In Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
edited Nov 27 at 9:37
LutzL
55.9k42054
asked Nov 27 at 6:46
Jacob S.
19010
add a comment |
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
edited Nov 27 at 9:37
LutzL
55.9k42054
asked Nov 27 at 6:46
Jacob S.
19010
add a comment |
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
edited Nov 27 at 9:37
LutzL
55.9k42054
asked Nov 27 at 6:46
Jacob S.
19010
Show that in the Newton-Raphson method $ g'(x^*) = 0 $ and $ g''(x^*)neq 0 $ for real roots not repeated, where $x^*$ is a fixed point.
Deriving the function $ g(x)=x-dfrac{f(x)}{f'(x)} $, and then evaluating in $ x^* $, I have proved that $g'(x^*) = 0$. I have also found that $ g''(x^*)=dfrac{f''(x^*)}{f'(x^*)} $.
Since the real roots are non-repetitive, $ f'(x^*) neq 0 $. Thus, for $ g''(x^*) neq 0 $, $ f''(x^*)$ must be different from zero. It's really $ f''(x^*) neq 0 $. If so, could you explain why?
numerical-methods
numerical-methods
edited Nov 27 at 9:37
LutzL
55.9k42054
asked Nov 27 at 6:46
Jacob S.
19010
edited Nov 27 at 9:37
LutzL
55.9k42054
asked Nov 27 at 6:46
Jacob S.
19010
edited Nov 27 at 9:37
LutzL
55.9k42054
edited Nov 27 at 9:37
LutzL
55.9k42054
edited Nov 27 at 9:37
LutzL
55.9k42054
55.9k42054
asked Nov 27 at 6:46
Jacob S.
19010
asked Nov 27 at 6:46
Jacob S.
19010
asked Nov 27 at 6:46
Jacob S.
19010
19010
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 at 9:15
LutzL
55.9k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
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%2f3015426%2fin-newton-raphson-method-gx-0-and-gx-neq-0-for-real-roots-n%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
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 at 9:15
LutzL
55.9k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
add a comment |
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 at 9:15
LutzL
55.9k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
add a comment |
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 at 9:15
LutzL
55.9k42054
Your observation is correct. You can try it out directly with something like $f(x)=x+x^3$ where $$g(x)=frac{x+3x^3-(x+x^3)}{1+3x^2}=frac{2x^3}{1+3x^2},$$ which has a triple root at $x^*=0$, thus $g''(x^*)=0$.
The removal of the curvature term, that is, the second derivative, is one of the motivations for the third order Halley method, you apply the Newton method to the function $F(x)=dfrac{f(x)}{sqrt{|f'(x)|}}$. Third order convergence means exactly your contradiction case, that not only is $g'(x^*)=0$, but also $g''(x^*)$ for $g$ the Newton step for $F$ at a simple root of $f$. This means that you can construct a counter-example $F$ to the claim in question from almost any function $f$.
answered Nov 27 at 9:15
LutzL
55.9k42054
edited Nov 27 at 9:39
answered Nov 27 at 9:15
LutzL
55.9k42054
answered Nov 27 at 9:15
LutzL
55.9k42054
answered Nov 27 at 9:15
LutzL
55.9k42054
55.9k42054
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
add a comment |
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
Why is $ f''(x^{*}) $ different from zero?
– Jacob S.
Nov 27 at 14:33
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
For a general function it will be. However nothing prevents it from occasionally also being zero. The claim that for simple roots you always have $f''(x^*)ne0$ is wrong. As you observed.
– LutzL
Nov 27 at 14:39
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3015426%2fin-newton-raphson-method-gx-0-and-gx-neq-0-for-real-roots-n%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
