$h(z)=|z-a|. |z-b|. |z-c|$, max value of $h$ is attained
$begingroup$
Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$
Then max value of $h$
is not attained at any point of $Delta$
is attained at an interior point of $Delta$
is attained at the centre of gravity of $Delta$
is attained at the boundary point of $Delta$
$4$ is correct due to Maximum Modulas Principle.Just confirm me please.
complex-analysis
$endgroup$
add a comment |
$begingroup$
Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$
Then max value of $h$
is not attained at any point of $Delta$
is attained at an interior point of $Delta$
is attained at the centre of gravity of $Delta$
is attained at the boundary point of $Delta$
$4$ is correct due to Maximum Modulas Principle.Just confirm me please.
complex-analysis
$endgroup$
1
$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44
4
$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47
1
$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51
add a comment |
$begingroup$
Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$
Then max value of $h$
is not attained at any point of $Delta$
is attained at an interior point of $Delta$
is attained at the centre of gravity of $Delta$
is attained at the boundary point of $Delta$
$4$ is correct due to Maximum Modulas Principle.Just confirm me please.
complex-analysis
$endgroup$
Let , $a,b,c$ be non-collinear points in complex plane, $Delta$ be the closed triangular region of the plane with vertices $a,b,c$. for $zinDelta$, let $$h(z)=|z-a|. |z-b|. |z-c|.$$
Then max value of $h$
is not attained at any point of $Delta$
is attained at an interior point of $Delta$
is attained at the centre of gravity of $Delta$
is attained at the boundary point of $Delta$
$4$ is correct due to Maximum Modulas Principle.Just confirm me please.
complex-analysis
complex-analysis
edited Oct 12 '15 at 20:15
Empty
8,10252660
8,10252660
asked Jan 6 '14 at 14:43
MarkovMarkov
17.2k957178
17.2k957178
1
$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44
4
$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47
1
$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51
add a comment |
1
$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44
4
$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47
1
$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51
1
1
$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44
$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44
4
4
$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47
$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47
1
1
$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51
$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Except your question, we can also answer to the maximum value of the function $h(x)$.
Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$
Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$
If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$
We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$
$endgroup$
add a comment |
$begingroup$
Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.
The specific triangular shape does not matter; the same holds for any domain.
I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.
$endgroup$
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
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%2f629120%2fhz-z-a-z-b-z-c-max-value-of-h-is-attained%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Except your question, we can also answer to the maximum value of the function $h(x)$.
Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$
Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$
If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$
We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$
$endgroup$
add a comment |
$begingroup$
Except your question, we can also answer to the maximum value of the function $h(x)$.
Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$
Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$
If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$
We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$
$endgroup$
add a comment |
$begingroup$
Except your question, we can also answer to the maximum value of the function $h(x)$.
Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$
Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$
If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$
We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$
$endgroup$
Except your question, we can also answer to the maximum value of the function $h(x)$.
Let $f(z)=(z-a)(z-b)(z-c), ; forall zinDelta.$ Then $f$ agrees to Maximum Modulus Principle, hence $$|f(z)|leqmax{|f(z)|vert zinpartialDelta},; forall zin IntDelta$$ i.e. $max{h(z)vert zinDelta}=max{h(z)vert zinpartialDelta}$ where $h(z)=|f(z)|.$
Since $f(a)=f(b)=f(c)=0,$ the maximum occurs at $zin(a,b)cup(b,c)cup(c,a).$ Since $Delta$ is an equilateral triamgle, wlog let $zin(a,b).$ Then $$z=lambda a+(1-lambda )b,; lambdain(0,1).$$
On that case, $h(z)=lambda(1-lambda)|a-b|^2|lambda a+(1-lambda )b-c|.$ Let $|a-b|=|b-c|=|c-a|=p.$ Then $h(z)=lambda(1-lambda)p^2|lambda a+(1-lambda )b-c|.$
If you look at the triangle $Delta,$ from the fact that the height of the triangle is $frac{psqrt{3}}{2},$ using the Pythagorean Theorem, we see that $|z-c|=|lambda a+(1-lambda )b-c|=psqrt{lambda^2-lambda+1},$ so $h(z)=lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}.$
We see that $max h(z)=max{lambda(1-lambda)p^3sqrt{lambda^2-lambda+1}|lambdain(0,1)}=frac{sqrt{3}}{2}p^3.$
answered Jan 7 at 15:55
EminEmin
1,36021330
1,36021330
add a comment |
add a comment |
$begingroup$
Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.
The specific triangular shape does not matter; the same holds for any domain.
I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.
$endgroup$
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
add a comment |
$begingroup$
Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.
The specific triangular shape does not matter; the same holds for any domain.
I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.
$endgroup$
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
add a comment |
$begingroup$
Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.
The specific triangular shape does not matter; the same holds for any domain.
I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.
$endgroup$
Yes, by the maximum principle applied to $(z-a)(z-b)(z-c)$, the maximum of $h$ is attained at a (not the) boundary point of the triangle.
The specific triangular shape does not matter; the same holds for any domain.
I'll add a similar, somewhat entertaining example: distribute $n$ points $z_1,dots,z_n$ uniformly along the unit circle. The maximum of the product $prod_{k=1}^n |z-z_k|$ over the disk $|z|le 1$ is equal to $2$ (and is attained at the boundary). It's not exactly intuitive that the maximum is independent of $n$.
answered Jan 6 '14 at 15:58
Post No BullsPost No Bulls
6,42321682
6,42321682
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
add a comment |
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
$begingroup$
@@Post No Bulls) Can you please tell me how maximum value is $2$ in your last answer ?
$endgroup$
– Empty
Oct 12 '15 at 20:13
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%2f629120%2fhz-z-a-z-b-z-c-max-value-of-h-is-attained%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
$begingroup$
$h$ is not holomorphic or harmonic, so the maximum modulus principle doesn't apply.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:44
4
$begingroup$
well what if I apply MMP on $g(z)=(z-a)(z-b)(z-c)$ ?
$endgroup$
– Markov
Jan 6 '14 at 14:47
1
$begingroup$
Ah, then you have a correct argument.
$endgroup$
– Daniel Fischer♦
Jan 6 '14 at 14:51