How does convergence of $z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$ relate to $z_{n+1} = frac{1}{2}...
$begingroup$
In class we looked at the following exercise:
Let $a ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{a}{z_n} right)$$
for $n ge 0$. For which $z_0 in mathbb{C}$ is the above sequence well defined and if it is, what is it's limit?
Someone said that it suffices to reduce the above exercise to the exercise below:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if
$x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$x_{0} < 0$ then $lim_{n to infty} z_n = -1$.
$x_{0} = 0$ and $y_{0} ne 0$ then $(z_n)_n$ is not defined or divergent.
I do not understand how such a reduction should work. Could you please explain that to me?
Note: Here I posted major parts of my solution to the second exercise.
sequences-and-series complex-analysis complex-numbers recursion
asked Jan 21 at 23:00
3nondatur3nondatur
399111
$endgroup$
add a comment |
$begingroup$
In class we looked at the following exercise:
Let $a ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{a}{z_n} right)$$
for $n ge 0$. For which $z_0 in mathbb{C}$ is the above sequence well defined and if it is, what is it's limit?
Someone said that it suffices to reduce the above exercise to the exercise below:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if
$x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$x_{0} < 0$ then $lim_{n to infty} z_n = -1$.
$x_{0} = 0$ and $y_{0} ne 0$ then $(z_n)_n$ is not defined or divergent.
I do not understand how such a reduction should work. Could you please explain that to me?
Note: Here I posted major parts of my solution to the second exercise.
sequences-and-series complex-analysis complex-numbers recursion
asked Jan 21 at 23:00
3nondatur3nondatur
399111
$endgroup$
1
$begingroup$
It appears it has been answered in this (now closed) question, math.stackexchange.com/questions/332993/… you should see if anything here helps.
$endgroup$
– LoveTooNap29
Jan 21 at 23:05
add a comment |
$begingroup$
In class we looked at the following exercise:
Let $a ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{a}{z_n} right)$$
for $n ge 0$. For which $z_0 in mathbb{C}$ is the above sequence well defined and if it is, what is it's limit?
Someone said that it suffices to reduce the above exercise to the exercise below:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if
$x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$x_{0} < 0$ then $lim_{n to infty} z_n = -1$.
$x_{0} = 0$ and $y_{0} ne 0$ then $(z_n)_n$ is not defined or divergent.
I do not understand how such a reduction should work. Could you please explain that to me?
Note: Here I posted major parts of my solution to the second exercise.
sequences-and-series complex-analysis complex-numbers recursion
asked Jan 21 at 23:00
3nondatur3nondatur
399111
$endgroup$
In class we looked at the following exercise:
Let $a ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{a}{z_n} right)$$
for $n ge 0$. For which $z_0 in mathbb{C}$ is the above sequence well defined and if it is, what is it's limit?
Someone said that it suffices to reduce the above exercise to the exercise below:
Let $z_0 = x_0+iy_0 ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = frac{1}{2} left( z_n+frac{1}{z_n} right)$$
for $n ge 0$. Show that if
$x_{0} > 0$ then $lim_{n to infty} z_n = 1$.
$x_{0} < 0$ then $lim_{n to infty} z_n = -1$.
$x_{0} = 0$ and $y_{0} ne 0$ then $(z_n)_n$ is not defined or divergent.
I do not understand how such a reduction should work. Could you please explain that to me?
Note: Here I posted major parts of my solution to the second exercise.
sequences-and-series complex-analysis complex-numbers recursion
sequences-and-series complex-analysis complex-numbers recursion
asked Jan 21 at 23:00
3nondatur3nondatur
399111
asked Jan 21 at 23:00
3nondatur3nondatur
399111
asked Jan 21 at 23:00
3nondatur3nondatur
399111
asked Jan 21 at 23:00
3nondatur3nondatur
399111
asked Jan 21 at 23:00
3nondatur3nondatur
399111
399111
1
$begingroup$
It appears it has been answered in this (now closed) question, math.stackexchange.com/questions/332993/… you should see if anything here helps.
$endgroup$
– LoveTooNap29
Jan 21 at 23:05
add a comment |
1
$begingroup$
It appears it has been answered in this (now closed) question, math.stackexchange.com/questions/332993/… you should see if anything here helps.
