Taylor and Maclaurin series for familiar functions

Understand the idea

A Taylor series extends the derivative-matching pattern of a Taylor polynomial indefinitely. A Maclaurin series uses center zero. To represent the original function, its remainder must approach zero on the region under discussion; the existence of derivatives alone is not enough.

Derive coefficients from derivative values or from a familiar series whose representation has been established. For exponential, sine, and cosine functions, the coefficient patterns and factorials produce convergence for every real input. State the convergence domain along with the expression.

Keep the summation index, exponent pattern, and factorial synchronized. Even powers occur in the cosine series, while odd powers occur in the sine series. Substituting a scaled input changes every power of that input, not just the first nonconstant term.

e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad \cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!},\qquad x\in\mathbb R

A method you can apply

Evaluate the derivative pattern at the center or start from an established familiar expansion.
Write several terms and a general term with a consistent starting index.
State where the expansion represents the function and verify the first few coefficients.

Worked example

Find the first four nonzero terms and the general Maclaurin series for $\displaystyle \cos x$ .

The derivative values at zero cycle through $\displaystyle 1,0,-1,0$ .
Only even powers remain, with alternating signs and factorial denominators.
The first four nonzero terms are $\displaystyle 1-x^2/2!+x^4/4!-x^6/6!$ .
The general term is $\displaystyle (-1)^n x^{2n}/(2n)!$ , starting at $\displaystyle n=0$ .

The series is

\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{2n}/(2n)!

and represents

\displaystyle \cos x

for every real

\displaystyle x

.

Try it yourself

Find the first four nonzero terms of the Maclaurin series for $\displaystyle e^{-2x}$ and state its radius of convergence.

Show solution and reasoning

Substitute $\displaystyle -2x$ into the exponential series: $\displaystyle 1-2x+2x^2-\tfrac43x^3+\cdots$ . The general term is $\displaystyle (-2x)^n/n!$ , and the radius is infinite.