Computation of Maclaurin Series












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    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


















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    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
















2












2








2


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
    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




















  • $begingroup$
    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
    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


















$begingroup$
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
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
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




$begingroup$
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












1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $x^4$



$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$






share|cite|improve this answer









$endgroup$













    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
    });


    }
    });














    draft saved

    draft discarded


















    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









    4












    $begingroup$

    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 $x^4$



    $$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      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 $x^4$



      $$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        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 $x^4$



        $$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$






        share|cite|improve this answer









        $endgroup$



        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 $x^4$



        $$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Will JagyWill Jagy

        102k5101199




        102k5101199






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            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





















































            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







            -

            Popular posts from this blog

            Accessing regular linux commands in Huawei's Dopra Linux

            Image
            4 Backstory: I have a Huawei HG8245 router in my house and I want to change the default username/password. I've tried following this guide but the config file always gets reset. My ISP isn't helping either so I'm looking for other ways to accomplish this. The router has telnet access so this seems like a way to change the credentials. The first shell I gain access to looks like this: WAP> Pressing "?" gives me a list of available commands which I've pasted here, the two interesting ones are "su" and "shell". I can't seem to get su access, even after logging in as root (challenge password prompt, and then nothing happens), but shell gives me a proper linux shell. WAP>shell BusyBox v1.18.4 (2017-12-26 17:06:34 CST) built-in shell (ash) Enter 'hel
            Read more

            Can't connect RFCOMM socket: Host is down

            Image
            up vote 3 down vote favorite I have bluez 5.4.3 and I am attempting to communicate with a ble device through minicom. These two deceives (my computer and a bluno nano) are alreadt paired and this has been my method so far. #bluetoothctl # pair xx:xx:xx:xx:xx:xx # exit Everything is working well so far then... #sudo rfcomm connect /dev/rfcomm0 xx:xx:xx:xx:xx:xx 1 & [1] 3665 Can't connect RFCOMM socket: Host is down Does anyone know how to resolve this issue? Any help would be very much appreciated. bluetooth bluez share | improve this question asked Feb 9 '17 at 18:46 Brandon Knape 46 2
            Read more

            Kernel panic - not syncing: Fatal Exception in Interrupt

            Image
            0 Every day this error occured on my pc, kernel panic error on interrupt, what will be the problem? linux-mint share asked 6 mins ago Vinod Selvin Vinod Selvin 1 1 New contributor Vinod Selvin is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment
            Read more