Mixing
Maclaurin's series for $tan(x+x^2)$
$begingroup$
I would like to find the Maclaurin's series for $tan(x+x^2)$ stopped at $x_0^3$.
I tried with:
$$tan(x+x^2)=1+frac{1+2x_0}{cos^2(x_0+x_0^2)}x+....,$$ but it seems like the derivation gets more complicated. I was wondering if there was a simpler way to solve the Maclaurin series. I was thinking about putting $t=x+x^2$, but it looks like wrong (why?).
Update:
I know that MacLaurin's expansion for tan(x): $$T(x) = x + frac {1}{3}x^{3} + frac { 2 } { 15 } x ^ { 5 } + frac { 17 } { 315 } x ^ { 7 } + ldots$$ But according to Mathematica, there should be a $x^2$ for the MacLaurin series for $tan(x+x^2)$, which I couldn't understand.
calculus derivatives taylor-expansion
asked Jan 21 at 14:18
KevinKevin
16211
$endgroup$
|
show 1 more comment
$begingroup$
I would like to find the Maclaurin's series for $tan(x+x^2)$ stopped at $x_0^3$.
I tried with:
$$tan(x+x^2)=1+frac{1+2x_0}{cos^2(x_0+x_0^2)}x+....,$$ but it seems like the derivation gets more complicated. I was wondering if there was a simpler way to solve the Maclaurin series. I was thinking about putting $t=x+x^2$, but it looks like wrong (why?).
Update:
I know that MacLaurin's expansion for tan(x): $$T(x) = x + frac {1}{3}x^{3} + frac { 2 } { 15 } x ^ { 5 } + frac { 17 } { 315 } x ^ { 7 } + ldots$$ But according to Mathematica, there should be a $x^2$ for the MacLaurin series for $tan(x+x^2)$, which I couldn't understand.
calculus derivatives taylor-expansion
asked Jan 21 at 14:18
KevinKevin
16211
$endgroup$
$begingroup$
If all you care about is the series and not its derivation, why not use Mathematica? I doubt there is any special trick for this.
$endgroup$
– Hamed
Jan 21 at 14:26
$begingroup$
Could you show me an example? Do you mean substituting $x_0$ with 0 and solving the derivatives?
$endgroup$
– Kevin
Jan 21 at 14:26
$begingroup$
@Hamed Because it is part of a question of my past exams, and we are not allowed to use Mathematica.
$endgroup$
– Kevin
Jan 21 at 14:28
2
$begingroup$
Setting $t=x+x^2$ in the series for $tan{t}$ will work fine. Why wouldn't it?
$endgroup$
– saulspatz
Jan 21 at 14:29
$begingroup$
Apparently, the expansion is around an arbitrary point $x_0$ not zero.
$endgroup$
– Hamed
Jan 21 at 14:33
|
show 1 more comment
$begingroup$
I would like to find the Maclaurin's series for $tan(x+x^2)$ stopped at $x_0^3$.
I tried with:
$$tan(x+x^2)=1+frac{1+2x_0}{cos^2(x_0+x_0^2)}x+....,$$ but it seems like the derivation gets more complicated. I was wondering if there was a simpler way to solve the Maclaurin series. I was thinking about putting $t=x+x^2$, but it looks like wrong (why?).
Update:
I know that MacLaurin's expansion for tan(x): $$T(x) = x + frac {1}{3}x^{3} + frac { 2 } { 15 } x ^ { 5 } + frac { 17 } { 315 } x ^ { 7 } + ldots$$ But according to Mathematica, there should be a $x^2$ for the MacLaurin series for $tan(x+x^2)$, which I couldn't understand.
calculus derivatives taylor-expansion
asked Jan 21 at 14:18
KevinKevin
16211
$endgroup$
I would like to find the Maclaurin's series for $tan(x+x^2)$ stopped at $x_0^3$.
I tried with:
$$tan(x+x^2)=1+frac{1+2x_0}{cos^2(x_0+x_0^2)}x+....,$$ but it seems like the derivation gets more complicated. I was wondering if there was a simpler way to solve the Maclaurin series. I was thinking about putting $t=x+x^2$, but it looks like wrong (why?).
Update:
I know that MacLaurin's expansion for tan(x): $$T(x) = x + frac {1}{3}x^{3} + frac { 2 } { 15 } x ^ { 5 } + frac { 17 } { 315 } x ^ { 7 } + ldots$$ But according to Mathematica, there should be a $x^2$ for the MacLaurin series for $tan(x+x^2)$, which I couldn't understand.
calculus derivatives taylor-expansion
calculus derivatives taylor-expansion
asked Jan 21 at 14:18
KevinKevin
16211
asked Jan 21 at 14:18
KevinKevin
16211
edited Jan 21 at 14:53
Kevin
asked Jan 21 at 14:18
KevinKevin
16211
asked Jan 21 at 14:18
KevinKevin
16211
asked Jan 21 at 14:18
KevinKevin
16211
16211
$begingroup$
If all you care about is the series and not its derivation, why not use Mathematica? I doubt there is any special trick for this.
$endgroup$
– Hamed
Jan 21 at 14:26
$begingroup$
Could you show me an example? Do you mean substituting $x_0$ with 0 and solving the derivatives?
$endgroup$
– Kevin
Jan 21 at 14:26
$begingroup$
@Hamed Because it is part of a question of my past exams, and we are not allowed to use Mathematica.
$endgroup$
– Kevin
Jan 21 at 14:28
2
$begingroup$
Setting $t=x+x^2$ in the series for $tan{t}$ will work fine. Why wouldn't it?
$endgroup$
– saulspatz
Jan 21 at 14:29
$begingroup$
Apparently, the expansion is around an arbitrary point $x_0$ not zero.
