Nth root

The Babylonians, as early as 1800 BCE, demonstrated numerical approximations of irrational quantities such as the square root of 2 on clay tablets, with an accuracy analogous to six decimal places, as in the tablet YBC 7289. Cuneiform tablets from Larsa include tables of square and cube roots of integers. The first to prove the irrationality of √2 was most likely the Pythagorean Hippasus. Plato in his Theaetetus, then describes how Theodorus of Cyrene (c. 400 BC) proved the irrationality of \sqrt3, \sqrt5, etc. up to \sqrt{17}. In the first century AD, Heron of Alexandria devised an iterative method to compute the square root, which is actually a special case of the more general Newton's method. The term surd traces back to Al-Khwarizmi (), who referred to rational and irrational numbers as "audible" and "inaudible", respectively. This later led to the Arabic word (, meaning "deaf" or "dumb") for "irrational number" being translated into Latin as (meaning "deaf" or "mute"). Gerard of Cremona (), Fibonacci (1202), and then Robert Recorde (1551) all used the term to refer to "unresolved irrational roots", that is, expressions of the form \sqrt[n]{r}, in which n and r are integer numerals and the whole expression denotes an irrational number. Irrational numbers of the form \pm\sqrt{a}, where a is rational, are called "pure quadratic surds"; irrational numbers of the form a \pm\sqrt{b}, where a and b are rational, are called mixed quadratic surds. An archaic term from the late 15th century for the operation of taking nth roots is radication, and an unresolved root is a radical. In 1665, Isaac Newton discovered the general binomial theorem, which can convert an nth root into an infinite series. Based on approach developed by François Viète, Newton devised an iterative method for solving a non-linear function of the form f(x) = 0, which can be used to extract an nth root. This technique was further refined by Joseph Raphson and became known as the Newton–Raphson method. In 1690, Michel Rolle introduced the notation \sqrt[n]{a} for the nth root of the value a. In 1629, Albert Girard proposed the fundamental theorem of algebra, but failed to produce a proof. This theorem states that every single-variable polynomial of degree n has n roots. Further, a polynomial with complex coefficients has at least one complex root. Equivalently, the theorem states that the field of complex numbers is algebraically closed. Among the notable mathematicians who worked on a proof during the 18th and 19th centuries were d'Alembert, Gauss, Bolzano, and Weierstrass, with Gauss usually being credited with the first correct proof. A consequence of this proof is that any nth root of a real or complex number will be on the complex plane. The ancient Greek mathematicians knew how to use compass and straightedge to construct a length equal to the square root of a given length, when an auxiliary line of unit length is given. In 1837 Pierre Wantzel proved that an nth root of a given length cannot be constructed if n is not a power of 2. ==Definition and notation==

An th root of a number x, where n is a positive integer, is any of the n real or complex numbers r whose nth power is x: r^n = x. Every positive real number x has a single positive nth root, called the principal nth root, which is written \sqrt[n]{x}. For n equal to 2 this is called the principal square root and the n is omitted. The nth root can also be represented using exponentiation as x. (In the case x is real, this count includes any real nth roots.) The only complex root of 0 is 0. The nth roots of almost all numbers (all integers except the nth powers, and all rationals except the quotients of two nth powers) are irrational. and is denoted with a radical sign: \sqrt{25} = 5. Since the square of every real number is nonnegative, negative numbers do not have real square roots. ==Identities and properties==

Expressing the degree of an nth root in its exponent form, as in x^{1/n}, makes it easier to manipulate powers and roots. If a is a non-negative real number, \sqrt[n]{a^m} = (a^m)^{1/n} = a^{m/n} = (a^{1/n})^m = (\sqrt[n]a)^m. Every non-negative number has exactly one non-negative real nth root, and so the rules for operations with surds involving non-negative radicands a and b are straightforward within the real numbers: ==Simplified form of a radical expression==

A non-nested radical expression is said to be in simplified form if no factor of the radicand can be written as a power greater than or equal to the index; there are no fractions inside the radical sign; and there are no radicals in the denominator. For example, to write the radical expression \textstyle \sqrt{32/5} in simplified form, we can proceed as follows. First, look for a perfect square under the square root sign and remove it: \sqrt{\frac{32}{5}} = \sqrt{\frac{16 \cdot 2}{5}} = \sqrt{16} \cdot \sqrt{\frac{2}{5}} = 4 \sqrt{\frac{2}{5}} Next, there is a fraction under the radical sign, which we change as follows: 4 \sqrt{\frac{2}{5}} = \frac{4 \sqrt{2}}{\sqrt{5}} Finally, we remove the radical from the denominator as follows: \frac{4 \sqrt{2}}{\sqrt{5}} = \frac{4 \sqrt{2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4 \sqrt{10}}{5} = \frac{4}{5}\sqrt{10} When there is a denominator involving surds it is always possible to find a factor to multiply both numerator and denominator by to simplify the expression. For instance using the factorization of the sum of two cubes: \frac{1}{\sqrt[3]{a} + \sqrt[3]{b}} = \frac{\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}}{\left(\sqrt[3]{a} + \sqrt[3]{b}\right)\left(\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}\right)} = \frac{\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}}{a + b} . Simplifying radical expressions involving nested radicals can be quite difficult. In particular, denesting is not always possible, and when possible, it may involve advanced Galois theory. Moreover, when complete denesting is impossible, there is no general canonical form such that the equality of two numbers can be tested by simply looking at their canonical expressions. For example, it is not obvious that \sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}. The above can be derived through: \sqrt{3 + 2\sqrt{2}} = \sqrt{1 + 2\sqrt{2} + 2} = \sqrt{1^2 + 2\sqrt{2} + \sqrt{2}^2} = \sqrt{\left(1 + \sqrt{2}\right)^2} = 1 + \sqrt{2} Let r=p/q, with and coprime and positive integers. Then \sqrt[n]r = \sqrt[n]{p}/\sqrt[n]{q} is rational if and only if both \sqrt[n]{p} and \sqrt[n]{q} are integers, which means that both and are nth powers of some integer. ==Infinite series==

The radical or root may be represented by the generalized binomial theorem: (1+x)^{s/t} = \sum_{m=0}^\infty \frac{x^m}{m!} \prod_{k=0}^{m-1} \left(\frac st - k\right) with |x|. This expression can be derived from the binomial series. For the nth root, this becomes (1+x)^\frac{1}{n} = \sum_{m=0}^\infty \frac{x^m}{m!} \prod_{k=0}^{m-1} \left(\frac{1}{n} - k\right) For numbers r \ge 2, choose a value p such that \frac{r}{p^n} - 1 = x', \text{ where } |x'| then per above, solve for r^\frac{1}{n} = p (1 + x')^\frac{1}{n} As an example, for r = 30 and n = 2, choose p = 5 ==Computing principal roots==

