MTH3051 Introduction to Computational Mathematics - User Web ...

School of Mathematical Sciences
Monash University
Yes, this is correct, it does converge in one iteration. But do not think that all functions
with a double root will converge so quickly. Here is another example, based on f(x) =
x 3 − 3x + 2 which has a double root at x = 1.
Newton-Raphson iterations x n+1 = x n − 2f n /f ′ n, f(x) = x 3 − 3x + 2
Iteration n Old guess x n New guess x n+1 (x n+1 − 1) 2 ɛ n+1 /ɛ 2 n
0 1.200000000000 1.280e-01
1 1.200000000000 1.006060606061 1.104e-04 1.515e-01
2 1.006060606061 1.000006103329 1.118e-10 1.662e-01
3 1.000006103329 1.000000000004 -2.220e-16 9.683e-02
Example 4.5
Repeat the above calculations for 7 iterations. What do you notice? Can you explain
this?
Example 4.6
Can you write out a formula for ɛ n as a function of n for the double root (easy) and for
the simple root (not so easy – hint, use the above tables and write out successive ɛ n ’s
and express each in terms of ɛ 1 ).
Example 4.7
For multiple roots we could apply the Newton-Raphson method to the function g(x) =
f m−1 (x), i.e the (m − 1) st derivative of f(x). What do you think would be the pros and
cons of this approach? (Hint – two words : efficiency, round-off).
4.3.3 Cycling
If we are unlucky we will find that the Newton-Raphson may cycle, just as we saw with
the fixed point algorithm.
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