompson5910
  • jim_thompson5910
idk if this will help, but it might http://www.cut-the-knot.org/arithmetic/algebra/HarlanBrothers.shtml
jim_thompson5910
  • jim_thompson5910
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
DLS
  • DLS
does anyone want options btw?
Jhannybean
  • Jhannybean
I think that would be helpful for the people solving the problem to cross-check their end result
DLS
  • DLS
\[\Large \frac{m^x}{x!}.e^-m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]
DLS
  • DLS
its e raise to power -m in the first option sory about that^^
jim_thompson5910
  • jim_thompson5910
I think I found it, but I hardly understand what half of it means lol http://mathworld.wolfram.com/PoissonDistribution.html
DLS
  • DLS
cool :O
jim_thompson5910
  • jim_thompson5910
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
DLS
  • DLS
its A as well as D
DLS
  • DLS
oh yeah same thing
jim_thompson5910
  • jim_thompson5910
hmm strangely enough, yeah, A and D are equivalent
jim_thompson5910
  • jim_thompson5910
so maybe it's not A afterall
Callisto
  • Callisto
If you are talking about binomial to poisson, then http://openstudy.com/users/rolypoly#/updates/518bc047e4b062a8d1d94aa1
Callisto
  • Callisto
lol, indeed it is... Where m/n = mu / n = p in that post.
experimentX
  • experimentX
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
experimentX
  • experimentX
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 \( e^{-x} \)
DLS
  • DLS
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
DLS
  • DLS
some misplaced n with x sorry about that :/

Looking for something else?

Not the answer you are looking for? Search for more explanations.