Series Expansions of Compound Functions
Substitute into a known Maclaurin series, split logarithms of quotients, or multiply two expansions together — three ways to avoid differentiating from scratch.
Substituting into a known series
Rather than differentiating a compound function from scratch, you can often substitute directly into one of the standard Maclaurin series. If you know the expansion of $f(x)$, then the expansion of $f(g(x))$ is found by replacing every $x$ in the known series with $g(x)$, then simplifying and collecting terms by power of $x$.
The standard expansion is $\sin\theta=\theta-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}-\dfrac{\theta^7}{7!}+\cdots$.
Substitute $\theta=3x^2$: $\sin(3x^2)=3x^2-\dfrac{(3x^2)^3}{3!}+\dfrac{(3x^2)^5}{5!}-\dfrac{(3x^2)^7}{7!}+\cdots$
$=3x^2-\dfrac{27x^6}6+\dfrac{243x^{10}}{120}-\cdots=3x^2-\dfrac{9}{2}x^6+\dfrac{81}{40}x^{10}-\cdots$
If the original series is only valid for $|x|<1$ (like the series for $\ln(1+x)$ or $\sqrt{1+x}$), then after substituting $g(x)$ for $x$, the new series is valid for $|g(x)|<1$ — you must re-solve this inequality for the new variable, not just copy across the old range unchanged.
Combining series with logarithm laws
For expressions like $\ln\left(\dfrac{1+ax}{1-bx}\right)$, it's usually far easier to split the logarithm using $\ln\dfrac PQ=\ln P-\ln Q$ before substituting into the series for $\ln(1+x)$, rather than trying to expand the whole quotient directly. Each piece is then a routine substitution.
$\ln\left(\dfrac{1+4x}{1-2x}\right)=\ln(1+4x)-\ln(1-2x)$.
Using $\ln(1+u)=u-\dfrac{u^2}2+\dfrac{u^3}3-\cdots$ with $u=4x$: $\ln(1+4x)=4x-\dfrac{16x^2}2+\dfrac{64x^3}3-\cdots=4x-8x^2+\dfrac{64}3x^3-\cdots$, valid for $|4x|<1$, i.e. $|x|<\dfrac14$.
Using the same series with $u=-2x$: $\ln(1-2x)=-2x-\dfrac{4x^2}2-\dfrac{8x^3}3-\cdots=-2x-2x^2-\dfrac83x^3-\cdots$, valid for $|-2x|<1$, i.e. $|x|<\dfrac12$.
Subtracting: $\ln(1+4x)-\ln(1-2x)=(4x-8x^2+\cdots)-(-2x-2x^2-\cdots)=6x-6x^2+24x^3-\cdots$
Both component series must be valid simultaneously, so the combined expansion is valid for the more restrictive condition, $|x|<\dfrac14$.
Products of two series
For a product like $e^x\cos x$, expand each factor as a separate Maclaurin series up to the power of $x$ you need, then multiply the two series together like polynomials, discarding any resulting terms beyond the power you were asked for (since higher powers would need more terms from each factor to compute correctly anyway).
$e^{-2x}=1-2x+\dfrac{4x^2}2-\dfrac{8x^3}6+\cdots=1-2x+2x^2-\dfrac43x^3+\cdots$ (up to $x^3$).
$(1+x)\left(1-2x+2x^2-\dfrac43x^3\right)$: multiply out, keeping only terms up to $x^3$:
$=1-2x+2x^2-\dfrac43x^3+x-2x^2+2x^3+O(x^4)$
$=1+(-2+1)x+(2-2)x^2+\left(-\dfrac43+2\right)x^3+\cdots=1-x+0x^2+\dfrac23x^3+\cdots$
So $(1+x)e^{-2x}=1-x+\dfrac23x^3+\cdots$ (the $x^2$ term vanishes entirely).
Series expansions of compound functions feed directly into later chapters — they're used to justify Euler's relation (Chapter 1), estimate the value of definite integrals with no elementary antiderivative, and approximate solutions to differential equations near a starting point (Chapters 7–8).
Exam-style questions
Question 1 5 marks
Find the first three non-zero terms of the series expansion of $\cos(x^2)-1$, and state the range of $x$ for which it's valid.
$\cos\theta=1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\cdots$, valid for all real $\theta$. (M1)
Substituting $\theta=x^2$: $\cos(x^2)=1-\dfrac{x^4}2+\dfrac{x^8}{24}-\cdots$. (M1, A1)
So $\cos(x^2)-1=-\dfrac{x^4}2+\dfrac{x^8}{24}-\cdots$, valid for all real $x$ (since the original series for $\cos\theta$ has no restriction). (A1, A1)
Question 2 7 marks
Find the series expansion of $e^x\ln(1+x)$ up to and including the term in $x^3$, and state the range of validity.
$e^x=1+x+\dfrac{x^2}2+\dfrac{x^3}6+\cdots$; $\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots$, the latter valid for $|x|<1$. (M1, A1)
Multiplying, keeping terms up to $x^3$: $\left(1+x+\dfrac{x^2}2\right)\left(x-\dfrac{x^2}2+\dfrac{x^3}3\right)=x-\dfrac{x^2}2+\dfrac{x^3}3+x^2-\dfrac{x^3}2+\dfrac{x^3}2+\cdots$ (M1, A1)
$=x+\left(-\dfrac12+1\right)x^2+\left(\dfrac13-\dfrac12+\dfrac12\right)x^3+\cdots=x+\dfrac12x^2+\dfrac13x^3+\cdots$ (M1, A1)
Valid for $|x|<1$ (the more restrictive of the two component ranges, since $e^x$'s series is valid everywhere). (A1)