Using Newton's method to find $M$, the unique positive zero of $x^2+x-1$
$begingroup$
Using Newton's method to find $M$, the unique positive zero of $x^2+x-1$. Obtain a recursive formula for the error term $e_n$ use it to prove $a_n rightarrow M$
Recursive formula for $a_n:quad$ $a_{n+1} = a_n - frac{a_n^2+a_n-1}{2a_n + 1} = frac{2a_n^2 + a_n - a_n^2-a_n+1}{2a_n+1}= frac{a_n^2+1}{2a_n+1}$
Recursive formula for $e_n:quad$ $e_{n+1} = frac{(e_n+M)^2+1}{2(e_n+M)+1}$
To show that $a_n rightarrow M$, it is sufficient to show that for every $epsilon > 0$ holds $|e_n| < epsilon$.
How should I do that?
limits recurrence-relations limits-without-lhopital recursion newton-raphson
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
$endgroup$
|
show 3 more comments
$begingroup$
Using Newton's method to find $M$, the unique positive zero of $x^2+x-1$. Obtain a recursive formula for the error term $e_n$ use it to prove $a_n rightarrow M$
Recursive formula for $a_n:quad$ $a_{n+1} = a_n - frac{a_n^2+a_n-1}{2a_n + 1} = frac{2a_n^2 + a_n - a_n^2-a_n+1}{2a_n+1}= frac{a_n^2+1}{2a_n+1}$
Recursive formula for $e_n:quad$ $e_{n+1} = frac{(e_n+M)^2+1}{2(e_n+M)+1}$
To show that $a_n rightarrow M$, it is sufficient to show that for every $epsilon > 0$ holds $|e_n| < epsilon$.
How should I do that?
limits recurrence-relations limits-without-lhopital recursion newton-raphson
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
$endgroup$
$begingroup$
@user376343 The one in the body is the right one. I made the changes.
$endgroup$
– Sargis Iskandaryan
Dec 13 '18 at 21:47
$begingroup$
Are you sure your simplification of $a_{n+1}$ is correct? It looks like you've got a vanishing $a_n^2$ in your last equality.
$endgroup$
– Jam
Dec 13 '18 at 22:31
$begingroup$
@Jam Thank you.
$endgroup$
– Sargis Iskandaryan
Dec 15 '18 at 12:23
1
$begingroup$
Please recheck the numerator of formula of $a_{n+1}$ you are getting. A one should be added I think.
$endgroup$
– Martund
Dec 15 '18 at 12:53
$begingroup$
Also an $M$ should be subtracted from the RHS of the last term since $e_{n+1}$ is error term.
$endgroup$
– Martund
Dec 15 '18 at 12:56
|
show 3 more comments
$begingroup$
Using Newton's method to find $M$, the unique positive zero of $x^2+x-1$. Obtain a recursive formula for the error term $e_n$ use it to prove $a_n rightarrow M$
Recursive formula for $a_n:quad$ $a_{n+1} = a_n - frac{a_n^2+a_n-1}{2a_n + 1} = frac{2a_n^2 + a_n - a_n^2-a_n+1}{2a_n+1}= frac{a_n^2+1}{2a_n+1}$
Recursive formula for $e_n:quad$ $e_{n+1} = frac{(e_n+M)^2+1}{2(e_n+M)+1}$
To show that $a_n rightarrow M$, it is sufficient to show that for every $epsilon > 0$ holds $|e_n| < epsilon$.
How should I do that?
limits recurrence-relations limits-without-lhopital recursion newton-raphson
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
$endgroup$
Using Newton's method to find $M$, the unique positive zero of $x^2+x-1$. Obtain a recursive formula for the error term $e_n$ use it to prove $a_n rightarrow M$
Recursive formula for $a_n:quad$ $a_{n+1} = a_n - frac{a_n^2+a_n-1}{2a_n + 1} = frac{2a_n^2 + a_n - a_n^2-a_n+1}{2a_n+1}= frac{a_n^2+1}{2a_n+1}$
Recursive formula for $e_n:quad$ $e_{n+1} = frac{(e_n+M)^2+1}{2(e_n+M)+1}$
To show that $a_n rightarrow M$, it is sufficient to show that for every $epsilon > 0$ holds $|e_n| < epsilon$.
How should I do that?
limits recurrence-relations limits-without-lhopital recursion newton-raphson
limits recurrence-relations limits-without-lhopital recursion newton-raphson
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
edited Dec 15 '18 at 16:05
Sargis Iskandaryan
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
asked Dec 13 '18 at 16:07
Sargis IskandaryanSargis Iskandaryan
560112
560112
$begingroup$
@user376343 The one in the body is the right one. I made the changes.
