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Question: Prove that if function has a root in [a,b] and no critical or inflection points in the interval, then the sequence of newton method and secant method approximate roots is a convergent sequence.

I know that the only time newtons method would not converge would be if the derivative is zero for one of the iteration terms, if there is no root to be found in the first place, or if the iterations enter a cycle and alternates back and forth between different values(I think those are the only times) As for secant method, it would not converge if when we pick our initials they are not close to the root. (not sure if there are other cases)

Sherien Hassan
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1 Answers1

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Assume that of the 4 cases, that the function is convex and monotonously increasing (the problem can always be converted to this via sign changes in argument and value).

  • Newton: Starting right of the root, the tangent root will be between function root and iteration point, thus moving step for step closer to the function root. However, starting left from the function root, the tangent root may lie outside the interval, where no guarantees of convexity are given.

  • Secant: This essentially has the same problem, if the current iteration points are left of the function root, the secant root can lie outside the interval.

If you implement the methods with a shrinking bracketing interval

  • using a bisection step if Newton fails to stay inside the interval, and
  • modifying the secant method to the false position method or Dekker's method,

respectively, you get a guarantee to find a root inside the interval, however the convergence will be slower in the cases where the original method also converges.

Lutz Lehmann
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