Whenever $y'$ is in $sin(y')$ or as a power, the degree of the polynomial equation is not defined, why?
$begingroup$
In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?
calculus ordinary-differential-equations
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
$endgroup$
add a comment |
$begingroup$
In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?
calculus ordinary-differential-equations
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
$endgroup$
2
$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47
$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05
add a comment |
$begingroup$
In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?
calculus ordinary-differential-equations
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
$endgroup$
In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
edited Nov 29 '18 at 14:46
Davide Giraudo
126k16150261
126k16150261
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
asked Sep 24 '17 at 5:13
WhiteHoleWhiteHole
216
216
2
$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47
$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05
add a comment |
2
$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47
$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05
2
2
$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47
$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47
$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05
$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
begin{array}{l}
y' + sin left( {y'} right) = 0\
{rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
{rm{Then,the first order differential equation can be written as }}\
y' + sin left( {y'} right) = 0\
y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
end{array}
Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
$endgroup$
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%2f2442701%2fwhenever-y-is-in-siny-or-as-a-power-the-degree-of-the-polynomial-equat%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
$begingroup$
begin{array}{l}
y' + sin left( {y'} right) = 0\
{rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
{rm{Then,the first order differential equation can be written as }}\
y' + sin left( {y'} right) = 0\
y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
end{array}
Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
$endgroup$
add a comment |
$begingroup$
begin{array}{l}
y' + sin left( {y'} right) = 0\
{rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
{rm{Then,the first order differential equation can be written as }}\
y' + sin left( {y'} right) = 0\
y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
end{array}
Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
$endgroup$
add a comment |
$begingroup$
begin{array}{l}
y' + sin left( {y'} right) = 0\
{rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
{rm{Then,the first order differential equation can be written as }}\
y' + sin left( {y'} right) = 0\
y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
end{array}
Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
$endgroup$
begin{array}{l}
y' + sin left( {y'} right) = 0\
{rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
{rm{Then,the first order differential equation can be written as }}\
y' + sin left( {y'} right) = 0\
y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
end{array}
Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
894
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.
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%2f2442701%2fwhenever-y-is-in-siny-or-as-a-power-the-degree-of-the-polynomial-equat%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
2
$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47
$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05