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The result is an equation with no fractions. The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0 , a mathematical truth.
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
The odd greedy algorithm cannot terminate when given a fraction with an even denominator, because these fractions do not have finite representations with odd denominators. Therefore, in this case, it produces an infinite series expansion of its input. For instance Sylvester's sequence can be viewed as generated by the odd greedy expansion of 1/2.
For divergent continued fractions, we can distinguish three cases: The two sequences {Τ 2n−1} and {Τ 2n} might themselves define two convergent continued fractions that have two different values, x odd and x even. In this case the continued fraction defined by the sequence {Τ n} diverges by oscillation between two distinct limit points.
Periodic continued fractions are in one-to-one correspondence with the real quadratic irrationals. The correspondence is explicitly provided by Minkowski's question-mark function. That article also reviews tools that make it easy to work with such continued fractions. Consider first the purely periodic part
Denoting the two roots by r 1 and r 2 we distinguish three cases. If the discriminant is zero the fraction converges to the single root of multiplicity two. If the discriminant is not zero, and |r 1 | ≠ |r 2 |, the continued fraction converges to the root of maximum modulus (i.e., to the root with the greater absolute value).
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
To find the number of negative roots, change the signs of the coefficients of the terms with odd exponents, i.e., apply Descartes' rule of signs to the polynomial = + + This polynomial has two sign changes, as the sequence of signs is (−, +, +, −) , meaning that this second polynomial has two or zero positive roots; thus the original ...