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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.
Trigonometric identities may help simplify the answer. [1] [2] Like other methods of integration by substitution, when evaluating a definite integral, ...
Since taking the square root is the same as raising to the power 1 / 2 , the following is also an algebraic expression: 1 − x 2 1 + x 2 {\displaystyle {\sqrt {\frac {1-x^{2}}{1+x^{2}}}}} An algebraic equation is an equation involving polynomials , for which algebraic expressions may be solutions .
Simplification is the process of replacing a mathematical expression by an equivalent one that is simpler (usually shorter), according to a well-founded ordering. Examples include:
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.
From top to bottom: x 1/8, x 1/4, x 1/2, x 1, x 2, x 4, x 8. If x is a nonnegative real number, and n is a positive integer, / or denotes the unique nonnegative real n th root of x, that is, the unique nonnegative real number y such that =.
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.).
Geometric intuition for the integral of 1/x. The three integrals from 1 to 2, from 2 to 4, and from 4 to 8 are all equal. Each region is the previous region halved vertically and doubled horizontally. Extending this, the integral from 1 to 2 k is k times the integral from 1 to 2, just as ln 2 k = k ln 2.