$endgroup$
– Hamed
Jan 21 at 14:33
|
show 1 more comment
$begingroup$
If all you care about is the series and not its derivation, why not use Mathematica? I doubt there is any special trick for this.
$endgroup$
– Hamed
Jan 21 at 14:26
$begingroup$
Could you show me an example? Do you mean substituting $x_0$ with 0 and solving the derivatives?
$endgroup$
– Kevin
Jan 21 at 14:26
$begingroup$
@Hamed Because it is part of a question of my past exams, and we are not allowed to use Mathematica.
$endgroup$
– Kevin
Jan 21 at 14:28
2
$begingroup$
Setting $t=x+x^2$ in the series for $tan{t}$ will work fine. Why wouldn't it?
$endgroup$
– saulspatz
Jan 21 at 14:29
$begingroup$
Apparently, the expansion is around an arbitrary point $x_0$ not zero.
$endgroup$
– Hamed
Jan 21 at 14:33
$begingroup$
If all you care about is the series and not its derivation, why not use Mathematica? I doubt there is any special trick for this.
$endgroup$
– Hamed
Jan 21 at 14:26
$begingroup$
If all you care about is the series and not its derivation, why not use Mathematica? I doubt there is any special trick for this.
$endgroup$
– Hamed
Jan 21 at 14:26
$begingroup$
Could you show me an example? Do you mean substituting $x_0$ with 0 and solving the derivatives?
$endgroup$
– Kevin
Jan 21 at 14:26
$begingroup$
Could you show me an example? Do you mean substituting $x_0$ with 0 and solving the derivatives?
$endgroup$
– Kevin
Jan 21 at 14:26
$begingroup$
@Hamed Because it is part of a question of my past exams, and we are not allowed to use Mathematica.
$endgroup$
– Kevin
Jan 21 at 14:28
$begingroup$
@Hamed Because it is part of a question of my past exams, and we are not allowed to use Mathematica.
$endgroup$
– Kevin
Jan 21 at 14:28
2
2
$begingroup$
Setting $t=x+x^2$ in the series for $tan{t}$ will work fine. Why wouldn't it?
$endgroup$
– saulspatz
Jan 21 at 14:29
$begingroup$
Setting $t=x+x^2$ in the series for $tan{t}$ will work fine. Why wouldn't it?
$endgroup$
– saulspatz
Jan 21 at 14:29
$begingroup$
Apparently, the expansion is around an arbitrary point $x_0$ not zero.
$endgroup$
– Hamed
Jan 21 at 14:33
$begingroup$
Apparently, the expansion is around an arbitrary point $x_0$ not zero.
$endgroup$
– Hamed
Jan 21 at 14:33
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Just substitute $x+x^2$ for $x$ in your formula of $T(x):$
$$T(x+x^2)=(x+x^2)+frac13(x+x^2)^3+cdots$$
Now you just have to expand and rearrange the terms. Since you only need powers up through $x^3$, there no need to be concerned with the higher-degree terms in $T(x)$
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
$endgroup$
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
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%2f3081921%2fmaclaurins-series-for-tanxx2%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
$begingroup$
Just substitute $x+x^2$ for $x$ in your formula of $T(x):$
$$T(x+x^2)=(x+x^2)+frac13(x+x^2)^3+cdots$$
Now you just have to expand and rearrange the terms. Since you only need powers up through $x^3$, there no need to be concerned with the higher-degree terms in $T(x)$
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
$endgroup$
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
add a comment |
$begingroup$
Just substitute $x+x^2$ for $x$ in your formula of $T(x):$
$$T(x+x^2)=(x+x^2)+frac13(x+x^2)^3+cdots$$
Now you just have to expand and rearrange the terms. Since you only need powers up through $x^3$, there no need to be concerned with the higher-degree terms in $T(x)$
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
$endgroup$
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
add a comment |
$begingroup$
Just substitute $x+x^2$ for $x$ in your formula of $T(x):$
$$T(x+x^2)=(x+x^2)+frac13(x+x^2)^3+cdots$$
Now you just have to expand and rearrange the terms. Since you only need powers up through $x^3$, there no need to be concerned with the higher-degree terms in $T(x)$
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
$endgroup$
Just substitute $x+x^2$ for $x$ in your formula of $T(x):$
$$T(x+x^2)=(x+x^2)+frac13(x+x^2)^3+cdots$$
Now you just have to expand and rearrange the terms. Since you only need powers up through $x^3$, there no need to be concerned with the higher-degree terms in $T(x)$
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
answered Jan 21 at 14:58
saulspatzsaulspatz
16.3k31332
16.3k31332
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
add a comment |
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
$begingroup$
Now I get it. Apparently, I had a lot of confusion in my mind. Thank you!
$endgroup$
– Kevin
Jan 21 at 15:09
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%2f3081921%2fmaclaurins-series-for-tanxx2%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
$begingroup$
If all you care about is the series and not its derivation, why not use Mathematica? I doubt there is any special trick for this.
$endgroup$
– Hamed
Jan 21 at 14:26
$begingroup$
Could you show me an example? Do you mean substituting $x_0$ with 0 and solving the derivatives?
$endgroup$
– Kevin
Jan 21 at 14:26
$begingroup$
@Hamed Because it is part of a question of my past exams, and we are not allowed to use Mathematica.
$endgroup$
– Kevin
Jan 21 at 14:28
2
$begingroup$
Setting $t=x+x^2$ in the series for $tan{t}$ will work fine. Why wouldn't it?
$endgroup$
– saulspatz
Jan 21 at 14:29
$begingroup$
Apparently, the expansion is around an arbitrary point $x_0$ not zero.
$endgroup$
– Hamed
Jan 21 at 14:33