# Fixed point iteration problem of $f(u)=u^3-u-1$

I was thinking about the following problem:

Let $f(u)=u^3-u-1$. Then I have to verify whether the following statements are true/false?

1.Starting with the initial guess $u^{(0)}=1.5,$ the fixed point iterates of the equation $u=g(u)$,where $g(u)=u^3-1$ converges

2.If $u^{\star}$ is a root of the equation $f(u)=0$ and $u^{\star} >1,$ then $u^{\star}$ is a stable fixed point of the equation $u=g(u)$

3.$f(u)=0$ has a root between $1$ and $2$

4.Staring with the initial guess $u^{(0)}=1.5,$ the fixed point iterates of the equation $u=\tilde{g}(u),\text{where} \space \tilde{g}(u)=\sqrt{1+u^3}$ converge.

Here,option 3 is true as we see that $f(1)<0 \space \text{whereas} f(2)>0$ .But I am not sure about the other options .How can I check other options?

Can someone point me in the right direction?

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For part 4, did you really want it like that or sqrt(u^3 - 1)? –  Amzoti May 16 '13 at 5:07

Hints

• There is a root at: $u = 1.3247179572447460260$ and the other two roots are imaginary.

Part 1.

Here are the iterates. What do you notice is happening from iterate to iterate?

• $u_1 = 2.375$
• $u_2 = 12.396484375$
• $u_3 = 1904.00277223$
• $u_4 = 6902441412.84$
• $u_5 = 3.28857830393E+29$
• $u_6 = 54608.3924665$
• $\ldots$
• $u_{40} = 5153.0115226$

So, what can you say about convergence?

Part 2

You just need to think about this one a bit (go back and look at the FP theory.

Part 3

Correct. We could have also plotted the function to see:

Part 4

• $u_1 = 2.09165006634$
• $u_2 = 3.18605854313$
• $u_3 = 5.77421697131$
• $u_4 = 13.9112014761$
• $u_5 = 51.8952497366$
• $u_6 = 373.846193564$
• $\ldots$
• $u_{40} = 1.5659769437E+15$

So, what can you say about convergence?

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Great work, Amzoti! +1 –  amWhy May 17 '13 at 0:26
@amWhy: Not sure the OP got it. Regards –  Amzoti May 17 '13 at 0:52

There is a condition you must check to verify that the $g(x)$ you've chosen converge:

$$|g'(u)| \leq \lambda <1$$

As for the suggested $g(x)$, apparently both of them diverge.

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