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

*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

*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

**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

*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},

*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

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

a.

b.

c. In(1 +

x)d.

e^{− x }^{2}e

. xcosxf. sin

xcosxg. arctan

xa. Replacing

xbyx^{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

xby −xin 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(wherecis some arbitrary constant), so strictly speaking, the equation above should have been written

However, substituting

x= 0 into this equation shows thatc= 0, so the expansion given above for In (1 +x) is indeed correct.d. Replacing

xby –x^{2}in the Taylor series expansion ofeyields the desired result:^{x}

e. Multiplying each term of the Taylor series for cos

xbyxgives

f. One way to find the series expansion for sin

xcosxis to multiply the expansions of sinxand cosx.A faster way, however, involves recalling the trigonometric identity sin 2x= 2 sinxcosxand then replacingxby 2xin the series expansion of sinx:

g. Since arctan

xis 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 arctanx+c, and the equation above should read

However, substituting

x= 0 into this equation shows thatc= 0, so the expansion given above for arctanxis indeed correct.