Question about lemma for $P(x,y) =Ax^2 + 2Bxy + Cy^2$.
up vote
0
down vote
favorite
Let $P(x,y)=Ax^2 + 2Bxy + Cy^2$, and set
$triangle = frac{(P_{12})^2 - P_{11}P_{22}}{4}=B^{2}-AC$
Then,
(i) If $triangle >0$, then there are two lines through the origin such that $P(x,y)>0$ for all $(x,y)$ on one, and $P(x,y)<0$ for all $(x,y)$ on the other, with the point $(0,0)$ omitted.
To prove (i), we assume that $triangle>0$ and compute
$P(B,-A) = AB^2 -2B^2A + CA^2= -A triangle$
$P(C,-B) = AC^2 -2B^2C + CB^2 = -Ctriangle$
Since $P$ is homogeneous of degree $2$, $P(lambda x,lambda y) = lambda^{2}P(x,y)$, it will be sufficient if we can find two points at which $P$ has opposite signs, since the same will then hold on the entire line joining these to the origin.
I was wondering if someone could further elaborate on the last sentence. Let's say that we find two points at which $P$ has opposite signs, why would the same hold on the entire line joining these to the origin? And why does this follow from homogeneity?
calculus multivariable-calculus
asked Nov 23 at 1:32
K.M
651312
add a comment |
up vote
0
down vote
favorite
Let $P(x,y)=Ax^2 + 2Bxy + Cy^2$, and set
$triangle = frac{(P_{12})^2 - P_{11}P_{22}}{4}=B^{2}-AC$
Then,
(i) If $triangle >0$, then there are two lines through the origin such that $P(x,y)>0$ for all $(x,y)$ on one, and $P(x,y)<0$ for all $(x,y)$ on the other, with the point $(0,0)$ omitted.
To prove (i), we assume that $triangle>0$ and compute
$P(B,-A) = AB^2 -2B^2A + CA^2= -A triangle$
$P(C,-B) = AC^2 -2B^2C + CB^2 = -Ctriangle$
Since $P$ is homogeneous of degree $2$, $P(lambda x,lambda y) = lambda^{2}P(x,y)$, it will be sufficient if we can find two points at which $P$ has opposite signs, since the same will then hold on the entire line joining these to the origin.
I was wondering if someone could further elaborate on the last sentence. Let's say that we find two points at which $P$ has opposite signs, why would the same hold on the entire line joining these to the origin? And why does this follow from homogeneity?
calculus multivariable-calculus
asked Nov 23 at 1:32
K.M
651312
Delta has undefined terms. Do you mean P(1,2)?
– William Elliot
Nov 23 at 1:48
Another problem is that it is possible to have positive discriminant with $A$ and $C$ having the same sign, that is $AC > 0.$ You have most likely left off a hypothesis. The one I would expect is Gauss-Lagrange reduced, which amounts to $AC < 0$ and, let's see, $2B > |A+C|$
– Will Jagy
Nov 23 at 1:52
I fixed it now. delta is equal to $B^{2}-AC$. thanks
– K.M
Nov 23 at 1:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $P(x,y)=Ax^2 + 2Bxy + Cy^2$, and set
$triangle = frac{(P_{12})^2 - P_{11}P_{22}}{4}=B^{2}-AC$
Then,
(i) If $triangle >0$, then there are two lines through the origin such that $P(x,y)>0$ for all $(x,y)$ on one, and $P(x,y)<0$ for all $(x,y)$ on the other, with the point $(0,0)$ omitted.
To prove (i), we assume that $triangle>0$ and compute
$P(B,-A) = AB^2 -2B^2A + CA^2= -A triangle$
$P(C,-B) = AC^2 -2B^2C + CB^2 = -Ctriangle$
Since $P$ is homogeneous of degree $2$, $P(lambda x,lambda y) = lambda^{2}P(x,y)$, it will be sufficient if we can find two points at which $P$ has opposite signs, since the same will then hold on the entire line joining these to the origin.
I was wondering if someone could further elaborate on the last sentence. Let's say that we find two points at which $P$ has opposite signs, why would the same hold on the entire line joining these to the origin? And why does this follow from homogeneity?
calculus multivariable-calculus
asked Nov 23 at 1:32
K.M
651312
Let $P(x,y)=Ax^2 + 2Bxy + Cy^2$, and set
$triangle = frac{(P_{12})^2 - P_{11}P_{22}}{4}=B^{2}-AC$
Then,
(i) If $triangle >0$, then there are two lines through the origin such that $P(x,y)>0$ for all $(x,y)$ on one, and $P(x,y)<0$ for all $(x,y)$ on the other, with the point $(0,0)$ omitted.
