Use the definition of limit to prove that $lim_{zto z_0} (az + b) = az_0 + b$, a,b belongs to complex numbers
Here what i already done:
begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}
If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:
$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$
Then:
$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$
So:
${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$
Then:
$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$
That proves:
$lim_{zto z_0} (az + b) = az_0 + b$
Is that correct?
complex-analysis limits proof-verification complex-numbers
edited Dec 12 '18 at 22:48
platty
3,370320
asked Dec 12 '18 at 22:44
RafaelRafael
12
add a comment |
Here what i already done:
begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}
If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:
$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$
Then:
$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$
So:
${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$
Then:
$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$
That proves:
$lim_{zto z_0} (az + b) = az_0 + b$
Is that correct?
complex-analysis limits proof-verification complex-numbers
edited Dec 12 '18 at 22:48
platty
3,370320
asked Dec 12 '18 at 22:44
RafaelRafael
12
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
– projectilemotion
Dec 12 '18 at 22:56
@projectilemotion You forget the case $a = 0$.
– Rebellos
Dec 12 '18 at 23:10
I didn't forget it, in that case you could just take any $delta>0$.
– projectilemotion
Dec 12 '18 at 23:15
add a comment |
Here what i already done:
begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}
If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:
$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$
Then:
$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$
So:
${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$
Then:
$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$
That proves:
$lim_{zto z_0} (az + b) = az_0 + b$
Is that correct?
complex-analysis limits proof-verification complex-numbers
edited Dec 12 '18 at 22:48
platty
3,370320
asked Dec 12 '18 at 22:44
RafaelRafael
12
Here what i already done:
begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\&=|a(z-z_0)|\&=|a||z-z_0|{le}|a|*{Delta}|z-z_0|
end{align*}
If ${Delta}=1$ and we know that $0{lt}|z-z_0|{lt}1$, then:
$||z|-|z_0||{le}|z-z_0|<1 \{to} |z|{lt}1+|z_o|$
Then:
$|z|<1+|z_0| {to} |z+z_0|{le}|z|+|z_0|{le}1+2|z_0|$
So:
${Delta}|a|*(1+2|z_0|)\ = {Delta}(|a|*(1+2|z_0|)={epsilon}\ {to} {Delta} = {epsilon}/(|a|(|1+2|z_0|)$
Then:
$|az+b - (az_0 + b)| {lt} |a|{Delta}|z-z_0|\ {le} {Delta}(|a|(1+2|(z_0)|\{lt}{epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {epsilon}$
That proves:
$lim_{zto z_0} (az + b) = az_0 + b$
Is that correct?
complex-analysis limits proof-verification complex-numbers
complex-analysis limits proof-verification complex-numbers
edited Dec 12 '18 at 22:48
platty
3,370320
asked Dec 12 '18 at 22:44
RafaelRafael
12
edited Dec 12 '18 at 22:48
platty
3,370320
asked Dec 12 '18 at 22:44
RafaelRafael
12
edited Dec 12 '18 at 22:48
platty
3,370320
edited Dec 12 '18 at 22:48
platty
3,370320
edited Dec 12 '18 at 22:48
platty
3,370320
3,370320
asked Dec 12 '18 at 22:44
RafaelRafael
12
asked Dec 12 '18 at 22:44
RafaelRafael
12
asked Dec 12 '18 at 22:44
RafaelRafael
12
12
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
– projectilemotion
Dec 12 '18 at 22:56
@projectilemotion You forget the case $a = 0$.
– Rebellos
Dec 12 '18 at 23:10
I didn't forget it, in that case you could just take any $delta>0$.
– projectilemotion
Dec 12 '18 at 23:15
add a comment |
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
– projectilemotion
Dec 12 '18 at 22:56
@projectilemotion You forget the case $a = 0$.
– Rebellos
Dec 12 '18 at 23:10
I didn't forget it, in that case you could just take any $delta>0$.
– projectilemotion
Dec 12 '18 at 23:15
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
– projectilemotion
Dec 12 '18 at 22:56
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
– projectilemotion
Dec 12 '18 at 22:56
@projectilemotion You forget the case $a = 0$.
– Rebellos
Dec 12 '18 at 23:10
@projectilemotion You forget the case $a = 0$.
– Rebellos
Dec 12 '18 at 23:10
I didn't forget it, in that case you could just take any $delta>0$.
– projectilemotion
Dec 12 '18 at 23:15
I didn't forget it, in that case you could just take any $delta>0$.
– projectilemotion
Dec 12 '18 at 23:15
add a comment |
1 Answer
1
active
oldest
votes
Seems okay in general (but your approach is more complex than it should), just some fixes.
First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.
But for a more fast and elegant approach, observe that
$$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$
considering $a=0$ and $aneq 0$ at the same time, thus you're finished.
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
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%2f3037332%2fuse-the-definition-of-limit-to-prove-that-lim-z-to-z-0-az-b-az-0-b%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
Seems okay in general (but your approach is more complex than it should), just some fixes.
First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.
But for a more fast and elegant approach, observe that
$$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$
considering $a=0$ and $aneq 0$ at the same time, thus you're finished.
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
add a comment |
Seems okay in general (but your approach is more complex than it should), just some fixes.
First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.
But for a more fast and elegant approach, observe that
$$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$
considering $a=0$ and $aneq 0$ at the same time, thus you're finished.
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
add a comment |
Seems okay in general (but your approach is more complex than it should), just some fixes.
First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.
But for a more fast and elegant approach, observe that
$$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$
considering $a=0$ and $aneq 0$ at the same time, thus you're finished.
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
Seems okay in general (but your approach is more complex than it should), just some fixes.
First of all, you need to consider the case $a=0 in mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $aneq 0 in mathbb C$.
But for a more fast and elegant approach, observe that
$$|az + b - (az_0 + b)| = |a||z-z_0| < varepsilon implies |z-z_0| < frac{varepsilon}{|a|+1}$$
considering $a=0$ and $aneq 0$ at the same time, thus you're finished.
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
answered Dec 12 '18 at 22:59
RebellosRebellos
14.5k31246
14.5k31246
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3037332%2fuse-the-definition-of-limit-to-prove-that-lim-z-to-z-0-az-b-az-0-b%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
For $aneq 0$, from the first three lines of your reasoning you could just select: $$0<|z-z_0|<delta=epsilon/|a| implies |az+b-(az_0+b)|<epsilon$$ And then consider the case where $a=0$.
– projectilemotion
Dec 12 '18 at 22:56
@projectilemotion You forget the case $a = 0$.
– Rebellos
Dec 12 '18 at 23:10
I didn't forget it, in that case you could just take any $delta>0$.
– projectilemotion
Dec 12 '18 at 23:15