A Category 2 or Category 3 power series in *x* defines a function *f* by setting

for any *x* in the series' interval of convergence.

The power series expansion for *f*( *x*) can be differentiated term by term, and the resulting series is a valid representation of *f*′( *x*) in the same interval:

Differentiating again gives

and so on. Substituting

and in general, substituting *x* = 0 in the power series expansion for the *n*th derivative of *f* yields

These are called the **Taylor coefficients** of *f*, and the resulting power series

is called the **Taylor series** of the function *f*.

Given a function *f*, its Taylor coefficients can be computed by the simple formula above, and the question arises, does the Taylor series of *f* actually converge to *f*( *x*)? If it does, that is, if

for all *x* in some **neighborhood of** (interval around) 0, then the function *f* is said to be **analytic** (at 0). [More generally, if you form the Taylor series of *f* about a point *x* = *x* _{0},

and if this series actually converges to *f*( *x*) for all *x* in some neighborhood of *x* _{0}, then *f* is said to be analytic at *x* _{0}.] Polynomials are analytic everywhere, and rational functions (quotients of polynomials) are analytic at all points where the denominator is not zero. Furthermore, the familiar **transcendental** (that is, nonalgebraic) functions *e *^{x} , sin *x*, and cos *x* are also analytic everywhere. The Taylor series in Table 1 are encountered so frequently that they are worth memorizing.

For a general power series, it is usually not possible to express it in closed form in terms of familiar functions.

**Example 1**: Use Table 1 to find the Taylor series expansion of each of the following functions:

a.

b.

c. In(1 + *x*)

d.* e* ^{− x }^{2}

e*. x* cos *x*

f. sin *x* cos *x*

g. arctan *x*

a. Replacing *x* by *x* ^{2} in the Taylor series expansion of 1/(1 – *x*) gives

since | *x*| < 1 is equivalent to | *x* ^{2}| < 1.

b. Differentiating 1/(1 – *x*) gives 1/(1 – *x*) ^{2}, so differentiating the Taylor series expansion of 1/(1 – *x*) term by term will give the series expansion of 1/(1 – *x*) ^{2}:

c. First, replacing *x* by − *x* in the Taylor series expansion of 1/(1 – *x*) gives the expansion of 1/(1 + *x*):

Now, since integrating 1/(1 + *x*) yields In(1 + *x*), integrating the Taylor series for 1/(1 + *x*) term by term gives the expansion for In(1 + *x*), valid for | *x*| < 1:

Technical note: Integrating 1/(1 + *x*) yields In (1 + *x*) + *c* (where *c* is some arbitrary constant), so strictly speaking, the equation above should have been written

However, substituting *x* = 0 into this equation shows that *c* = 0, so the expansion given above for In (1 + *x*) is indeed correct.

d. Replacing *x* by – *x* ^{2} in the Taylor series expansion of *e *^{x} yields the desired result:

e. Multiplying each term of the Taylor series for cos *x* by *x* gives

f. One way to find the series expansion for sin *x* cos *x* is to multiply the expansions of sin *x* and cos *x.* A faster way, however, involves recalling the trigonometric identity sin 2 *x* = 2 sin *x* cos *x* and then replacing *x* by 2 *x* in the series expansion of sin *x*:

g. Since arctan *x* is the integral of 1/(1 + *x* ^{2}), integrate the series expansion of 1/(1 + *x* ^{2}) term by term:

Recall the technical note accompanying part c which also involved the term‐by‐term integration of a power series). The integral of 1/(1 + *x* ^{2}) is actually arctan *x* + *c*, and the equation above should read

However, substituting *x* = 0 into this equation shows that *c* = 0, so the expansion given above for arctan *x* is indeed correct.