## DLS 2 years ago Limit question()2

1. DLS

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

2. DLS

@terenzreignz @yrelhan4 @shubhamsrg

3. terenzreignz

Is that a combination?

4. DLS

yo!

5. DLS

summon @ParthKohli

6. terenzreignz

I'm thinking an approximation is due here?

7. DLS

what kind of?

8. terenzreignz

Stirling's

9. terenzreignz

I'm just tossing ideas :D

10. DLS

yeah I've heard about that :P

11. terenzreignz

but it sort of resembles a binomial approximation thingy

12. terenzreignz

So I'm thinking 0.

13. DLS

14. DLS

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

15. terenzreignz

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

16. terenzreignz

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

17. terenzreignz

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

18. DLS

there must be some sort of trick here

19. terenzreignz

Not that I've heard of :D

20. terenzreignz

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

21. DLS

I don't think stirling would work out.

22. terenzreignz

Then we need a new plan :D

23. DLS

We need to summon people :O

24. terenzreignz

Proceed, summoner :)

25. DLS

I don't want to further torment people :P

26. terenzreignz

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

27. DLS

lol xD

28. DLS

@mathslover

29. mathslover

No idea , sorry!

30. mathslover

@mathstudent55

31. DLS

@ParthKohli

32. RolyPoly

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

33. terenzreignz

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

34. RolyPoly

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

35. terenzreignz

Stirling's approximation for factorials?

36. RolyPoly

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

37. joemath314159

Whats the question exactly? Is there a summation involved?

38. terenzreignz

oh right.. typo sorry about that.

39. terenzreignz

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

40. RolyPoly

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

41. Jhannybean

@jim_thompson5910 @saifoo

42. Jhannybean

@saifoo.khan *

43. joemath314159

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.

44. Jhannybean

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

45. joemath314159

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

46. joemath314159

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

47. joemath314159

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.

48. RolyPoly

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

49. DLS

Answer is neither 0 nor 2

50. jim_thompson5910

idk if this will help, but it might http://www.cut-the-knot.org/arithmetic/algebra/HarlanBrothers.shtml

51. 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

52. DLS

does anyone want options btw?

53. Jhannybean

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

54. DLS

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

55. DLS

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

56. jim_thompson5910

I think I found it, but I hardly understand what half of it means lol http://mathworld.wolfram.com/PoissonDistribution.html

57. DLS

cool :O

58. 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

59. DLS

its A as well as D

60. DLS

oh yeah same thing

61. jim_thompson5910

hmm strangely enough, yeah, A and D are equivalent

62. jim_thompson5910

so maybe it's not A afterall

63. Callisto

64. Callisto

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

65. 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

66. 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 $$e^{-x}$$

67. 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

68. DLS

some misplaced n with x sorry about that :/