# Math Help - Numerical analysis

1. ## Numerical analysis

I want just how to solve it give me a hint

The question:-

Use fixed-point iteration method to determine a solution accurate to within $10^{-2}$ for $x^3-x-1=0$ on [1,2] use $p_{0}=1$

here is my solution (correct me if I was wrong )

first we should dtermine F(x)=x so

$x^3-x-1=0 \Rightarrow x^3=x+1 \Rightarrow x=\sqrt[3]{x+1}\Rightarrow F(x)=\sqrt[3]{x+1}$

$F'(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}$

$\mid F'(x) \mid <1$ so F(x) is converge now

$F(p_{0})=\sqrt[3]{2}=1.2599210498949=p_1$

$F(p_1)=1.3122938366833=p_2$

$F(p_2)=1.3223538191388=p_3$

..................etc

when I should stop ??
my solution is correct or it is wrong ??

Thanks for the help

2. You should stop when the first 2 digits after the coma don't change with newer iterations.
For the rest I can't help you because I have to revise my Numerical Analysis course first.

3. Why not use Newton's Iteration?

4. Originally Posted by Random Variable
Why not use Newton's Iteration?
Because the question ask to use fixed-point iteration.

5. Originally Posted by Random Variable
Why not use Newton's Iteration?
I know it but I should understand how this work to understand fixed point iteration and newton is a special function F(x)

I can use it to find F(x) but I do not know how to work with fixed point iteration in general

6. Say you decide that the answer should be $x=1.32$ correct to 2~decimal places. How can you check your answer? Use the sign change method.

If $1.32$ is correct to 2 d.p. then the root must lie in the interval $(1.315,1.325)$.

$f(x)=x^3-x-1$, $f(1.315)=-0.0411$ and $f(1.325)=0.0012$ (4 d.p.).

Since $f(x)$ changes sign and is continuous on $(1.315,1.325)$, the root is $1.32$ correct to 2 d.p.

7. Originally Posted by Amer
I want just how to solve it give me a hint

The question:-

Use fixed-point iteration method to determine a solution accurate to within $10^{-2}$ for $x^3-x-1=0$ on [1,2] use $p_{0}=1$
The iteration we will consider is generated from the rearrangement:

$x=f(x)=(1+x)^{1/3}$

then:

$x_n=f(x_{n-1})$

is our iteration.

Now we expand $f(x)$ as a Taylor series about the root:

$f(x+\varepsilon)=f(x)+\varepsilon f'(x)+ \varepsilon^2 R(x)$

where $\varepsilon^2 R(x)$ is the remainder term, and as the serier is alternating from the second term onwards we know that this is bound by the absolute value of the first neglected term: $|\varepsilon^2 f''(x)/2!|< 0.2a; \varepsilon^2$ (please justify this last inequality yourself, or something tighter if you like)

Hence as on $[1,2]$ $|\varepsilon|<1$ we have $\varepsilon^2 R(x)<0.2\;\varepsilon$

and so:

$
|f(x)-f(x+\varepsilon)|<0.5 \varepsilon.
$

Hence in the iteration the error more than halves at each pass (in fact it does better than this I leave it to the reader to find tighter bounds here).

Hence the error after $n$ passes through the iteration is less than $2^{-n}$ and we need to find $n$ such that:

$2^{-n}<10^{-2}$

this will guarantee that our error is less than the required limit, which looks like 6 iterations to me.

(in fact this is pessimistic since I have been rather generous in estimating the bounds)

CB