Recently I've been studying for the AP Calculus BC test. As part of the test, you need to understand how to deal with Taylor series generally, and Maclaurin series specifically. While I think I understand most of it, there is one bit that keeps coming up which I'm having trouble understanding.

A lot of problems I see on practice tests have you find the Maclaurin polynomial of a common series (e.g., $\sin (x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + ...$) times something else (such as $x^2$). For example, from the practice questions for the 2022 AP Live Review videos:

Write the first three nonzero terms and the general term of the Maclaurin series for $x^2 \cos(x)$.

Unfortunately, they don't solve this in class like they do with some of the other problems, but the answer sheet quotes the Maclaurin series for cos(x) and then gives the following answer:

$$ x^2 \cos(x) = x^2 - \frac{x^2 x^2}{2!} + \frac{x^2 x^2}{4!} + ... + \frac{x^2 (-1)^n x^2n}{(2n!)} $$

If I'm reading that correctly, they're multiplying each term of the Maclaurin series for $\cos{x}$ by $x^2$. Is that correct? Is there any way to solve such problems as a class (for example, say if I want to find the Maclaurin series for $f(x) = x^3 \sqrt{x} \cos{x}$)?

  • $\begingroup$ That $\sqrt{x}$ will cause issues around $x=0$ if you want integer powers. Otherwise, you could say $x^3 \sqrt{x} \cos{x} = x^{5/2}-\frac12 x^{9/2}+\frac{1}{24}x^{13/2}-\cdots$ $\endgroup$
    – Henry
    Apr 24 at 22:12

1 Answer 1


In general you'll want to take the Cauchy product of the two series. Given $$ f(x) = \sum_{n=0}^\infty a_n x^n \qquad g(x) = \sum_{n=0}^\infty b_n x^n $$ then $$ f(x)g(x) = \sum_{n=0}^\infty c_n x^n \text{ where } c_n = \sum_{i=0}^n a_i b_{n-i} = \sum_{\substack{i+j=n \\ i,j \ge 0}} a_i b_j $$

In your particular example of $x^2 \cos(x)$, the above works perfectly fine, if trivially, since

$$x^2 = \sum_{n=0}^\infty a_n x^n \text{ for } a_n = \begin{cases} 0, & n \ne 2 \\ 1, & n = 2 \end{cases}$$

Of course, more simply, you should be able to recognize that multiplying a power series by $x^2$ (or more generally, $x^m$ for some nonnegative integer $m$) generates just another power series, with the coefficients "shuffled along" a bit. That is, for $f$ as above,

$$x^2 f(x) = x^2 \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty a_n x^2 x^n = \sum_{n=0}^\infty a_n x^{n+2} = \sum_{n=2}^\infty a_{n-2} x^n$$

as an example. (Note: The formula for the product of series is only valid for the points $x$ where both series converge, so be mindful of that detail.)

You can't do this with, on the other hand, $\sqrt x \cos(x)$. For the same $f$,

$$\sqrt x f(x) = \sum_{n=0}^\infty a_n x^{n+\frac 1 2}$$

but this is not a power series, because the powers of $x$ need to be nonnegative integers to be a power series. You would have to, instead, find a power series expansion for $\sqrt x$ and then use the Cauchy product as stated previously.


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