Mixing
Maclaurin expansion
$begingroup$
$ f(x,y)=sin(x-2y)$
I have to expand this into Maclaurin series.
Of course I would like to use known series
$sinx = sum_{n=0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$
First of all, i treat $x$ as variable, and $y$ as parameter. We have that
$frac{d}{dx}f(x,y) = cos(x-2y)$
$frac{d^2}{dx^2}f(x,y) = -sin(x-2y)$
$frac{d^3}{dx^3}f(x,y) = -cos(x-2y)$
$frac{d^4}{dx^4}f(x,y) = sin(x-2y)$
...
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
...
$y$ as variable, and $x$ as parameter. We have that
$frac{d}{dy}f(x,y) = -2cos(x-2y)$
$frac{d^2}{dy^2}f(x,y) = -4sin(x-2y)$
$frac{d^3}{dy^3}f(x,y) = 8cos(x-2y)$
$frac{d^4}{dy^4}f(x,y) = 16sin(x-2y)$
...
$frac{d}{dy}f(0,0) = -2$
$frac{d}{dy}f(0,0) = 0$
$frac{d}{dy}f(0,0) = 8$
$frac{d}{dy}f(0,0) = 0$
And what now? How I should write my series?
I think this
$sin(x-2y) = (sum_{k=0}^infty (-1)^kfrac{x^{2k+1}}{(2k+1)!})$$( sum_{l=0}^infty (-1)^lfrac{y^{2l+1}}{(2l+1)!})$
is completly wrong. Give me some advice or just show me how to make it, thanks.
analysis
asked Jan 27 at 9:24
VictorVictor
464
$endgroup$
add a comment |
$begingroup$
$ f(x,y)=sin(x-2y)$
I have to expand this into Maclaurin series.
Of course I would like to use known series
$sinx = sum_{n=0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$
First of all, i treat $x$ as variable, and $y$ as parameter. We have that
$frac{d}{dx}f(x,y) = cos(x-2y)$
$frac{d^2}{dx^2}f(x,y) = -sin(x-2y)$
$frac{d^3}{dx^3}f(x,y) = -cos(x-2y)$
$frac{d^4}{dx^4}f(x,y) = sin(x-2y)$
...
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
...
$y$ as variable, and $x$ as parameter. We have that
$frac{d}{dy}f(x,y) = -2cos(x-2y)$
$frac{d^2}{dy^2}f(x,y) = -4sin(x-2y)$
$frac{d^3}{dy^3}f(x,y) = 8cos(x-2y)$
$frac{d^4}{dy^4}f(x,y) = 16sin(x-2y)$
...
$frac{d}{dy}f(0,0) = -2$
$frac{d}{dy}f(0,0) = 0$
$frac{d}{dy}f(0,0) = 8$
$frac{d}{dy}f(0,0) = 0$
And what now? How I should write my series?
I think this
$sin(x-2y) = (sum_{k=0}^infty (-1)^kfrac{x^{2k+1}}{(2k+1)!})$$( sum_{l=0}^infty (-1)^lfrac{y^{2l+1}}{(2l+1)!})$
is completly wrong. Give me some advice or just show me how to make it, thanks.
analysis
asked Jan 27 at 9:24
VictorVictor
464
$endgroup$
add a comment |
$begingroup$
$ f(x,y)=sin(x-2y)$
I have to expand this into Maclaurin series.
Of course I would like to use known series
$sinx = sum_{n=0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$
First of all, i treat $x$ as variable, and $y$ as parameter. We have that
$frac{d}{dx}f(x,y) = cos(x-2y)$
$frac{d^2}{dx^2}f(x,y) = -sin(x-2y)$
$frac{d^3}{dx^3}f(x,y) = -cos(x-2y)$
$frac{d^4}{dx^4}f(x,y) = sin(x-2y)$
...
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
...
$y$ as variable, and $x$ as parameter. We have that
$frac{d}{dy}f(x,y) = -2cos(x-2y)$
$frac{d^2}{dy^2}f(x,y) = -4sin(x-2y)$
$frac{d^3}{dy^3}f(x,y) = 8cos(x-2y)$
$frac{d^4}{dy^4}f(x,y) = 16sin(x-2y)$
...
$frac{d}{dy}f(0,0) = -2$
$frac{d}{dy}f(0,0) = 0$
$frac{d}{dy}f(0,0) = 8$
$frac{d}{dy}f(0,0) = 0$
And what now? How I should write my series?
I think this
$sin(x-2y) = (sum_{k=0}^infty (-1)^kfrac{x^{2k+1}}{(2k+1)!})$$( sum_{l=0}^infty (-1)^lfrac{y^{2l+1}}{(2l+1)!})$
is completly wrong. Give me some advice or just show me how to make it, thanks.
analysis
asked Jan 27 at 9:24
VictorVictor
464
$endgroup$
$ f(x,y)=sin(x-2y)$
I have to expand this into Maclaurin series.
