An example by Wilkinson [1963] shows that minute alterations in the coefficients of a polynomial may have massive effects on the roots. Let
f (x) = (x − 1)(x − 2) · · · (x − 20)
which has become known as the Wilkinson polynomial. The zeros of f are, of course, the integers 1, 2, . . . , 20. Try to determine what happens to the zero r = 20 when the function is altered to f (x) − 10−8x19.
Hint: The secant method in double precision will locate a zero in the interval [20, 21].
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