# Approximation of Roots

Christoph Reineke
Dortmund, Germany
April 2010

For sufficiently close a and b,

\frac{b^\frac{1}{n}}{a^{\frac{1}{n}}}\approx\frac{(n-1)a + (n+1)b}{(n-1)b + (n+1)a}

Proof

By the trapezoidal rule,

\int^{b}_{a}{x^\frac{1}{n}}dx \approx \frac{a^{\frac{1}{n}} + b^{\frac{1}{n}}}{2}(b-a),

n,a,b \in R, b>a, b\approx{a}. The error is in the order \frac{1-n}{n^{2}}{b}^{(1-2n) / {n^{2}}} (b-a). We have successively

\frac{n}{n+1}b^{\frac{n+1}{n}}-\frac{n}{n+1}a^{\frac{n+1}{n}} \approx \frac{ba^{\frac{1}{n}}-a^{\frac{n+1}{n}}+b^{\frac{n+1}{n}}-ab^{\frac{1}{n}}}{2}
\frac{2n}{n+1}b^{\frac{n+1}{n}}-b^{\frac{n+1}{n}}-\frac{2n}{n+1}a^{\frac{n+1}{n}}+a^{\frac{n+1}{n}} \approx ba^{\frac{1}{n}}-ab^{\frac{1}{n}}
\frac{n-1}{n+1}b^{\frac{n+1}{n}}-\frac{n-1}{n+1}a^{\frac{n+1}{n}} \approx ba^{\frac{1}{n}}-ab^{\frac{1}{n}}
\frac{n-1}{n+1}b^{\frac{n+1}{n}}+ab^{\frac{1}{n}} \approx \frac{n-1}{n+1}a^{\frac{n+1}{n}}+ba^{\frac{1}{n}}
b^{\frac{1}{n}}(\frac{n-1}{n+1}{b}+a) \approx a^{\frac{1}{n}}(\frac{n-1}{n+1}{a}+b)
\frac{b^\frac{1}{n}}{a^{\frac{1}{n}}}\approx\frac{(n-1)a + (n+1)b}{(n-1)b + (n+1)a}