Mixing
![How to use dice to randomly select from 10 choices? [duplicate]](https://lh6.googleusercontent.com/-7Mng4mN1X2s/AAAAAAAAAAI/AAAAAAAAAAA/AAN31DW_XDkvv3xrhJACH5mIjz7MZXrO3Q/s72-c-mo/photo.jpg?sz=32)
Newton's finding root method - n digits accuracy
$begingroup$
I have read about the Newton's method in Wikipedia.
https://en.wikipedia.org/wiki/Newton%27s_method
By checking the pseudocode in the thread,
I have noticed a tolerance is defined (in that case $10^{-7}$).
and if
$|x_n-x_{n-1}|le tolerancecdot|x_n|$
is satisfied then the root is found with the same accuracy (or better) of the tolerance (7 digits in our case).
I'd appreciate if anyone could explain why is that true?
For the sake of my question, let's assume that the method converges.
numerical-methods roots newton-raphson
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
$endgroup$
add a comment |
$begingroup$
I have read about the Newton's method in Wikipedia.
https://en.wikipedia.org/wiki/Newton%27s_method
By checking the pseudocode in the thread,
I have noticed a tolerance is defined (in that case $10^{-7}$).
and if
$|x_n-x_{n-1}|le tolerancecdot|x_n|$
is satisfied then the root is found with the same accuracy (or better) of the tolerance (7 digits in our case).
I'd appreciate if anyone could explain why is that true?
For the sake of my question, let's assume that the method converges.
numerical-methods roots newton-raphson
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
$endgroup$
add a comment |
$begingroup$
I have read about the Newton's method in Wikipedia.
https://en.wikipedia.org/wiki/Newton%27s_method
By checking the pseudocode in the thread,
I have noticed a tolerance is defined (in that case $10^{-7}$).
and if
$|x_n-x_{n-1}|le tolerancecdot|x_n|$
is satisfied then the root is found with the same accuracy (or better) of the tolerance (7 digits in our case).
I'd appreciate if anyone could explain why is that true?
For the sake of my question, let's assume that the method converges.
numerical-methods roots newton-raphson
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
$endgroup$
I have read about the Newton's method in Wikipedia.
https://en.wikipedia.org/wiki/Newton%27s_method
By checking the pseudocode in the thread,
I have noticed a tolerance is defined (in that case $10^{-7}$).
and if
$|x_n-x_{n-1}|le tolerancecdot|x_n|$
is satisfied then the root is found with the same accuracy (or better) of the tolerance (7 digits in our case).
I'd appreciate if anyone could explain why is that true?
For the sake of my question, let's assume that the method converges.
numerical-methods roots newton-raphson
numerical-methods roots newton-raphson
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
edited Jan 11 at 17:22
Um Shmum
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
asked Jan 11 at 17:06
Um ShmumUm Shmum
1298
1298
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Close to the root you may assume that $|x_n-x_*|<Ccdot|x_n-x_{n-1}|^2=ϵ|x_n-x_{n-1}|$ with some reasonably small $ϵll 1$. Thus it is safe to claim that more than $7$ leading digits are correct if the relative step size is $10^{-7}$.
For this to fail the factor $C$ would have to be excessively large, which could happen with closely clustered roots (like for example $x^7+10^{-9}$) or an otherwise wildly oscillating function. Most of the examples you see that do not involve large numbers or small constants do not fall in that class.
answered Jan 11 at 22:00
LutzLLutzL
58k42054
$endgroup$
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
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%2f3070078%2fnewtons-finding-root-method-n-digits-accuracy%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
$begingroup$
Close to the root you may assume that $|x_n-x_*|<Ccdot|x_n-x_{n-1}|^2=ϵ|x_n-x_{n-1}|$ with some reasonably small $ϵll 1$. Thus it is safe to claim that more than $7$ leading digits are correct if the relative step size is $10^{-7}$.
For this to fail the factor $C$ would have to be excessively large, which could happen with closely clustered roots (like for example $x^7+10^{-9}$) or an otherwise wildly oscillating function. Most of the examples you see that do not involve large numbers or small constants do not fall in that class.
answered Jan 11 at 22:00
LutzLLutzL
58k42054
$endgroup$
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
add a comment |
$begingroup$
Close to the root you may assume that $|x_n-x_*|<Ccdot|x_n-x_{n-1}|^2=ϵ|x_n-x_{n-1}|$ with some reasonably small $ϵll 1$. Thus it is safe to claim that more than $7$ leading digits are correct if the relative step size is $10^{-7}$.
For this to fail the factor $C$ would have to be excessively large, which could happen with closely clustered roots (like for example $x^7+10^{-9}$) or an otherwise wildly oscillating function. Most of the examples you see that do not involve large numbers or small constants do not fall in that class.
answered Jan 11 at 22:00
LutzLLutzL
58k42054
$endgroup$
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
add a comment |
$begingroup$
Close to the root you may assume that $|x_n-x_*|<Ccdot|x_n-x_{n-1}|^2=ϵ|x_n-x_{n-1}|$ with some reasonably small $ϵll 1$. Thus it is safe to claim that more than $7$ leading digits are correct if the relative step size is $10^{-7}$.
For this to fail the factor $C$ would have to be excessively large, which could happen with closely clustered roots (like for example $x^7+10^{-9}$) or an otherwise wildly oscillating function. Most of the examples you see that do not involve large numbers or small constants do not fall in that class.
answered Jan 11 at 22:00
LutzLLutzL
58k42054
$endgroup$
Close to the root you may assume that $|x_n-x_*|<Ccdot|x_n-x_{n-1}|^2=ϵ|x_n-x_{n-1}|$ with some reasonably small $ϵll 1$. Thus it is safe to claim that more than $7$ leading digits are correct if the relative step size is $10^{-7}$.
For this to fail the factor $C$ would have to be excessively large, which could happen with closely clustered roots (like for example $x^7+10^{-9}$) or an otherwise wildly oscillating function. Most of the examples you see that do not involve large numbers or small constants do not fall in that class.
answered Jan 11 at 22:00
LutzLLutzL
58k42054
answered Jan 11 at 22:00
LutzLLutzL
58k42054
answered Jan 11 at 22:00
LutzLLutzL
58k42054
answered Jan 11 at 22:00
LutzLLutzL
58k42054
58k42054
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
add a comment |
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Why can I assume that inequality?
$endgroup$
– Galush Balush
Jan 12 at 17:29
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
$begingroup$
Because of the quadratic convergence. $|f(x_n)|le frac{|f''|}2|x_n-x_{n-1}|^2$, and $f(x_n)approx f'(x_*)(x_n-x_*)$.
$endgroup$
– LutzL
Jan 12 at 17:32
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%2f3070078%2fnewtons-finding-root-method-n-digits-accuracy%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
