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What would be the example of a function for which a Secant Method fails but Bisection Method converges (to the root). In particular, if we are checking the interval $[a,b]$, then starting points for the Secant Method are $a$ and $b$.

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  • $\begingroup$ Both methods converge. Do they not? I took starting points for the Secant Method as (0,-1) and (1,1). $\endgroup$
    – Ekber
    Feb 27, 2018 at 23:26
  • $\begingroup$ it is the same as (0,-1) and (1,1) (for the Secant Method). Bisection converges for sure, since the function is continuous and changes sign in the interval [0,1]. But, Secant Method converges as well, there is no reason why it shouldn't. I don't see how it diverges with these starting points. $\endgroup$
    – Ekber
    Feb 27, 2018 at 23:43

1 Answer 1

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Try to find a continuously differentiable function with the following properties:

  • $f(a)$ and $f(b)$ have opposite signs and
  • $f'(\xi) = 0$ for a $\xi \in [a,b]$

The first point ensures that the bisection methods converges. Whereas if $f'(\xi)=0$, the secant method can fail. See these lecture notes (page 101) for an example.

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  • $\begingroup$ Do you mean $f'(a)$ or $f'(r)$? $\endgroup$
    – Ian
    Feb 28, 2018 at 16:01
  • $\begingroup$ I mean $f'(a)=0$ (or $f'(b)=0$). Functions where the derivative vanishes at the border can cause problems for the secant method. But any $f'(y)=0$ for $y \in [a,b]$ can cause problems. $\endgroup$
    – TheWaveLad
    Feb 28, 2018 at 16:03
  • $\begingroup$ Look at the figure from the lectures notes for example. There we have $f'(x_0)=0$, which in this case causes the secant method to go into the opposite direction of where the root is $\endgroup$
    – TheWaveLad
    Feb 28, 2018 at 16:07

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