To prove (i), we assume that $triangle>0$ and compute
$P(B,-A) = AB^2 -2B^2A + CA^2= -A triangle$
$P(C,-B) = AC^2 -2B^2C + CB^2 = -Ctriangle$
Since $P$ is homogeneous of degree $2$, $P(lambda x,lambda y) = lambda^{2}P(x,y)$, it will be sufficient if we can find two points at which $P$ has opposite signs, since the same will then hold on the entire line joining these to the origin.
I was wondering if someone could further elaborate on the last sentence. Let's say that we find two points at which $P$ has opposite signs, why would the same hold on the entire line joining these to the origin? And why does this follow from homogeneity?
calculus multivariable-calculus
calculus multivariable-calculus
asked Nov 23 at 1:32
K.M
651312
asked Nov 23 at 1:32
K.M
651312
edited Nov 23 at 1:52
asked Nov 23 at 1:32
K.M
651312
asked Nov 23 at 1:32
K.M
651312
asked Nov 23 at 1:32
K.M
651312
651312
Delta has undefined terms. Do you mean P(1,2)?
– William Elliot
Nov 23 at 1:48
Another problem is that it is possible to have positive discriminant with $A$ and $C$ having the same sign, that is $AC > 0.$ You have most likely left off a hypothesis. The one I would expect is Gauss-Lagrange reduced, which amounts to $AC < 0$ and, let's see, $2B > |A+C|$
– Will Jagy
Nov 23 at 1:52
I fixed it now. delta is equal to $B^{2}-AC$. thanks
– K.M
Nov 23 at 1:53
add a comment |
Delta has undefined terms. Do you mean P(1,2)?
– William Elliot
Nov 23 at 1:48
Another problem is that it is possible to have positive discriminant with $A$ and $C$ having the same sign, that is $AC > 0.$ You have most likely left off a hypothesis. The one I would expect is Gauss-Lagrange reduced, which amounts to $AC < 0$ and, let's see, $2B > |A+C|$
– Will Jagy
Nov 23 at 1:52
I fixed it now. delta is equal to $B^{2}-AC$. thanks
– K.M
Nov 23 at 1:53
Delta has undefined terms. Do you mean P(1,2)?
– William Elliot
Nov 23 at 1:48
Delta has undefined terms. Do you mean P(1,2)?
– William Elliot
Nov 23 at 1:48
Another problem is that it is possible to have positive discriminant with $A$ and $C$ having the same sign, that is $AC > 0.$ You have most likely left off a hypothesis. The one I would expect is Gauss-Lagrange reduced, which amounts to $AC < 0$ and, let's see, $2B > |A+C|$
– Will Jagy
Nov 23 at 1:52
Another problem is that it is possible to have positive discriminant with $A$ and $C$ having the same sign, that is $AC > 0.$ You have most likely left off a hypothesis. The one I would expect is Gauss-Lagrange reduced, which amounts to $AC < 0$ and, let's see, $2B > |A+C|$
– Will Jagy
Nov 23 at 1:52
I fixed it now. delta is equal to $B^{2}-AC$. thanks
– K.M
Nov 23 at 1:53
I fixed it now. delta is equal to $B^{2}-AC$. thanks
– K.M
Nov 23 at 1:53
add a comment |
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',
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%2f3009876%2fquestion-about-lemma-for-px-y-ax2-2bxy-cy2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
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.
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%2f3009876%2fquestion-about-lemma-for-px-y-ax2-2bxy-cy2%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
Delta has undefined terms. Do you mean P(1,2)?
– William Elliot
Nov 23 at 1:48
Another problem is that it is possible to have positive discriminant with $A$ and $C$ having the same sign, that is $AC > 0.$ You have most likely left off a hypothesis. The one I would expect is Gauss-Lagrange reduced, which amounts to $AC < 0$ and, let's see, $2B > |A+C|$
– Will Jagy
Nov 23 at 1:52
I fixed it now. delta is equal to $B^{2}-AC$. thanks
– K.M
Nov 23 at 1:53