Topic 10.14 Finding Taylor or Maclaurin Series for a Function: write the full Taylor or Maclaurin series of a function and recall the standard series for e^x, sin x, cos x and 1/(1-x) (BC).

What this topic is asking

The College Board (Topic 10.14, BC only) extends the Taylor polynomial to the full Taylor series, the infinite sum that the polynomials approximate. It also expects you to know, cold, the standard Maclaurin series for $e^x$, $\sin x$, $\cos x$, and $\frac{1}{1 - x}$, and to build new series from them by substitution and other manipulations.

The series and the four to memorize

Building series by substitution

The fastest way to find most series on the exam is not to compute derivatives but to substitute into a known series. To find the series for $e^{-x^2}$, take the $e^x$ series and replace $x$ with $-x^2$: $e^{-x^2} = \sum\frac{(-x^2)^n}{n!} = 1 - x^2 + \frac{x^4}{2!} - \cdots$. To find $\cos(2x)$, replace $x$ with $2x$ in the cosine series. To find $\frac{1}{1 + x}$, replace $x$ with $-x$ in the geometric series, giving $\sum(-1)^n x^n$. This substitution method is legitimate and expected; computing the derivatives of $e^{-x^2}$ directly would be far more work and error-prone. The skill is recognizing which standard series the target is a disguised version of.

A worked series from the geometric series

Differentiating and integrating series term by term

Within their interval of convergence, power series can be differentiated and integrated term by term, which generates new series for free. Differentiating the geometric series $\frac{1}{1-x} = \sum x^n$ gives $\frac{1}{(1-x)^2} = \sum n x^{n-1}$. Integrating $\frac{1}{1+x} = \sum(-1)^n x^n$ gives the series for $\ln(1 + x) = \sum\frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$. Likewise, differentiating the sine series term by term yields the cosine series, a nice consistency check. This term-by-term calculus is a powerful exam shortcut for functions like $\ln(1+x)$ and $\arctan x$, whose series are most easily obtained by integrating a geometric series rather than by repeated differentiation.

Computing series directly when needed

When a function is not a disguised standard series, you fall back on the definition: compute $f(a), f'(a), f''(a), \ldots$ and assemble $\sum\frac{f^{(n)}(a)}{n!}(x - a)^n$. This is the method when the center is not $0$ or the function has no neat substitution form. The work is to find a pattern in the derivatives so you can write the general term, not just the first few. For instance, the derivatives of $\frac{1}{x}$ at $x = 1$ alternate in sign and grow factorially, giving the series $\sum(-1)^n(x - 1)^n$. Recognizing the pattern and expressing the general $n$-th term, with correct sign and factorial, is what distinguishes a full series from a finite polynomial.