how many corners is in a trapizoid — Preceding unsigned comment added by 24.10.95.44 (talk) 00:14, 25 January 2013 (UTC)[reply]

See trapezoid. --Trovatore (talk) 00:17, 25 January 2013 (UTC)[reply]

${\displaystyle \lim _{n\rightarrow \infty }{\frac {n^{n+1/2}}{(n-1)^{n-1/2}}}{\frac {(n/2-1)^{n-1}}{(n/2-1/2)^{n}}}}$. --AnalysisAlgebra (talk) 10:36, 25 January 2013 (UTC)[reply]

You know, I know you've said in the past that these are not homework questions, but we have no way of knowing that. If you're going to post things that look like homework questions, then please follow the procedure for asking homework questions, even if they are not. That is, show how far you've gotten, and where you're stuck. --Trovatore (talk) 10:57, 25 January 2013 (UTC)[reply]

${\displaystyle =\lim _{n\rightarrow \infty }{\frac {n^{n+1/2}}{(n-1)^{n-1/2}}}{\frac {(n-2)^{n-1}}{(n-1)^{n}}}2}$

${\displaystyle =\lim _{n\rightarrow \infty }{\frac {n^{n+1/2}(n-2)^{n-1}}{(n-1)^{2n-1/2}}}2}$

--AnalysisAlgebra (talk) 11:11, 25 January 2013 (UTC)[reply]

Is it okay to say ${\displaystyle n\approx n-1\approx n-2}$ for ${\displaystyle n\rightarrow \infty }$ and conclude ${\displaystyle \lim _{n\rightarrow \infty }{\frac {n^{n+1/2}(n-2)^{n-1}}{(n-1)^{2n-1/2}}}2=2}$? --AnalysisAlgebra (talk) 11:15, 25 January 2013 (UTC)[reply]

Informally, yes, I think so, unless I've made a mistake. If you want to give an actual proof, you should estimate the error from that approximation and show that it goes to zero. For example, you could take the logarithm of the part that you're claiming goes to 1, and show that it's bounded by some constant times 1/n. --Trovatore (talk) 11:20, 25 January 2013 (UTC)[reply]

(Wrong crap removed.. see below) (ー_ー)!! Ssscienccce bows his head in shame Ssscienccce (talk) 12:42, 26 January 2013 (UTC)[reply]

The work above is not correct. You need to make use of the fact that ${\displaystyle \lim _{n\rightarrow \infty }(1+{\frac {x}{n}})^{n}=e^{x}}$. The factor of n in the exponent means that this is not a limit of rational functions. Looie496 (talk) 17:47, 25 January 2013 (UTC)[reply]

You caught me speeding. Comes of trying to do things in one's head late at night. --Trovatore (talk) 21:52, 26 January 2013 (UTC)[reply]

Let's see... Once you got rid of halves in parentheses I'd suggest to exclude also fractions from exponents, and then subtractions, too:

${\displaystyle {\frac {n^{n+1/2}(n-2)^{n-1}}{(n-1)^{2n-1/2}}}\times 2=2\,{\frac {n^{n}\,{\sqrt {n}}\,(n-2)^{n-1}}{(n-1)^{2n}:{\sqrt {n-1}}}}=2\,{\frac {\sqrt {n(n-1)}}{n-2}}\,{\frac {n^{n}}{(n-1)^{n}}}\,{\frac {(n-2)^{n}}{(n-1)^{n}}}}$

Now divide numerators and denumerators by ${\displaystyle {\cancel {n^{2}}}\ n}$ and ${\displaystyle n^{n}}$:

${\displaystyle =2\,{\frac {\sqrt {1-{\frac {1}{n}}}}{1-{\frac {2}{n}}}}\,{\frac {1}{\left(1-{\frac {1}{n}}\right)^{n}}}\,{\frac {\left(1-{\frac {2}{n}}\right)^{n}}{\left(1-{\frac {1}{n}}\right)^{n}}}}$

As we know (see Exponential function)

${\displaystyle \exp(x)=\lim _{n\to \infty }\left(1+{\frac {x}{n}}\right)^{n},}$

so the limit of your expression is

${\displaystyle \lim _{n\to \infty }2\,{\frac {\sqrt {1-{\frac {1}{n}}}}{1-{\frac {2}{n}}}}\,{\frac {1}{\left(1-{\frac {1}{n}}\right)^{n}}}\,{\frac {\left(1-{\frac {2}{n}}\right)^{n}}{\left(1-{\frac {1}{n}}\right)^{n}}}=2\,{\frac {1}{1}}\,{\frac {1}{e}}\,{\frac {e^{2}}{e}}=2}$

CiaPan (talk) 20:39, 25 January 2013 (UTC)[reply]

If all you care about is the answer, Wolfram Alpha can do problems like this: http://www.wolframalpha.com/input/?i=limit+as+n+goes+to+infinity+of+%28n^%28n%2B1%2F2%29%2F%28n-1%29^%28n-1%2F2%29+*+%28n%2F2+-+1%29^%28n-1%29%2F%28n%2F2-1%2F2%29^n%29 Staecker (talk) 02:56, 26 January 2013 (UTC)[reply]

Assume a random population of 10,000 US residents. What is the probability that at least two of them will have the same date of birth AND the same last four digits of social security number? — Preceding unsigned comment added by 205.156.136.229 (talk) 17:53, 25 January 2013 (UTC)[reply]

That's a seriously hard problem. A back-of-the-envelope calculation yields a value around 0.5, which means that it isn't safe to make simplifying assumptions. Looie496 (talk) 18:12, 25 January 2013 (UTC)[reply]

Let's state some of those assumptions. If we assume each of the last 4 digits in the social security number is truly random (which is reasonable), then we get a 1/10⁴ probability for each set of two there. The birth date is trickier. If we assume everyone has a birthday evenly distributed over the last 100 years (a very rough approximation, neglecting the few who are over 100), then the chance of each set of two having a common birth date is 1/(100*365.25). Before proceeding further, we need to know if these assumptions are good enough, or if you want to use other assumptions. StuRat (talk) 18:07, 27 January 2013 (UTC)[reply]

To be fair, SSNs cannot end in "0000". It probably is fair to assume that the birthdate and last four digits are independent, although that isn't precisely accurate for earlier birthdates. Finally, though, the actual distribution of birthdates cannot be assumed to be uniform. Even with the simplifying assumptions, though, it's a difficult problem. — Arthur Rubin (talk) 16:17, 30 January 2013 (UTC)[reply]

See the Birthday Problem for details, but basically you want to solve the same problem, only with 9999 * 366 = 3659654 "birthdays". The probability that at least 2 people share the same birthday and last four digits of SSN is the same as 1 - p, where p is the probability that all of them have different birthdays. This assumes, of course, that all 366 birthdays and all 9999 final-four-digits are equally likely, and independent of each-other. I quickly coded the needed values in an excel spreadsheet and got p to be about .00000115, so the probability of 2 matching people is very close to 1 (99.999885%).--144.81.85.9 (talk) 20:04, 30 January 2013 (UTC)[reply]