numerical-methods, Fixed point theorem.

Question

I am just looking to apply a result, so can someone confirm the following for me.

Let say I have a equation of the form below:

$V_1(x) = a + bV_0(x)$,

where in theory $V_1(x) = V_0(x)$.

I have an algorithm as follow: Start with initial guess $V_0(x)=0$ and proceed to obtain $V_1(x)=a$. Using '$a$' as the new guess, and repeat this until say the difference is less than $0.1\%$.

What sort of regularity conditions or things to look out for to ensure I obtained the true solution?

Thank you community once again.

No problem. Did you try it? Take, say, $a=3,b=2$. Of course the fixed point is $\frac a{1-b}$ which in this case is $-3$. Start with a guess of $0$. What happens? — lulu, Jul 16 '16 at 17:15
It appears that I don't get convergence, and the value keep increasing. — chuck, Jul 16 '16 at 17:18
Indeed. You can never get a negative value by iterating that expression (with those values)...so, sadly, the method fails. — lulu, Jul 16 '16 at 17:19
Would you mind telling me, under what conditions this method fails? — chuck, Jul 16 '16 at 17:20
By contrast, look what happens if, say, $a=7$ and $b=.5$. Here, the fixed point is $14$. — lulu, Jul 16 '16 at 17:20
You need the process to contract, not expand. $|b|<1$ should do the trick. — lulu, Jul 16 '16 at 17:21
I see, this is very helpful. I suppose this is why it's called the contraction mapping theorem. Thank you kind user. — chuck, Jul 16 '16 at 17:24

score 1 · Accepted Answer · answered Jul 16 '16 at 17:54

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ You have a sort of recurrence likes $\ds{\xi_{n + 1} = a + b\xi_{n}}$ which can be rewritten, for $\ds{b \not= 1}$, as \begin{align} \xi_{n + 1} - {a \over 1 - b} & = b\pars{\xi_{n} - {a \over 1 - b}} = b^{2}\pars{\xi_{n - 1} - {a \over 1 - b}} = \cdots = b^{n}\pars{\xi_{1} - {a \over 1 - b}} \end{align}

$$ \mbox{If}\ \verts{b} < 1\,,\ \mbox{it converges to}\ \color{#f00}{{a \over 1 - b}} $$

numerical-methods, Fixed point theorem.

1 Answers1