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DLS

  • 2 years ago

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  1. DLS
    • 2 years ago
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    \[\LARGE ^{n}C_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\] where n->infinity

  2. DLS
    • 2 years ago
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    @terenzreignz @yrelhan4 @shubhamsrg

  3. terenzreignz
    • 2 years ago
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    Is that a combination?

  4. DLS
    • 2 years ago
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    yo!

  5. DLS
    • 2 years ago
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    summon @ParthKohli

  6. terenzreignz
    • 2 years ago
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    I'm thinking an approximation is due here?

  7. DLS
    • 2 years ago
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    what kind of?

  8. terenzreignz
    • 2 years ago
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    Stirling's

  9. terenzreignz
    • 2 years ago
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    I'm just tossing ideas :D

  10. DLS
    • 2 years ago
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    yeah I've heard about that :P

  11. terenzreignz
    • 2 years ago
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    but it sort of resembles a binomial approximation thingy

  12. terenzreignz
    • 2 years ago
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    So I'm thinking 0.

  13. DLS
    • 2 years ago
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    TBH,no freaking idea about this one. we can try stirlings

  14. DLS
    • 2 years ago
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    I have options,but 0 isn't one of them. D:

  15. terenzreignz
    • 2 years ago
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    That's a shame :D okay, let's see...

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

  17. terenzreignz
    • 2 years ago
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    \[\Large = \frac{\left(\frac{n}e\right)^n}{\left(\frac{n-x}e\right)^{n-x}\left(\frac{x}e\right)^x\sqrt{2\pi n}}\]

  18. DLS
    • 2 years ago
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    there must be some sort of trick here

  19. terenzreignz
    • 2 years ago
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    Not that I've heard of :D

  20. terenzreignz
    • 2 years ago
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    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
    • 2 years ago
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    I don't think stirling would work out.

  22. terenzreignz
    • 2 years ago
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    Then we need a new plan :D

  23. DLS
    • 2 years ago
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    We need to summon people :O

  24. terenzreignz
    • 2 years ago
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    Proceed, summoner :)

  25. DLS
    • 2 years ago
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    I don't want to further torment people :P

  26. terenzreignz
    • 2 years ago
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    Well, we'll just stare mindlessly at this limit :D

  27. DLS
    • 2 years ago
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    lol xD

  28. DLS
    • 2 years ago
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    @mathslover

  29. mathslover
    • 2 years ago
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    No idea , sorry!

  30. mathslover
    • 2 years ago
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    @mathstudent55

  31. DLS
    • 2 years ago
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    @ParthKohli

  32. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    Approximations. It led to a dead end, pay no heed to it (unless you can make use of it, of course :) )

  34. RolyPoly
    • 2 years ago
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    Why is it \[\sqrt{2 \pi n}\] for the denominator?

  35. terenzreignz
    • 2 years ago
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    Stirling's approximation for factorials?

  36. RolyPoly
    • 2 years ago
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    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
    • 2 years ago
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    Whats the question exactly? Is there a summation involved?

  38. terenzreignz
    • 2 years ago
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    oh right.. typo sorry about that.

  39. terenzreignz
    • 2 years ago
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    But even then, I'm not sure this was the correct way to do it :)

  40. RolyPoly
    • 2 years ago
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    @joemath314159 The question is \[\LARGE \lim_{n\rightarrow \infty} \ ^nC_x (\frac{m}{n})^x(1-\frac{m}{n})^{n-x}\]

  41. Jhannybean
    • 2 years ago
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    @jim_thompson5910 @saifoo

  42. Jhannybean
    • 2 years ago
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    @saifoo.khan *

  43. joemath314159
    • 2 years ago
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    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
    • 2 years ago
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    Roly, how did you get your font so big.....

  45. joemath314159
    • 2 years ago
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    \[(a+b)^n=\sum_{x=0}^{n}\left(\begin{matrix}n \\ x\end{matrix}\right)a^xb^{n-x}\]let:\[a=\frac{m}{n}, b=1-\frac{m}{n}\]

  46. joemath314159
    • 2 years ago
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    Your nCx is my:\[\left(\begin{matrix}n \\ x\end{matrix}\right)\]

  47. joemath314159
    • 2 years ago
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    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
    • 2 years ago
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    @bookylovingmeh Maybe you'd love to explain why you think it is two?

  49. DLS
    • 2 years ago
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    Answer is neither 0 nor 2

  50. jim_thompson5910
    • 2 years ago
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    idk if this will help, but it might http://www.cut-the-knot.org/arithmetic/algebra/HarlanBrothers.shtml

  51. jim_thompson5910
    • 2 years ago
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    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
    • 2 years ago
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    does anyone want options btw?

  53. Jhannybean
    • 2 years ago
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    I think that would be helpful for the people solving the problem to cross-check their end result

  54. DLS
    • 2 years ago
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    \[\Large \frac{m^x}{x!}.e^-m\] \[\Large \frac{m^x}{x!}.e^m\] \[\Large e^0\] \[\Large \frac{m^{x+1}}{me^m x!}\]

  55. DLS
    • 2 years ago
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    its e raise to power -m in the first option sory about that^^

  56. jim_thompson5910
    • 2 years ago
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    I think I found it, but I hardly understand what half of it means lol http://mathworld.wolfram.com/PoissonDistribution.html

  57. DLS
    • 2 years ago
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    cool :O

  58. jim_thompson5910
    • 2 years ago
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    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
    • 2 years ago
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    its A as well as D

  60. DLS
    • 2 years ago
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    oh yeah same thing

  61. jim_thompson5910
    • 2 years ago
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    hmm strangely enough, yeah, A and D are equivalent

  62. jim_thompson5910
    • 2 years ago
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    so maybe it's not A afterall

  63. Callisto
    • 2 years ago
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    If you are talking about binomial to poisson, then http://openstudy.com/users/rolypoly#/updates/518bc047e4b062a8d1d94aa1

  64. Callisto
    • 2 years ago
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    lol, indeed it is... Where m/n = mu / n = p in that post.

  65. experimentX
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    some misplaced n with x sorry about that :/

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