Proof using Cauchy's Integral Formula
$begingroup$
let $f: Omega rightarrow mathbb{C}$ be analytic and $z_0 in mathbb{C}$.
Define
$$g(z) = begin{cases}
frac{f(z)-f(z_0)}{z- z_0} & z not = z_0 \
f'(z_0) & z = z_0
end{cases}$$
now pick $varepsilon$ small enough so that $overline{D(z_0, varepsilon)} subset Omega$
Show that whenever $z in D(z_0, varepsilon)$
$$frac{g(z) - g(z_0)}{z-z_0} = frac{1}{2pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta$$
My question is if my proof below, which makes use of the Cauchy integral formula is correct
We can write $frac{g(z) - g(z_0)}{z-z_0} = frac{frac{f(z)-f(z_0)}{z- z_0}-f'(z_0)}{z-z_0} = frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0}$
now as epsilon gets arbitrarily small, $z rightarrow z_0$ and so now making use of the Cauchy integral formula and the fact that $lim_{z rightarrow z_0} frac{f(z_0)}{z-z_0} = 0$ we should have
$$lim_{z rightarrow z_0} frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0} = frac{1}{2 pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta - 0$$
which gives us the result
complex-analysis cauchy-integral-formula
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
$endgroup$
add a comment |
$begingroup$
let $f: Omega rightarrow mathbb{C}$ be analytic and $z_0 in mathbb{C}$.
Define
$$g(z) = begin{cases}
frac{f(z)-f(z_0)}{z- z_0} & z not = z_0 \
f'(z_0) & z = z_0
end{cases}$$
now pick $varepsilon$ small enough so that $overline{D(z_0, varepsilon)} subset Omega$
Show that whenever $z in D(z_0, varepsilon)$
$$frac{g(z) - g(z_0)}{z-z_0} = frac{1}{2pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta$$
My question is if my proof below, which makes use of the Cauchy integral formula is correct
We can write $frac{g(z) - g(z_0)}{z-z_0} = frac{frac{f(z)-f(z_0)}{z- z_0}-f'(z_0)}{z-z_0} = frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0}$
now as epsilon gets arbitrarily small, $z rightarrow z_0$ and so now making use of the Cauchy integral formula and the fact that $lim_{z rightarrow z_0} frac{f(z_0)}{z-z_0} = 0$ we should have
$$lim_{z rightarrow z_0} frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0} = frac{1}{2 pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta - 0$$
which gives us the result
complex-analysis cauchy-integral-formula
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
$endgroup$
$begingroup$
The problem has been solved here: math.stackexchange.com/questions/3037532/…
$endgroup$
– Richard Villalobos
Dec 13 '18 at 22:00
add a comment |
$begingroup$
let $f: Omega rightarrow mathbb{C}$ be analytic and $z_0 in mathbb{C}$.
Define
$$g(z) = begin{cases}
frac{f(z)-f(z_0)}{z- z_0} & z not = z_0 \
f'(z_0) & z = z_0
end{cases}$$
now pick $varepsilon$ small enough so that $overline{D(z_0, varepsilon)} subset Omega$
Show that whenever $z in D(z_0, varepsilon)$
$$frac{g(z) - g(z_0)}{z-z_0} = frac{1}{2pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta$$
My question is if my proof below, which makes use of the Cauchy integral formula is correct
We can write $frac{g(z) - g(z_0)}{z-z_0} = frac{frac{f(z)-f(z_0)}{z- z_0}-f'(z_0)}{z-z_0} = frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0}$
now as epsilon gets arbitrarily small, $z rightarrow z_0$ and so now making use of the Cauchy integral formula and the fact that $lim_{z rightarrow z_0} frac{f(z_0)}{z-z_0} = 0$ we should have
$$lim_{z rightarrow z_0} frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0} = frac{1}{2 pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta - 0$$
which gives us the result
complex-analysis cauchy-integral-formula
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
$endgroup$
let $f: Omega rightarrow mathbb{C}$ be analytic and $z_0 in mathbb{C}$.
Define
$$g(z) = begin{cases}
frac{f(z)-f(z_0)}{z- z_0} & z not = z_0 \
f'(z_0) & z = z_0
end{cases}$$
now pick $varepsilon$ small enough so that $overline{D(z_0, varepsilon)} subset Omega$
Show that whenever $z in D(z_0, varepsilon)$
$$frac{g(z) - g(z_0)}{z-z_0} = frac{1}{2pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta$$
My question is if my proof below, which makes use of the Cauchy integral formula is correct
We can write $frac{g(z) - g(z_0)}{z-z_0} = frac{frac{f(z)-f(z_0)}{z- z_0}-f'(z_0)}{z-z_0} = frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0}$
now as epsilon gets arbitrarily small, $z rightarrow z_0$ and so now making use of the Cauchy integral formula and the fact that $lim_{z rightarrow z_0} frac{f(z_0)}{z-z_0} = 0$ we should have
$$lim_{z rightarrow z_0} frac{f(z)-f(z_0)}{(z- z_0)^2} - frac{f'(z_0)}{z-z_0} = frac{1}{2 pi i}int_{D(z_0, varepsilon)}frac{f(zeta)}{(zeta-z)(zeta-z_0)^2}dzeta - 0$$
which gives us the result
complex-analysis cauchy-integral-formula
complex-analysis cauchy-integral-formula
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
edited Dec 12 '18 at 17:34
Richard Villalobos
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 17:27
Richard VillalobosRichard Villalobos
1757
1757
$begingroup$
The problem has been solved here: math.stackexchange.com/questions/3037532/…
$endgroup$
– Richard Villalobos
Dec 13 '18 at 22:00
add a comment |
$begingroup$
The problem has been solved here: math.stackexchange.com/questions/3037532/…
$endgroup$
– Richard Villalobos
Dec 13 '18 at 22:00
$begingroup$
The problem has been solved here: math.stackexchange.com/questions/3037532/…
$endgroup$
– Richard Villalobos
Dec 13 '18 at 22:00
$begingroup$
The problem has been solved here: math.stackexchange.com/questions/3037532/…
$endgroup$
– Richard Villalobos
Dec 13 '18 at 22:00
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%2f3036981%2fproof-using-cauchys-integral-formula%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%2f3036981%2fproof-using-cauchys-integral-formula%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
$begingroup$
The problem has been solved here: math.stackexchange.com/questions/3037532/…
$endgroup$
– Richard Villalobos
Dec 13 '18 at 22:00