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My Professor told the class that we could earn extra credit on the test by finding out what the Chebyshev Theorem formula was and how to use it. I'm trying to get as many extra credit points as I can since you never know what's going to come up in the future.

2007-09-10 14:33:14 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

Wiki says Chebyshev's Theorem is seperated into three theorems:

- Bertrand's postulate
- Chebyshev's inequality
- Chebyshev's sum inequality

Where: The statement that if the function [pi(x)ln(x)]/x has a limit at infinity, then the limit is 1 (where pi is the prime counting function). This result has been superseded by the Prime Number Theorem.

So, just find a way to prove that lim [pi(x)ln(x)]/x = 1
http://en.wikipedia.org/wiki/Prime_Number_Theorem

Reply if you need help in understanding the proof.

2007-09-10 14:42:40 · answer #1 · answered by Anonymous · 0 0

???
Chebyshev's Theorem? I'm guessing that you mean Chebyshev's Inequality.
It's used to provide an upper bound on the probability that a random variable will exceed some given deviance from it's mean value.
Lots of sites on the web have information on it.

HTH

Doug

2007-09-10 14:42:57 · answer #2 · answered by doug_donaghue 7 · 0 0

a minimum of a million-a million/ok^2 of the observations could fall interior of two familiar deviations of the mean. a million-a million/ok^2 = .seventy 5 -a million/ok^2 =-.25 a million/ok^2=.25 ok^2 = a million/.25 = 4 ok = +/-2 we choose 2 familiar deviations of the mean (the two area) one hundred fifty-2(15) = one hundred twenty one hundred fifty+2(15) =a hundred and eighty a minimum of seventy 5% of observations will fall between one hundred twenty and a hundred and eighty.

2016-11-14 21:48:31 · answer #3 · answered by gracely 4 · 0 0

See this website

2007-09-10 14:43:02 · answer #4 · answered by frank 7 · 0 0

Is it what you look for?

This formulat gives you approximate number of prime numbers below x.

2007-09-10 14:44:59 · answer #5 · answered by Alexey V 5 · 0 0

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