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).

  • $\begingroup$ 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

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 $x^4$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.