Of course I would like to use known series
$sinx = sum_{n=0}^infty (-1)^nfrac{x^{2n+1}}{(2n+1)!}$
First of all, i treat $x$ as variable, and $y$ as parameter. We have that
$frac{d}{dx}f(x,y) = cos(x-2y)$
$frac{d^2}{dx^2}f(x,y) = -sin(x-2y)$
$frac{d^3}{dx^3}f(x,y) = -cos(x-2y)$
$frac{d^4}{dx^4}f(x,y) = sin(x-2y)$
...
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
$frac{d}{dx}f(0,0) = 1$
$frac{d}{dx}f(0,0) = 0$
...
$y$ as variable, and $x$ as parameter. We have that
$frac{d}{dy}f(x,y) = -2cos(x-2y)$
$frac{d^2}{dy^2}f(x,y) = -4sin(x-2y)$
$frac{d^3}{dy^3}f(x,y) = 8cos(x-2y)$
$frac{d^4}{dy^4}f(x,y) = 16sin(x-2y)$
...
$frac{d}{dy}f(0,0) = -2$
$frac{d}{dy}f(0,0) = 0$
$frac{d}{dy}f(0,0) = 8$
$frac{d}{dy}f(0,0) = 0$
And what now? How I should write my series?
I think this
$sin(x-2y) = (sum_{k=0}^infty (-1)^kfrac{x^{2k+1}}{(2k+1)!})$$( sum_{l=0}^infty (-1)^lfrac{y^{2l+1}}{(2l+1)!})$
is completly wrong. Give me some advice or just show me how to make it, thanks.
analysis
analysis
asked Jan 27 at 9:24
VictorVictor
464
asked Jan 27 at 9:24
VictorVictor
464
edited Jan 27 at 14:22
Victor
asked Jan 27 at 9:24
VictorVictor
464
asked Jan 27 at 9:24
VictorVictor
464
asked Jan 27 at 9:24
VictorVictor
464
464
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Better way is to remember that
$$ sin x = sum frac{ (-1)^n x^{2n+1} }{(2n+1)!} $$
and so
$$ sin(x-2y) = sum frac{ (-1)^n (x-2y)^{2n+1} }{(2n+1)!} $$
One can expand further if using the binomial
$$ (a+b)^n = sum_{i=0}^n {n choose i} a^i b^{n-i} $$
Only a masochist would compute the partials..
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
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%2f3089324%2fmaclaurin-expansion%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
$begingroup$
Better way is to remember that
$$ sin x = sum frac{ (-1)^n x^{2n+1} }{(2n+1)!} $$
and so
$$ sin(x-2y) = sum frac{ (-1)^n (x-2y)^{2n+1} }{(2n+1)!} $$
One can expand further if using the binomial
$$ (a+b)^n = sum_{i=0}^n {n choose i} a^i b^{n-i} $$
Only a masochist would compute the partials..
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
add a comment |
$begingroup$
Better way is to remember that
$$ sin x = sum frac{ (-1)^n x^{2n+1} }{(2n+1)!} $$
and so
$$ sin(x-2y) = sum frac{ (-1)^n (x-2y)^{2n+1} }{(2n+1)!} $$
One can expand further if using the binomial
$$ (a+b)^n = sum_{i=0}^n {n choose i} a^i b^{n-i} $$
Only a masochist would compute the partials..
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
add a comment |
$begingroup$
Better way is to remember that
$$ sin x = sum frac{ (-1)^n x^{2n+1} }{(2n+1)!} $$
and so
$$ sin(x-2y) = sum frac{ (-1)^n (x-2y)^{2n+1} }{(2n+1)!} $$
One can expand further if using the binomial
$$ (a+b)^n = sum_{i=0}^n {n choose i} a^i b^{n-i} $$
Only a masochist would compute the partials..
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
$endgroup$
Better way is to remember that
$$ sin x = sum frac{ (-1)^n x^{2n+1} }{(2n+1)!} $$
and so
$$ sin(x-2y) = sum frac{ (-1)^n (x-2y)^{2n+1} }{(2n+1)!} $$
One can expand further if using the binomial
$$ (a+b)^n = sum_{i=0}^n {n choose i} a^i b^{n-i} $$
Only a masochist would compute the partials..
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
edited Jan 27 at 9:36
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
answered Jan 27 at 9:31
Jimmy SabaterJimmy Sabater
2,539325
2,539325
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
add a comment |
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
LOL, and that is it? Why we don't use a double sum here? Edit: Because there is only one function, right?
$endgroup$
– Victor
Jan 27 at 9:35
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
$begingroup$
Use binomial with $a=x$ and $b=-2y$
$endgroup$
– Jimmy Sabater
Jan 27 at 9:36
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%2f3089324%2fmaclaurin-expansion%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
