- #1
evinda
Gold Member
MHB
- 3,741
- 0
Hello! ;) I have a question.
Let $\varphi:[-1,1] \to [-1,1]$ with $L=0.8$ at $[-1,1]$, $\varphi$ has a unique fixed point $x^{*}$ and the sequence $(x_{n})$ with $x_{n+1}=\varphi(x_{n}) ,n=0,1,2,...$ is well defined and coverges to $x^{*}$ for any $x_{0} \in [-1,1]$.Then if the 9th approximation is $x_{9}=0.37282$ and the 10th is $x_{10}=0.37382$,we can say for sure that:
1) $|x_{10}-x^{*}|<0.005$
2) $|x_{10}-x^{*}|<0.001$
3) $|x_{10}-x^{*}|<0.002$
I used the formula $|x_{10}-x^{*}|\leq\frac{L}{1-L}|x_{n}-x_{n-1}|$ and found that $|x_{10}-x^{*}|\leq 0.004$.Is this right?So,is 1) the right answer? (Thinking)
Let $\varphi:[-1,1] \to [-1,1]$ with $L=0.8$ at $[-1,1]$, $\varphi$ has a unique fixed point $x^{*}$ and the sequence $(x_{n})$ with $x_{n+1}=\varphi(x_{n}) ,n=0,1,2,...$ is well defined and coverges to $x^{*}$ for any $x_{0} \in [-1,1]$.Then if the 9th approximation is $x_{9}=0.37282$ and the 10th is $x_{10}=0.37382$,we can say for sure that:
1) $|x_{10}-x^{*}|<0.005$
2) $|x_{10}-x^{*}|<0.001$
3) $|x_{10}-x^{*}|<0.002$
I used the formula $|x_{10}-x^{*}|\leq\frac{L}{1-L}|x_{n}-x_{n-1}|$ and found that $|x_{10}-x^{*}|\leq 0.004$.Is this right?So,is 1) the right answer? (Thinking)