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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. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.
One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
The factor (2 −1) is a right arithmetic shift, a (0) results in no operation (since 2 0 = 1 is the multiplicative identity element), and a (2 1) results in a left arithmetic shift. The multiplication product can now be quickly calculated using only arithmetic shift operations, addition and subtraction.
D 1 is x + 1; set it equal to zero. This gives the residue for A when x = −1. Next, substitute this value of x into the fractional expression, but without D 1. Put this value down as the value of A. Proceed similarly for B and C. D 2 is x + 2; For the residue B use x = −2. D 3 is x + 3; For residue C use x = −3.
A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
The Barth surface, shown in the figure is the geometric representation of the solutions of a polynomial system reduced to a single equation of degree 6 in 3 variables. Some of its numerous singular points are visible on the image. They are the solutions of a system of 4 equations of degree 5 in 3 variables.
As another example, in radix 5, a string of digits such as 132 denotes the (decimal) number 1 × 5 2 + 3 × 5 1 + 2 × 5 0 = 42. This representation is unique. Let b be a positive integer greater than 1. Then every positive integer a can be expressed uniquely in the form
The solution = is in fact a valid solution to the original equation; but the other solution, =, has disappeared. The problem is that we divided both sides by x {\displaystyle x} , which involves the indeterminate operation of dividing by zero when x = 0. {\displaystyle x=0.}