Question

Limit question()2

Answer 1

\[\LARGE ^{n}C_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\]
where n->infinity

Answer 2

@terenzreignz @yrelhan4 @shubhamsrg

Answer 3

Is that a combination?

Answer 4

yo!

Answer 5

summon @ParthKohli

Answer 6

I'm thinking an approximation is due here?

Answer 7

what kind of?

Answer 8

Stirling's

Answer 9

I'm just tossing ideas :D

Answer 10

yeah I've heard about that :P

Answer 11

but it sort of resembles a binomial approximation thingy

Answer 12

So I'm thinking 0.

Answer 13

TBH,no freaking idea about this one.
we can try stirlings

Answer 14

I have options,but 0 isn't one of them. D:

Answer 15

That's a shame :D okay, let's see...

Answer 16

\[\Large _nC_x = \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]

Answer 17

\[\Large = \frac{\left(\frac{n}e\right)^n}{\left(\frac{n-x}e\right)^{n-x}\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]

Answer 18

there must be some sort of trick here

Answer 19

Not that I've heard of :D

Answer 20

The e's probably cancel out...
\[\Large = \frac{\left(n\right)^n}{\left(n-x\right)^{n-x}\left(x\right)^x\sqrt{2\pi n}}\]

Answer 21

I don't think stirling would work out.

Answer 22

Then we need a new plan :D

Answer 23

We need to summon people :O

Answer 24

Proceed, summoner :)

Answer 25

I don't want to further torment people :P

Answer 26

Well, we'll just stare mindlessly at this limit :D

Answer 27

lol xD

Answer 28

@mathslover

Answer 29

No idea , sorry!

Answer 30

@mathstudent55

Answer 31

@ParthKohli

Answer 32

Why is \[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi n}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]?

Answer 33

Approximations. It led to a dead end, pay no heed to it (unless you can make use of it, of course :) )

Answer 34

Why is it \[\sqrt{2 \pi n}\] for the denominator?

Answer 35

Stirling's approximation for factorials?

Answer 36

Shouldn't it be\[\Large \frac{n!}{(n-x)!x!}\approx\frac{\left(\frac{n}e\right)^n\sqrt{2\pi n}}{\left(\frac{n-x}e\right)^{n-x}\sqrt{2\pi (n-x)}\cdot\left(\frac{x}e\right)^x\sqrt{2\pi x}}\]?

Answer 37

Whats the question exactly? Is there a summation involved?

Answer 38

oh right.. typo sorry about that.

Answer 39

But even then, I'm not sure this was the correct way to do it :)

Answer 40

@joemath314159 The question is
\[\LARGE \lim_{n\rightarrow \infty} \ ^nC_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\]

Answer 41

@jim_thompson5910 @saifoo

Answer 42

@saifoo.khan *

Answer 43

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.

Answer 44

Roly, how did you get your font so big.....

Answer 45

\[(a+b)^n=\sum_{x=0}^{n}\left(\begin{matrix}n \\ x\end{matrix}\right)a^xb^{n-x}\]let:\[a=\frac{m}{n}, b=1-\frac{m}{n}\]

Answer 46

Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]

Answer 47

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.

Answer 48

@bookylovingmeh Maybe you'd love to explain why you think it is two?

Answer 49

Answer is neither 0 nor 2

Answer 50

idk if this will help, but it might

http://www.cut-the-knot.org/arithmetic/algebra/HarlanBrothers.shtml

Answer 51

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

Answer 52

does anyone want options btw?

Answer 53

I think that would be helpful for the people solving the problem to cross-check their end result

Answer 54

\[\Large \frac{m^x}{x!}.e^-m\]
\[\Large \frac{m^x}{x!}.e^m\]
\[\Large e^0\]
\[\Large \frac{m^{x+1}}{me^m x!}\]

Answer 55

its e raise to power -m in the first option sory about that^^

Answer 56

I think I found it, but I hardly understand what half of it means lol

http://mathworld.wolfram.com/PoissonDistribution.html

Answer 57

cool :O

Answer 58

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

Answer 59

its A as well as D

Answer 60

oh yeah same thing

Answer 61

hmm strangely enough, yeah, A and D are equivalent

Answer 62

so maybe it's not A afterall

Answer 63

If you are talking about binomial to poisson, then
http://openstudy.com/users/rolypoly#/updates/518bc047e4b062a8d1d94aa1

Answer 64

lol, indeed it is...
Where m/n = mu / n = p in that post.

Answer 65

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

Answer 66

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 \( e^{-x} \)

Answer 67

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

Answer 68

some misplaced n with x sorry about that :/