Computation of Maclaurin Series












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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







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$endgroup$












  • $begingroup$
    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






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asked 4 hours ago









Turan NəsibliTuran Nəsibli

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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
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$begingroup$

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









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    $begingroup$

    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






























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