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6 1 2 1 1 −1 4 5 9. and would be written in modern notation as 6 1 / 4 , 1 1 / 5 , and 2 − 1 / 9 (i.e., 1 8 / 9 ). The horizontal fraction bar is first attested in the work of Al-Hassār (fl. 1200), a Muslim mathematician from Fez, Morocco, who specialized in Islamic inheritance jurisprudence.
1/4 + 1/16 + 1/64 + 1/256 + ⋯. In mathematics, the infinite series 1 4 + 1 16 + 1 64 + 1 256 + ⋯ is an example of one of the first infinite series to be summed in the history of mathematics; it was used by Archimedes circa 250–200 BC. [1] As it is a geometric series with first term 1 4 and common ...
Calculus. In mathematics, the harmonic series is the infinite series formed by summing all positive unit fractions : The first terms of the series sum to approximately , where is the natural logarithm and is the Euler–Mascheroni constant. Because the logarithm has arbitrarily large values, the harmonic series does not have a finite limit: it ...
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.
1/2 + 1/4 + 1/8 + 1/16 + ⋯. First six summands drawn as portions of a square. The geometric series on the real line. In mathematics, the infinite series 1 2 + 1 4 + 1 8 + 1 16 + ··· is an elementary example of a geometric series that converges absolutely. The sum of the series is 1. In summation notation ...
It is unknown whether these constants are transcendental in general, but Γ( 1 / 3 ) and Γ( 1 / 4 ) were shown to be transcendental by G. V. Chudnovsky. Γ( 1 / 4 ) / 4 √ π has also long been known to be transcendental, and Yuri Nesterenko proved in 1996 that Γ( 1 / 4 ), π, and e π are algebraically independent.
The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc.The nth partial sum is given by a simple formula: = = (+). This equation was known ...
However, even Ahmes' answer here is inconsistent with the problem's other information. Happily the context of 51 and 52, together with the base, mid-line, and smaller triangle area (which are given as 4 + 1/2, 2 + 1/4 and 7 + 1/2 + 1/4 + 1/8, respectively) make it possible to interpret the problem and its solution as has been done here. The ...