Computation of Maclaurin Series

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).

• 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. – Evan William Chandra Jan 21 at 2:37
• For a composition of two functions with known power series you can use the Cauchy product. – Ian Jan 21 at 2:49
• (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). – 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 $$x^4$$
$$\log(1+2x+2x^2) = 2 x - \frac{4x^3}{3} + 2 x^4 \cdots$$
• Why is the substitution no longer correct after $x^4$? – Alex Jan 21 at 9:19