$endgroup$
– Sargis Iskandaryan
Dec 13 '18 at 21:47
$begingroup$
Are you sure your simplification of $a_{n+1}$ is correct? It looks like you've got a vanishing $a_n^2$ in your last equality.
$endgroup$
– Jam
Dec 13 '18 at 22:31
$begingroup$
@Jam Thank you.
$endgroup$
– Sargis Iskandaryan
Dec 15 '18 at 12:23
1
$begingroup$
Please recheck the numerator of formula of $a_{n+1}$ you are getting. A one should be added I think.
$endgroup$
– Martund
Dec 15 '18 at 12:53
$begingroup$
Also an $M$ should be subtracted from the RHS of the last term since $e_{n+1}$ is error term.
$endgroup$
– Martund
Dec 15 '18 at 12:56
|
show 3 more comments
$begingroup$
@user376343 The one in the body is the right one. I made the changes.
$endgroup$
– Sargis Iskandaryan
Dec 13 '18 at 21:47
$begingroup$
Are you sure your simplification of $a_{n+1}$ is correct? It looks like you've got a vanishing $a_n^2$ in your last equality.
$endgroup$
– Jam
Dec 13 '18 at 22:31
$begingroup$
@Jam Thank you.
$endgroup$
– Sargis Iskandaryan
Dec 15 '18 at 12:23
1
$begingroup$
Please recheck the numerator of formula of $a_{n+1}$ you are getting. A one should be added I think.
$endgroup$
– Martund
Dec 15 '18 at 12:53
$begingroup$
Also an $M$ should be subtracted from the RHS of the last term since $e_{n+1}$ is error term.
$endgroup$
– Martund
Dec 15 '18 at 12:56
$begingroup$
@user376343 The one in the body is the right one. I made the changes.
$endgroup$
– Sargis Iskandaryan
Dec 13 '18 at 21:47
$begingroup$
@user376343 The one in the body is the right one. I made the changes.
$endgroup$
– Sargis Iskandaryan
Dec 13 '18 at 21:47
$begingroup$
Are you sure your simplification of $a_{n+1}$ is correct? It looks like you've got a vanishing $a_n^2$ in your last equality.
$endgroup$
– Jam
Dec 13 '18 at 22:31
$begingroup$
Are you sure your simplification of $a_{n+1}$ is correct? It looks like you've got a vanishing $a_n^2$ in your last equality.
$endgroup$
– Jam
Dec 13 '18 at 22:31
$begingroup$
@Jam Thank you.
$endgroup$
– Sargis Iskandaryan
Dec 15 '18 at 12:23
$begingroup$
@Jam Thank you.
$endgroup$
– Sargis Iskandaryan
Dec 15 '18 at 12:23
1
1
$begingroup$
Please recheck the numerator of formula of $a_{n+1}$ you are getting. A one should be added I think.
$endgroup$
– Martund
Dec 15 '18 at 12:53
$begingroup$
Please recheck the numerator of formula of $a_{n+1}$ you are getting. A one should be added I think.
$endgroup$
– Martund
Dec 15 '18 at 12:53
$begingroup$
Also an $M$ should be subtracted from the RHS of the last term since $e_{n+1}$ is error term.
$endgroup$
– Martund
Dec 15 '18 at 12:56
$begingroup$
Also an $M$ should be subtracted from the RHS of the last term since $e_{n+1}$ is error term.
$endgroup$
– Martund
Dec 15 '18 at 12:56
|
show 3 more comments
0
active
oldest
votes
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%2f3038209%2fusing-newtons-method-to-find-m-the-unique-positive-zero-of-x2x-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3038209%2fusing-newtons-method-to-find-m-the-unique-positive-zero-of-x2x-1%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
$begingroup$
@user376343 The one in the body is the right one. I made the changes.
$endgroup$
– Sargis Iskandaryan
Dec 13 '18 at 21:47
$begingroup$
Are you sure your simplification of $a_{n+1}$ is correct? It looks like you've got a vanishing $a_n^2$ in your last equality.
$endgroup$
– Jam
Dec 13 '18 at 22:31
$begingroup$
@Jam Thank you.
$endgroup$
– Sargis Iskandaryan
Dec 15 '18 at 12:23
1
$begingroup$
Please recheck the numerator of formula of $a_{n+1}$ you are getting. A one should be added I think.
$endgroup$
– Martund
Dec 15 '18 at 12:53
$begingroup$
Also an $M$ should be subtracted from the RHS of the last term since $e_{n+1}$ is error term.
$endgroup$
– Martund
Dec 15 '18 at 12:56