Constructing Taylor Series

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1 Constructing Taylor Series The Taylor series for fx at x = c is fc + f cx c + f c 2! x c 2 + f c x c 3 + = 3! f n c x c n. By convention, f 0 = f. When c = 0, the series is called a Maclaurin series. You can construct the series on the right provided that f is infinitely differentiable on an interval containing c. You already know how to determine the interval of convergence of the series. However, the fact that the series converges at x does not imply that the series converges to fx. Example. The function fx = { e /x2 if x 0 0 if x = 0 is infinitely differentiable everywhere. In particular, all the derivatives of f at 0 vanish, and the Maclaurin series for f is identically 0. Hence, the Maclaurin series for f converges for all x, but only converges to fx at x = 0. The following result [], page 8 gives a sufficient condition for the Taylor series of a function to converge to the function: Theorem. Let fx be infinitely differentiable on a x b, and let a c b. Suppose there is a constant M such that f n x M for all n, and for all x in N [a, b], where N is a neighborhood of c. Then for all x N [a, b], f n c fx = x c n. In other words, under reasonable conditions: You can construct a Taylor series by computing the derivatives of f. The series will converge to f on an interval around the expansion point. You can find the interval of convergence as usual. Example. Find the first four nonzero terms and the general term of the Taylor series for fx = e x at a = 0 and at a = by computing the derivatives of f. fx = e x, f x = e x, and in general f n x = e x. For a = 0, f n 0 = e 0 = for all n. The Taylor series at a = 0 is fx = + x + 2! x2 + 3! x3 + + xn +. For a =, f n = e = e for all n. The Taylor series at a = is fx = e + ex + e 2! x ! x x n +.

2 Terminology. If you truncate the series after the n th -degree term, what s left is the n th -degree Taylor polynomial. For example, the third degree polynomial of e x at a = 0 is p 3 x = + x + 2! x2 + 3! x3. Note that the n here refers to the largest power of x, not the number of terms. For example, the Taylor series for at a = 0 is x2 x 2 = + x2 + x + + x 2n +. The 2-nd degree Taylor polynomial and the 3-rd degree Taylor polynomial are equal: p 2 x = p 3 x = + x 2. Example. Find the 3 rd degree Taylor polynomial for fx = tanx at x = π. fx = tanx, f x = sec x 2, f x = 2sec x 2 tan x, f x = 2sec x + sec x 2 tanx 2. Thus, π f =, f π The 3 rd degree Taylor polynomial is = 2, f π =, f π = 6. p 3 x; π = + 2 x π + 2 x π x π 3. 3 It s tedious to have to compute lots of derivatives, and in many cases you can derive a series from another, known series. Here are the series expansions for several important functions: u = u n = + u + u u n + < u < e u u n = = + u + u2 2! + + un + < u < + cosu = n u 2n 2n! = u2 2! + u! + u 2n n + 2n! < u < + sin u = n u2n+ 2n +! = u u3 3! + u u2n+ + n +! 2n +! < u < + n+ un ln + u = n = u u2 2 + u3 3 + n+un n + < u n= + u a = + n= 2 aa a n + u n < u <

3 Example. Find the Taylor series for lnx at a =. What is its interval of convergence? Use ln + u = n= n+ un n = u u2 2 + u3 3 + n+ un n +. I m expanding at a =, so I want the result to come out in powers of x. This is easy just set u = x : lnx = x 2 x x n+ n x n. The u-series converges for < u, so the x-series converges for < x, or 0 < x 2. /2 Example. The quantity v2 occurs in special relativity. v is the velocity of an object, and c is c 2 /2 the speed of light. Approximate v2 using the first two nonzero terms of the binomial series. so for a = 2, c 2 + u a = + au + aa u 2 +, 2! + u /2 = 2 u u2. Take u = v2 c 2 : /2 v2 c 2 = + v 2 2 c v 8 c + + v 2 2 c 2. The approximation is good as long as v is small compared to c. Example. Find the Taylor series for x + 3 at a = 2. What is its interval of convergence? I want things to come out in powers of x 2, so I ll write the function in terms of x 2: I ll use the series for x + 3 = + x 2.. To do this, I need u on the bottom: u + x 2 = + x 2 = x 2. Let u = x 2 in the series for u. Then x 2 = x x 2 x

4 Hence, [ x + 3 = x x 2 x 2 + ]. The u-series converges for < u <, so the x-series converges for < x 2 <, or 3 < x < 7. Example. Find the Taylor series for x 2 + x at a =. Since I m expanding at a =, the answer should have the form b 0 + b x + + b 2 x where the b s are numbers. That is, the answer must come out in terms of powers of x +. Start with the function you re trying to expand. To get x + s in the answer, write the given function in terms of x + : x x + = 2 + x + x +. Notice that the work has to be legal algebra. I ll break up the fraction and do the pieces separately. x + + x + = x + + x + + x +. I want to match each piece against the standard series Expand x + + x + = x + [ x + ]. [ x + ] by setting u = x + in u :. Here s the first piece: u x+ [ x + ] = x+ x + + x + 2 x = x+ x+ 2 +x+ 3. Here s the second piece: + x + = [ x + ] = x + + x + 2 x Put the two pieces together: [ x + x x + 3 ] [ x + + x + 2 x ] = x + x x x + x x + 3 = That is, + 2x + 2x x + 3. x 2 + x = + 2x + 2x x + 3.

5 Example. What is the Maclaurin series for fx = 7x 2 3x + 3? What is the Taylor series for fx = 7x 2 3x + 3 at a =? The Maclaurin series for a polynomial is the polynomial: fx = 7x 2 3x + 3. To obtain the Taylor expansion at a =, write the function in terms of x + : 7x 2 3x + 3 = 7x + 2 7x + 6 = 7x + 2 7x It s also possible to construct power series by integrating or differentiating other power series. A power series may be integrated or differentiated term-by-term in the interior of its interval of convergence. You will need to check convergence at the endpoints separately. Example. The Maclaurin series for + x is + x = x + x2 x 3 +. Put u = x in the series for. It converges for < x <. u Integrate the series from 0 to u: ln + u = u 0 x + x 2 x 3 + dx = u u2 2 + u3 3 u +. This series will converge for < u <. The left side blows up at u =. On the other hand, if u =, ln2 = The right side does converges by the Alternating Series Test, so the ln + u series converges for < u. Example. Find the Taylor series for ln x at a = 2. I ll use the fact that a Taylor series can be integrated term-by-term on the interval where it converges absolutely. x x t dt = [ ln t]x 2 = ln x + ln 3, so ln x = ln 3 2 t dt. 2 I integrated from 2 to x because I want the expansion at a = 2. Now find the series at a = 2 for t = 3 t 2 = 3 t 2 3 = 3 Plug this series back into the integral and integrate term-by-term: x ln x = ln 3 2 t dt = ln3 x 3 2 t 2 n t 2 n 3 n. 3 n dt = ln 3 3 [ t 2 n+ 3 n n + ] x 2 = t :

6 ln3 3 x 2 n+ 3 n n + = ln 3 x 2 n+ 3 n+ n +. Example. Find f 00 0 for fx = 3 x. 3 x = 3 x 3 = 3 x3 x xn 3 n + = 3 + x 3 + x xn n+ The 00 th degree term is x On the other hand, Taylor s formula says that the 00th degree term is f 00 0 x 00. Equating the coefficients, I get 00! 3 0 = f00 0, so f 00 0 = 00! 00! 3 0. [] Tom M. Apostol, Mathematical Analysis. Reading, Massachusetts: Addision-Wesley Publishing Company, Inc., 97. c 200 by Bruce Ikenaga 6

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