Using Newton's method The th root of a positive real number can be computed with Newton's method, which starts with an initial guess , which is also a positive real number, and then iterates using the recurrence relation x_{k+1} = x_k-\frac{x_k^n-A}{nx_k^{n-1}} until the desired precision is reached. For computational efficiency, the recurrence relation can be rewritten For large values of n and higher requirements for precision, a more rapid algorithm than Newton's method for finding the nth root is to use a truncated Taylor series with a Padé approximant. Using the Viète technique showing P(4,1) = 4. The technique of François Viète, published c. 1600, can be used to perform digit-by-digit calculation of principal roots of decimal (base 10) numbers. This method is based on the binomial theorem and is essentially an inverse algorithm solving (10 x+y)^n = \sum_{k=0}^n P(n, k) (10 x)^{n-k} y^k where P(n, k), the binomial coefficient, is the kth entry on the nth row of Pascal's triangle. To compute the root of a number C, choose a series of approximations x_i^n, i = 0, 1, \ldots \text{ with } x_0 = 0 that satisfy x_i^n \le C, where the difference between x_{i+1} and x_i is the next digit in the approximation. The decimal fraction y_i is chosen to be the largest number with a single significant digit that satisfies 10 x_i + y_i = 10 x_{i+1}, \text{ where } x_{i+1}^n \le C then per the binomial theorem (10 x_i+y)^n - (10 x_i)^n = \sum_{k=0}^{n-1} P(n, k) (10 x_i)^{n-k} y_i^k \le 10^n(C - x_i^n) The term 10^n(C - x_i^n) is just a 10^n multiple of the ith remainder, C - x_i^n. Using this expression, any positive principal root can be computed, digit-by-digit, as follows. Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into groups of digits equating to the root being taken, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the radicand. One digit of the root will appear above each group of digits of the original number. Beginning with the left-most group of digits, do the following procedure for each group: • Starting on the left, bring down the most significant (leftmost) group of digits not yet used (if all the digits have been used, write "0" the number of times required to make a group) and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by 10^n and add the digits from the next group. This will be the 'current value c'''''. • Find p and x, as follows: • Let p be the part of the root found so far, ignoring any decimal point. (For the first step, p = 0 and 0^0 = 1). • Determine the greatest digit x such that y \le c. • Place the digit x as the next digit of the root, i.e., above the group of digits you just brought down. Thus the next p will be the old p times 10 plus x. • Subtract y from c to form a new remainder. • If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration. Examples Find the square root of 152.2756 For clarity, the value of the chosen digit x is in red while the current digital tally is in blue. Algorithm terminates: Answer is 12.34 Find the cube root of 4192 truncated to the nearest thousandth The desired precision is achieved. The cube root of 4,192 is 16.124... Logarithmic calculation The principal nth root of a positive number can be computed using logarithms. Starting from the equation that defines r as an nth root of x, namely r^n=x, with x positive and therefore its principal root r also positive, one takes logarithms of both sides (any base of the logarithm will do) to obtain n \log_b r = \log_b x \quad \quad \text{hence} \quad \quad \log_b r = \frac{\log_b x}{n}. The root r is recovered from this by taking the antilog: r = b^{\frac{1}{n}\log_b x}. (Note: That formula shows b raised to the power of the result of the division, not b multiplied by the result of the division.) For the case in which x is negative and n is odd, there is one real root r which is also negative. This can be found by first multiplying both sides of the defining equation by −1 to obtain |r|^n = |x|, then proceeding as before to find |r|, and using . ==Complex roots==

Every complex number other than 0 has n different nth roots. For example, the square roots of are and , and the square roots of are \tfrac{1}{\sqrt{2}}(1 + i) \quad\text{and}\quad -\tfrac{1}{\sqrt{2}}(1 + i). If we express a complex number in polar form, then the square root can be obtained by taking the square root of the radius and halving the angle: \sqrt{re^{i\theta}} = \pm\sqrt{r} \cdot e^{i\theta/2}. A principal root of a complex number may be chosen in various ways, for example \sqrt{re^{i\theta}} = \sqrt{r} \cdot e^{i\theta/2} which introduces a branch cut in the complex plane along the positive real axis with the condition , or along the negative real axis with . Using the first(last) branch cut the principal square root \scriptstyle \sqrt z maps \scriptstyle z to the half plane with non-negative imaginary(real) part. The last branch cut is presupposed in mathematical software like Matlab or Scilab. Roots of unity The number 1 has n different nth roots in the complex plane, \sum_{k=0}^{n-1} e^{\frac{2\pi i}{n} k} = 0 nth roots Every complex number has n different nth roots in the complex plane. z^\frac{1}{n} = \sqrt[n]{re^{i\theta}} = \sqrt[n]{r} \cdot e^{i\theta/n} = r^\frac{1}{n} \cdot \left( \cos\left(\frac{\theta}{n}\right) + i \sin\left(\frac{\theta}{n}\right) \right) Here r is the magnitude (the modulus, also called the absolute value) of the number whose root is to be taken; if the number can be written as a + i b then r=\sqrt{a^2+b^2}. The \theta is the angle formed as one pivots on the origin counterclockwise from the positive horizontal axis to a ray going from the origin to the number; it has the properties that : \cos \theta = \frac{a}{r}, \sin \theta = \frac{b}{r}, \text{ and } \tan \theta = \frac{b}{a}. Thus finding nth roots in the complex plane can be segmented into two steps. First, the magnitude of all the nth roots is the nth root of the magnitude of the original number. Second, the angle between the positive horizontal axis and a ray from the origin to one of the nth roots is \theta / n, where \theta is the angle defined in the same way for the number whose root is being taken. Furthermore, all n of the nth roots are at equally spaced angles from each other, as proven by the nth root theorem :\sqrt[n]{r} \cdot \left( \cos\left(\frac{\theta + 2\pi k}{n}\right) + i \sin\left(\frac{\theta + 2\pi k}{n}\right) \right) \text{ for } k = 0, 1, 2, \ldots, n - 1. If n is even, a complex number's nth roots, of which there are an even number, come in additive inverse pairs, so that if a number r1 is one of the nth roots then r2 = −r1 is another. This is because raising the latter's coefficient −1 to the nth power for even n yields 1: that is, (−r1) = (−1) × r1 = r1. As with square roots, the formula above does not define a continuous function over the entire complex plane, but instead has a branch cut at points where θ / n is discontinuous. ==Polynomial roots==