$endgroup$
– LoveTooNap29
Jan 21 at 23:05
1
1
$begingroup$
It appears it has been answered in this (now closed) question, math.stackexchange.com/questions/332993/… you should see if anything here helps.
$endgroup$
– LoveTooNap29
Jan 21 at 23:05
$begingroup$
It appears it has been answered in this (now closed) question, math.stackexchange.com/questions/332993/… you should see if anything here helps.
$endgroup$
– LoveTooNap29
Jan 21 at 23:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Exploit the homogeneity of the sequence to make a convenient rescaling. Suppose that $z_n$ satisfies the recurrence
$$z_{n + 1} = frac 1 2 left(z_n + frac{1}{z_n}right)$$
and define a new sequence $w_n := sqrt{a} z_n$. Notice that
$$frac 1 2 left(w_n + frac{a}{w_n}right) = frac 1 2 left(sqrt a z_n + frac{a}{sqrt a z_n}right) = frac {sqrt a} 2 left(z_n + frac{1}{z_n}right) = w_{n + 1}.$$
From here, the reduction is immediate.
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
$endgroup$
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%2f3082529%2fhow-does-convergence-of-z-n1-frac12-left-z-n-frac1z-n-right%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
$begingroup$
Exploit the homogeneity of the sequence to make a convenient rescaling. Suppose that $z_n$ satisfies the recurrence
$$z_{n + 1} = frac 1 2 left(z_n + frac{1}{z_n}right)$$
and define a new sequence $w_n := sqrt{a} z_n$. Notice that
$$frac 1 2 left(w_n + frac{a}{w_n}right) = frac 1 2 left(sqrt a z_n + frac{a}{sqrt a z_n}right) = frac {sqrt a} 2 left(z_n + frac{1}{z_n}right) = w_{n + 1}.$$
From here, the reduction is immediate.
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
$endgroup$
add a comment |
$begingroup$
Exploit the homogeneity of the sequence to make a convenient rescaling. Suppose that $z_n$ satisfies the recurrence
$$z_{n + 1} = frac 1 2 left(z_n + frac{1}{z_n}right)$$
and define a new sequence $w_n := sqrt{a} z_n$. Notice that
$$frac 1 2 left(w_n + frac{a}{w_n}right) = frac 1 2 left(sqrt a z_n + frac{a}{sqrt a z_n}right) = frac {sqrt a} 2 left(z_n + frac{1}{z_n}right) = w_{n + 1}.$$
From here, the reduction is immediate.
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
$endgroup$
add a comment |
$begingroup$
Exploit the homogeneity of the sequence to make a convenient rescaling. Suppose that $z_n$ satisfies the recurrence
$$z_{n + 1} = frac 1 2 left(z_n + frac{1}{z_n}right)$$
and define a new sequence $w_n := sqrt{a} z_n$. Notice that
$$frac 1 2 left(w_n + frac{a}{w_n}right) = frac 1 2 left(sqrt a z_n + frac{a}{sqrt a z_n}right) = frac {sqrt a} 2 left(z_n + frac{1}{z_n}right) = w_{n + 1}.$$
From here, the reduction is immediate.
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
$endgroup$
Exploit the homogeneity of the sequence to make a convenient rescaling. Suppose that $z_n$ satisfies the recurrence
$$z_{n + 1} = frac 1 2 left(z_n + frac{1}{z_n}right)$$
and define a new sequence $w_n := sqrt{a} z_n$. Notice that
$$frac 1 2 left(w_n + frac{a}{w_n}right) = frac 1 2 left(sqrt a z_n + frac{a}{sqrt a z_n}right) = frac {sqrt a} 2 left(z_n + frac{1}{z_n}right) = w_{n + 1}.$$
From here, the reduction is immediate.
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
answered Jan 21 at 23:06
T. BongersT. Bongers
23.5k54762
23.5k54762
add a comment |
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%2f3082529%2fhow-does-convergence-of-z-n1-frac12-left-z-n-frac1z-n-right%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
1
$begingroup$
It appears it has been answered in this (now closed) question, math.stackexchange.com/questions/332993/… you should see if anything here helps.
$endgroup$
– LoveTooNap29
Jan 21 at 23:05