Fixed-point method for $x=x+wf(x)$
$begingroup$
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
asked Jan 4 at 10:56
HB khaledHB khaled
12410
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 10:56
HB khaledHB khaled
12410
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 4 at 10:56
HB khaledHB khaled
12410
$endgroup$
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
asked Jan 4 at 10:56
HB khaledHB khaled
12410
asked Jan 4 at 10:56
HB khaledHB khaled
12410
edited Jan 4 at 11:29
HB khaled
asked Jan 4 at 10:56
HB khaledHB khaled
12410
asked Jan 4 at 10:56
HB khaledHB khaled
12410
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.
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.
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
$endgroup$
$begingroup$
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
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.
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.
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.
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.
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.
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.
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
$endgroup$
You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.
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.
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
edited Jan 4 at 14:08
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
answered Jan 4 at 13:13
LutzLLutzL
59.6k42057
59.6k42057
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
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