A root of a polynomial is a number such that . An th root of a number is by definition a root of the polynomial . Algebraic numbers are the numbers that are polynomial roots. The quadratic formula expresses the roots of quadratic polynomials in terms of square roots. During the 16th century, Gerolamo Cardano and other Italian mathematicians discovered that, similarly, the roots of the polynomials of degree 3 and 4 can always be expressed in terms of th roots (see Cubic equation and Quartic equation). During the two next centuries, a considerable effort was devoted to the question of whether every algebraic number can be expressed in terms of radicals. In 1824, the proof of the Abel–Ruffini theorem showed that there is no general formula for the degree 5. This did not completely exclude the possibility of expressing polynomial roots in terms of radicals with formulas depending on each specific polynomial. For example, the quintic polynomial : p(x) = (x - a_1)(x - a_2)(x - a_3)(x - a_4)(x - a_5) = 0 has radical roots a_1, a_2, ..., a_5. Galois theory, introduced in 1830 showed that there are polynomials of degree 5 and higher whose roots cannot be expressed in terms of radicals, the simplest example being . See and . In summary, radicals are not always sufficient for expressing polynomial roots. In spite of this obstacle, Demoivre's theorem demonstrates that an nth root of a number can always be extracted, even for a quintic function x^5 - a. It may be unclear why any number a^n has n roots rather than just a primary root. To demonstrate this, for the principal root a of the variable x taken to the nth power, the following polynomial relation holds: p(x) = x^n - a^n = 0 This polynomial can be factored as follows: \begin{align} p(x) & = x^n - a^n \\ & = (x-a) (x^{n-1} + a x^{n-2} + a^2 x^{n-3} + \cdots + a^{n-1}) \\ & = (x-a) \left( \sum_{k=0}^{n-1} x^{n-k-1} a^k \right) \\ \end{align} Thus, the polynomial p(x) is zero for x equal to a, or for any x that solves the equation: \sum_{k=0}^{n-1} x^{n-k-1} a^k = 0 By the fundamental theorem of algebra, this series has n-1 roots, for a combined total of n. As an example, let n = 3 and a = 1, then find the cube roots of 1 p(x) = x^3 - 1^3 = (x - 1) (x^2 + x^1 + 1) = 0 Thus the first root is x = 1, and the other two roots can be derived using the quadratic equation with a = b = c = 1 x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} = \frac{-1 \pm \sqrt{1^2 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2} == Proof of irrationality for non-perfect nth power x ==

Assume that \sqrt[n]{x} is rational. That is, it can be reduced to a fraction \frac{a}{b}, where and are integers without a common factor. This means that x = \frac{a^n}{b^n}. Since x is an integer, a^nand b^nmust share a common factor if b \neq 1. This means that if b \neq 1, \frac{a^n}{b^n} is not in simplest form. Thus b should equal 1. Since 1^n = 1 and \frac{n}{1} = n, \frac{a^n}{b^n} = a^n. This means that x = a^n and thus, \sqrt[n]{x} = a. This implies that \sqrt[n]{x} is an integer. Since is not a perfect th power, this is impossible. Thus \sqrt[n]{x} is irrational. ==See also==