Computation of Maclaurin Series
$begingroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
$endgroup$
add a comment |
$begingroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
$endgroup$
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
4 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
add a comment |
$begingroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
$endgroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
calculus logarithms taylor-expansion
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
asked 4 hours ago
Turan NəsibliTuran Nəsibli
564
564
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
4 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
add a comment |
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
4 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
4 hours ago
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
4 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
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%2f3081410%2fcomputation-of-maclaurin-series%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
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
answered 3 hours ago
Will JagyWill Jagy
102k5101199
answered 3 hours ago
Will JagyWill Jagy
102k5101199
answered 3 hours ago
Will JagyWill Jagy
102k5101199
102k5101199
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%2f3081410%2fcomputation-of-maclaurin-series%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
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
4 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago