Fixed-point method for $x=x+wf(x)$












0












$begingroup$


The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19


















0












$begingroup$


The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19
















0












0








0





$begingroup$


The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$




The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems




numerical-methods fixed-point-theorems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:29







HB khaled

















asked Jan 4 at 10:56









HB khaledHB khaled

12410




12410












  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19




















  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19


















$begingroup$
And your take on this would be, what?
$endgroup$
– Did
Jan 4 at 10:59




$begingroup$
And your take on this would be, what?
$endgroup$
– Did
Jan 4 at 10:59












$begingroup$
The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
$endgroup$
– HB khaled
Jan 4 at 11:04




$begingroup$
The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
$endgroup$
– HB khaled
Jan 4 at 11:04












$begingroup$
Yes. And? $ $ $ $
$endgroup$
– Did
Jan 4 at 11:05




$begingroup$
Yes. And? $ $ $ $
$endgroup$
– Did
Jan 4 at 11:05












$begingroup$
I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
$endgroup$
– HB khaled
Jan 4 at 11:19






$begingroup$
I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
$endgroup$
– HB khaled
Jan 4 at 11:19












1 Answer
1






active

oldest

votes


















2












$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061520%2ffixed-point-method-for-x-xwfx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02


















2












$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02
















2












2








2





$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$



You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 14:08

























answered Jan 4 at 13:13









LutzLLutzL

59.2k42057




59.2k42057












  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02




















  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02


















$begingroup$
Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
$endgroup$
– HB khaled
Jan 4 at 13:44




$begingroup$
Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
$endgroup$
– HB khaled
Jan 4 at 13:44












$begingroup$
For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
$endgroup$
– LutzL
Jan 4 at 14:02






$begingroup$
For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
$endgroup$
– LutzL
Jan 4 at 14:02




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061520%2ffixed-point-method-for-x-xwfx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more