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The approximation can be proven several ways, and is closely related to the binomial theorem. By Bernoulli's inequality , the left-hand side of the approximation is greater than or equal to the right-hand side whenever x > − 1 {\displaystyle x>-1} and α ≥ 1 {\displaystyle \alpha \geq 1} .
Meissel already found that for k ≥ 3, P k (x, a) = 0 if a = π(x 1/3).He used the resulting equation for calculations of π(x) for big values of x. [1]Meissel calculated π(x) for values of x up to 10 9, but he narrowly missed the correct result for the biggest value of x.
Any improper rational fraction can be expressed as the sum of a polynomial (possibly constant) and a proper rational fraction. In the first example of an improper fraction one has x 3 + x 2 + 1 x 2 − 5 x + 6 = ( x + 6 ) + 24 x − 35 x 2 − 5 x + 6 , {\displaystyle {\frac {x^{3}+x^{2}+1}{x^{2}-5x+6}}=(x+6)+{\frac {24x-35}{x^{2}-5x+6}},}
Unit fractions can also be expressed using negative exponents, as in 2 −1, which represents 1/2, and 2 −2, which represents 1/(2 2) or 1/4. A dyadic fraction is a common fraction in which the denominator is a power of two, e.g. 1 / 8 = 1 / 2 3 . In Unicode, precomposed fraction characters are in the Number Forms block.
and can be found by examination of the coefficient of in the expansion of (1 + x) m (1 + x) n−m = (1 + x) n using equation . When m = 1, equation reduces to equation . In the special case n = 2m, k = m, using , the expansion becomes (as seen in Pascal's triangle at right)
By applying the fundamental recurrence formulas we may easily compute the successive convergents of this continued fraction to be 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, ..., where each successive convergent is formed by taking the numerator plus the denominator of the preceding term as the denominator in the next term, then adding in the ...
The 1/3–2/3 conjecture states that, at each step, one may choose a comparison to perform that reduces the remaining number of linear extensions by a factor of 2/3; therefore, if there are E linear extensions of the partial order given by the initial information, the sorting problem can be completed in at most log 3/2 E additional comparisons.
If the car is behind door 1, the host can open either door 2 or door 3, so the probability that the car is behind door 1 and the host opens door 3 is 1 / 3 × 1 / 2 = 1 / 6 . If the car is behind door 2 – with the player having picked door 1 – the host must open door 3, such the probability that the car is behind door ...