root sequence

x^p=a has a unique root x>0.
There exists an x_1>0 such that x_1^p>a.
Define a sequence x_n>0 by

    \[x_{n+1}=\frac{(p-1)x_n+a/x_n^{p-1}}{p}\]

    \[y=\frac{(p-1)x^p+a}{px^{p-1}}\]

When p=2,

    \[2x_{n+1}=x_n+a/x_n\]

With a=2 and x_1=3/2, x^2=17/12, and x^3=(17/12+24/17)/2. “This value was once found on a Babylonian tablet of the 18th C BC.”

Graph of y=\frac{x+2/x}{2}, the average of y=x and y=2/x:
root 2 seq

references

Roger Godement – Analysis I, p108-9

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