Show that for a function $f$ bounded by $M$ on a disk $D_{r}$ show that $|f^{n}(z)|leq n!M/delta^{n}$ for...
$begingroup$
Suppose that a function $f$ is analytic in the open disk
$$D_{r}={zin mathbb{C}:|z|<r}$$
where $r>0$, and there is a number $Minmathbb{R} $ such that $|f(z)| leq M$ for all $z in D_{r}$.
Let $delta in (0, r)$. Then show that:
$$|f^{n}(z)|leq frac{n!M}{delta^{n}}$$
for all $zin D_{r−δ}$ where $D_{r−δ} = {zinmathbb{C} : |z| < r − δ}$.
My try:
The initial disk $D_{r}$ is centered at $z_{0}=0$, the Cauchy formula reduces to
$$|f^{n}(z)|=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{(zeta-z_{0})^{n+1}} dzeta$$
$$=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{zeta^{n+1}} dzeta$$
$$leq frac{n!}{2pi }frac{M}{delta^{n+1}}2pi delta=frac{n!M}{delta^{n}}$$
and then extending this theorem, the new disk $D_{r-delta}$ is still centered at $z_{0}=0$ and the radius is still $R=delta$. So, you can show that the inequality is true for the new disk by replacing $M=M_{1}$ and show that $M= M_{1}$ equal each other. This doesn't seem like a very complete evaluation of the problem in hand.
complex-analysis upper-lower-bounds cauchy-integral-formula
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
$endgroup$
add a comment |
$begingroup$
Suppose that a function $f$ is analytic in the open disk
$$D_{r}={zin mathbb{C}:|z|<r}$$
where $r>0$, and there is a number $Minmathbb{R} $ such that $|f(z)| leq M$ for all $z in D_{r}$.
Let $delta in (0, r)$. Then show that:
$$|f^{n}(z)|leq frac{n!M}{delta^{n}}$$
for all $zin D_{r−δ}$ where $D_{r−δ} = {zinmathbb{C} : |z| < r − δ}$.
My try:
The initial disk $D_{r}$ is centered at $z_{0}=0$, the Cauchy formula reduces to
$$|f^{n}(z)|=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{(zeta-z_{0})^{n+1}} dzeta$$
$$=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{zeta^{n+1}} dzeta$$
$$leq frac{n!}{2pi }frac{M}{delta^{n+1}}2pi delta=frac{n!M}{delta^{n}}$$
and then extending this theorem, the new disk $D_{r-delta}$ is still centered at $z_{0}=0$ and the radius is still $R=delta$. So, you can show that the inequality is true for the new disk by replacing $M=M_{1}$ and show that $M= M_{1}$ equal each other. This doesn't seem like a very complete evaluation of the problem in hand.
complex-analysis upper-lower-bounds cauchy-integral-formula
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
$endgroup$
add a comment |
$begingroup$
Suppose that a function $f$ is analytic in the open disk
$$D_{r}={zin mathbb{C}:|z|<r}$$
where $r>0$, and there is a number $Minmathbb{R} $ such that $|f(z)| leq M$ for all $z in D_{r}$.
Let $delta in (0, r)$. Then show that:
$$|f^{n}(z)|leq frac{n!M}{delta^{n}}$$
for all $zin D_{r−δ}$ where $D_{r−δ} = {zinmathbb{C} : |z| < r − δ}$.
My try:
The initial disk $D_{r}$ is centered at $z_{0}=0$, the Cauchy formula reduces to
$$|f^{n}(z)|=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{(zeta-z_{0})^{n+1}} dzeta$$
$$=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{zeta^{n+1}} dzeta$$
$$leq frac{n!}{2pi }frac{M}{delta^{n+1}}2pi delta=frac{n!M}{delta^{n}}$$
and then extending this theorem, the new disk $D_{r-delta}$ is still centered at $z_{0}=0$ and the radius is still $R=delta$. So, you can show that the inequality is true for the new disk by replacing $M=M_{1}$ and show that $M= M_{1}$ equal each other. This doesn't seem like a very complete evaluation of the problem in hand.
complex-analysis upper-lower-bounds cauchy-integral-formula
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
$endgroup$
Suppose that a function $f$ is analytic in the open disk
$$D_{r}={zin mathbb{C}:|z|<r}$$
where $r>0$, and there is a number $Minmathbb{R} $ such that $|f(z)| leq M$ for all $z in D_{r}$.
Let $delta in (0, r)$. Then show that:
$$|f^{n}(z)|leq frac{n!M}{delta^{n}}$$
for all $zin D_{r−δ}$ where $D_{r−δ} = {zinmathbb{C} : |z| < r − δ}$.
My try:
The initial disk $D_{r}$ is centered at $z_{0}=0$, the Cauchy formula reduces to
$$|f^{n}(z)|=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{(zeta-z_{0})^{n+1}} dzeta$$
$$=frac{n!}{2pi i}int_{D_{r}} frac{f(zeta)}{zeta^{n+1}} dzeta$$
$$leq frac{n!}{2pi }frac{M}{delta^{n+1}}2pi delta=frac{n!M}{delta^{n}}$$
and then extending this theorem, the new disk $D_{r-delta}$ is still centered at $z_{0}=0$ and the radius is still $R=delta$. So, you can show that the inequality is true for the new disk by replacing $M=M_{1}$ and show that $M= M_{1}$ equal each other. This doesn't seem like a very complete evaluation of the problem in hand.
complex-analysis upper-lower-bounds cauchy-integral-formula
complex-analysis upper-lower-bounds cauchy-integral-formula
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
asked Dec 8 '18 at 14:26
Cassidy HazelCassidy Hazel
92
92
add a comment |
add a comment |
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%2f3031166%2fshow-that-for-a-function-f-bounded-by-m-on-a-disk-d-r-show-that-fn%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%2f3031166%2fshow-that-for-a-function-f-bounded-by-m-on-a-disk-d-r-show-that-fn%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
