Suppose $\varphi \colon [0,1] \to \Bbb R$ is three times continuously differentiable function

Question

I was looking at old exam papers and was stuck on the following problem:

Suppose $\,\,\varphi \colon [0,1] \to \Bbb R$ is three times continuously differentiable function. Suppose further that the iterates defined by $x_{n+1}=\varphi(x_n), n \ge 0$ converge to the fixed point $\xi$ of $\varphi$ . If the order of convergence is three then which of the following options is correct?

$\varphi'(\xi)=0,\,\,\varphi''(\xi)=0$

$\varphi'(\xi) \ne 0,\,\,\varphi''(\xi)=0$

$\varphi'(\xi)=0,\,\,\varphi''(\xi)\ne0$

$\varphi'(\xi)\ne 0,\,\,\varphi''(\xi)\ne 0$

Can someone point me in the right direction? Thanks in advance for your time.

I vote for option 1. Start with the Taylor polynomial of degree $2$ or $3$ at $\xi$ and the definition of order of convergence. It makes notation a bit easier to assume $\xi=0$. — Ted Shifrin, Jun 02 '13 at 15:40

score 1 · Answer 1 · answered Jun 03 '13 at 04:45

1

In order to have $\varphi(x)-\xi=O((x-\xi)^3)$, one needs the Taylor expansion of $\varphi$ at $\xi$ to be of the form $\xi+0(x-\xi)+0(x-\xi)^2+\dots$. Hence, 1) is correct

answered Jun 03 '13 at 04:45

ˈjuː.zɚ79365

3,004

Suppose $\varphi \colon [0,1] \to \Bbb R$ is three times continuously differentiable function

1 Answers1