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 3 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

            flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

            Image
            0 I recently upgraded an Ubuntu 16.04 virtual machine to 18.04. I regularly use this VM to run apt-mirror at school on their high-speed internet, then rsync these files to my rack-mount server at home for updating multiple computers at home without wasting the tiny amount of bandwidth I have there. Since updating I can't get apt-mirror to run on my VM. It keeps giving the error: flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror line 214 immediately followed by: apt-mirror is already running, exiting at /usr/bin/apt-mirror line 217 apt-mirror is NOT running, and does not have a cron job. I've tried rebooting, and even looked for the old-style lockfile to delete, but I guess that method is no longer used. Is there some way to force unlock this so I can run apt-mirror? UPDATE: So... I f...
            Read more

            Mangá

            Image
              Nota:  Não confundir com Manga. Mangás e Animes Mangá Mangaká • Tankōbon Film comic La nouvelle manga Mangá original em inglês Anime História Animação influenciada por animes • Filler • OVA • Seiyū Públicos Kodomo • Shōjo • Shōnen Josei • Seinen Gêneros Mahō shōjo • Mecha • Harém Ecchi • Hentai ( Yaoi • Yuri Futanari • Shotacon • Lolicon Toddlercon ) Termos Bishōjo • Chibi • Fan service Gaiden • Kawaii • Kaitō • Light novel Mahō shōnen • Moe (antropomorfismo • Kemonomimi Nekomimi ) • Super deformed Omake • Tsundere • Yonkoma Listas Mangás • Animes • Ecchi • Mangaka • Estúdios • Revistas Fãs ( otaku ) Anime Music Video Convenções • Cosplay Dōjinshi • Fansub • Fandub • Fujoshi OS-tan • Scanlation Portal de Anime e mangá v d e O mangá (português brasileiro) ou manga (português europeu) [ 1 ] (em japonês: 漫画 , manga ? , lit. “história em quadrinhos”) , é a ...
            Read more

            Eduardo VII do Reino Unido

            Image
            Eduardo VII Rei do Reino Unido da Grã-Bretanha e Irlanda e dos Domínios Britânicos de Além-mar Imperador da Índia Rei do Reino Unido e Imperador da Índia Reinado 22 de janeiro de 1901 a 6 de maio de 1910 Coroação 9 de agosto de 1902 Delhi Durbar 1 de janeiro de 1903 Predecessora Vitória Sucessor Jorge V   Esposa Alexandra da Dinamarca Descendência Alberto Vitor, Duque de Clarence e Avondale Jorge V do Reino Unido Luísa, Princesa Real Vitória de Gales Maud de Gales Alexandre João de Gales Casa Saxe-Coburgo-Gota Nome completo Alberto Eduardo Nascimento 9 de novembro de 1841   Palácio de Buckingham, Londres, Reino Unido Morte 6 de maio de 1910 (68 anos)   Palácio de Buckingham, Londres, Reino Unido Enterro Capela de São Jorge, Windsor, Berkshire, Reino Unido 20 de maio de 1910 Pai Alberto de Saxe-Coburgo-Gota Mãe Vitória do Reino Unido Assinatura ...
            Read more