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A root of degree 2 is called a square root and a root of degree 3, a cube root. Roots of higher degree are referred by using ordinal numbers, as in fourth root, twentieth root, etc. The computation of an n th root is a root extraction. For example, 3 is a square root of 9, since 3 2 = 9, and −3 is also a square root of 9, since (−3) 2 = 9.
For example, for Newton's method as applied to a function f to oscillate between 0 and 1, it is only necessary that the tangent line to f at 0 intersects the x-axis at 1 and that the tangent line to f at 1 intersects the x-axis at 0. [19] This is the case, for example, if f(x) = x 3 − 2x + 2.
Thus the fraction 3 / 4 can be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division 3 ÷ 4 (three divided by four). We can also write negative fractions, which represent the opposite of a positive fraction. For example, if 1 / 2 represents a half-dollar profit, then − 1 / 2 represents ...
A continued fraction is an expression of the form = + + + + + where the a n (n > 0) are the partial numerators, the b n are the partial denominators, and the leading term b 0 is called the integer part of the continued fraction.
The simplest fraction 3 / y with a three-term expansion is 3 / 7 . A fraction 4 / y requires four terms in its greedy expansion if and only if y ≡ 1 or 17 (mod 24), for then the numerator −y mod x of the remaining fraction is 3 and the denominator is 1 (mod 6). The simplest fraction 4 / y with a four-term ...
An obvious necessary condition is that the starting fraction x / y have an odd denominator y, and it is conjectured but not known that this is also a sufficient condition. It is known [20] that every x / y with odd y has an expansion into distinct odd unit fractions, constructed using a different method than the greedy algorithm.
In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30.
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x ( x + 1)( x + 2) – namely x = 0 , x = −1 , and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.