Positive value of function and root
$begingroup$
Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.
The polynomial
$$
f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
$$
has exactly one positive root $x_0$ by Descartes' rule.
Do we also have that
$$
f(x)>0implies x>x_0?
$$
real-analysis roots
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
$endgroup$
add a comment |
$begingroup$
Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.
The polynomial
$$
f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
$$
has exactly one positive root $x_0$ by Descartes' rule.
Do we also have that
$$
f(x)>0implies x>x_0?
$$
real-analysis roots
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
$endgroup$
add a comment |
$begingroup$
Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.
The polynomial
$$
f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
$$
has exactly one positive root $x_0$ by Descartes' rule.
Do we also have that
$$
f(x)>0implies x>x_0?
$$
real-analysis roots
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
$endgroup$
Suppose $j,kinmathbb{N}$ and $1leqslant jleqslant k$.
The polynomial
$$
f(x)=x^{2k}-sum_{i=0}^{k-j-1}x^i
$$
has exactly one positive root $x_0$ by Descartes' rule.
Do we also have that
$$
f(x)>0implies x>x_0?
$$
real-analysis roots
real-analysis roots
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
asked Dec 18 '18 at 10:10
SalamoSalamo
359412
359412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.
However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
1
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
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%2f3044989%2fpositive-value-of-function-and-root%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
$begingroup$
Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.
However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
1
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
add a comment |
$begingroup$
Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.
However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
1
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
add a comment |
$begingroup$
Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.
However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
$endgroup$
Since $x_0$ is the only positive root of $f$, it is the largest one. Then, there is no sign change of $f$ on $(x_0, infty)$ and since $underset{xto + infty}{lim} f = + infty$. You may indeed assert that $ x > x_0 Rightarrow f(x) > 0 $.
However $ f(x) > 0 Rightarrow x > x_0 $ does not generally hold. Take $j=k=1$, then $f(x) = x^2$ and $f(-1)>0$ but $-1 < 0$.
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
answered Dec 18 '18 at 10:21
Bill O'HaranBill O'Haran
2,5431418
2,5431418
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
1
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
add a comment |
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
1
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
Thanks. But I guess that $f(x)>0$ and $xgeq 0$ imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 10:42
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
That depends on the multiplicity of $x_0$. If it is odd then yes. Otherwise, for $xin (0,x_0)$, $f(x)>0$.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 10:54
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
$begingroup$
And here, $x_0$ is a simple root (hence odd), hence my statement holds, i.e. $f(x)>0$ and $xgeq 0$ indeed imply $x>x_0$.
$endgroup$
– Salamo
Dec 18 '18 at 13:13
1
1
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
$begingroup$
Sure. I forgot that Descartes' rule took multiplicity into account.
$endgroup$
– Bill O'Haran
Dec 18 '18 at 13:27
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%2f3044989%2fpositive-value-of-function-and-root%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
