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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 ${1\over1+x}$ or ${1\over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).

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  • $\begingroup$ 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 Jan 21 at 2:37
  • $\begingroup$ For a composition of two functions with known power series you can use the Cauchy product. $\endgroup$ – Ian Jan 21 at 2:49
  • $\begingroup$ (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 Jan 21 at 2:58
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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 $x^4$

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

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  • $\begingroup$ Why is the substitution no longer correct after $x^4$? $\endgroup$ – Alex Jan 21 at 9:19

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