We consider only real numbers. The p-th root of a number N is a number b satisfying bp = N.
A real number N has p p-th roots. They are (b)e2pki/p, where i = √-1 and k = 0,1,..,p-1. If p is odd, the case k = 0 gives the one and only real root, the others being complex roots.
If p is even, there are two real roots which differ only in sign; the others being complex roots. We will be concerned here only with the real root b since the other (complex) roots are (b)e2pki/p.
Notice first that, when p = 4, we are looking at the 4-th root which is the square-root of the square-root and hence the methods for square-roots can be applied twice to obtain the approximation.
When p > 4, we have to resort to other methods. The long-division-type algorithms presented elsewhere on these web pages become complicated for finding higher roots.
However, there are a number of methods that still work. Some of these are the Bisection Method, Regula Falsi, the Secant Method, and the Newton-Raphson method.They use the function f(x) = xp - N. We will briefly describe the Newton-Raphson Method, the method of logarithms and the procedure employed to find the p-th root in calculators.
p-th roots with Newton-Raphson Method
The Newton-Raphson method finds close approximations to solutions of functional equations f(x) = 0, with the only restriction that the function must be differentiable. You will learn about these matters in calculus.
If you already had an introduction to calculus up to the Taylor formula for the approximation of functions possessing derivates of many orders, you will find useful an elementary introduction on the Newton-Raphson method.
When f(x) = xp - N, we get the formula
x1 = ((p-1)x0 + N/(x0p-1))/p
That is, to find N1/p, you substitute your guess x0 of N1/p into the formula. The value, x1, thus obtained is the new approximation x1 which you plug into the formula again and continue to obtain N1/p to greater and greater precision.
So, another way of writing the formula is
xn+1 = ((p-1)xn + N/(xnp-1))/p
You will notice that the square-root case is p = 2, and the cube-root is p = 3.
The first guess, x0, is not critical. So long as you choose a value x0 > 0, you will eventually get close to the p-th root.
p-th root via logarithms
A rule of logarithms in any base b is that log b Nt = t log b N, which holds for all real numbers t and positive real numbers b and N. Thus in common logarithms too, log 10 Nt = t log 10 N.
When t = 1/p, we have log b N1/p = (1/p) log b N.
Let us find the 5-th root of 2. We take log 21/5 which is (1/5) log 2 = (1/5) 0.3010 = 0.0602 to four decimal digits. Then 21/5 = antilog 0.0602 = 100.0602 = 1.1486..., to four decimal digits. Observe that antilog x in base 10 is 10x.
The one disadvantage of the method is clear - the approximation we get is limited by the number of digits available for logarithm and antilogarithm.
p-th root on a calculator
Scientific calculators have a key that gives p-th roots. It is the yx key. Let us find the 5th root of 2. Since the 5th root of 2 is 21/5, we take
y = 2 and x = 1/5.
Press the 1/x key.
Press the yx key.
Answer: 21/5 = 1.1486...
But how does the calculator find yx?
It uses three principles. They are:
(1) x = elnx
(2) ln(xt) = t(ln x)
(3) (antilog x) = ex
Let us apply these.
Let z = yx.
First, z = eln z.
But ln(z) = ln(yx).
Applying the second principle,
ln(yx) = x(ln(y)).
So ln(z) = x(ln(y))
Applying the third principle, antilog (ln(z)) = eln(z) = ex(ln(y))
But z = antilog (ln(z))
So z = ex(ln(y))
Thus, to find y1/p, the calculator looks for e(1/p)(ln(y)).
Let us compute 21/5 the way the calculator does it.
21/5 = e(1/5)ln2 = e(1/5)0.6931 = e0.1386 = 1.1486...
The p-th root is obtained up to the number of decimal digits available on the calculator screen.
You will notice that the way of the calculator is really the method of finding using logarithms discussed earlier where we used common logarithms as opposed to the use of natural logarithms by the caculator. Of'course, in the method of logarithms, we could have used natural logarithms using the "ln" key on the calculator. Then antilog(lnz) = ez and the two methods would coincide.
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Last updated - November 8, 2003