Suppose we're working with a function f(x) that is continuous and has (n+1) continuous derivatives on an interval about x = 0. We can approximate f near 0 by a polynomial P_n(x) of degree n:

• For n = 0, the best constant approximation near 0 is
  
  

  P0(x) = f(0)

  which matches f at 0.

• For n = 1, the best linear approximation near 0 is
  
  

  P1(x) = f(0)+f¢(0)x.

  Note that P₁ matches f at 0 and P₁¢ matches f¢ at 0.

• For n = 2, the best quadratic approximation near 0 is
  
  

  P2(x) = f(0)+f¢(0)x+

  f¢¢(0) 2!

  x2.

  Note that P₂, P₂¢, and P₂¢¢ match f, f¢, and f¢¢, respectively, at 0.

Continuing this process,

This is the Taylor polynomial of degree n about 0 (also called the Maclaurin series of degree n). More generally, if f has n+1 continuous derivatives at x = a, the Taylor series of degree n about a is

This formula approximates f(x) near a. (If the value of a is not specified, we take a=0.) Taylor's Theorem gives bounds for the error in this approximation:

Taylor's Theorem

Suppose f has n+1 continuous derivatives on an open interval containing a. Then for each x in the interval,

This form for the error R_n+1(x), derived in 1797 by Joseph Lagrange, is called the Lagrange formula for the remainder. The infinite Taylor series converges to f,

Examples of Taylor Series about 0

1. For f(x) = e^x,

   f(k)(x) = ex    Þ    f(k)(0) = 1.

   So

   ex	=	1 + x + x2 2! + x3 3! + ¼
   	=	¥ å k = 0  xk k!

   which converges for all x since
   
   

   lim n®¥

   Rn(x)

   =

   lim n®¥

   ecx(n+1) / (n+1)! = 0

   for all c between 0 and x.

2. For f(x) = ln(1+x),

   f¢(x) = 1/(1+x) f¢¢(x) = -1/(1+x)2 f¢¢¢(x) = 2/(1+x)3 f(4)(x) = (-3·2)/(1+x)4 · · ·

   ü ï ï ï ý ï ï ï þ

   Þ

   f(0) = 0 f¢(0) = 1 f¢¢(0) = -1 f¢¢¢(0) = 2 f(4)(x) = -6

   So

   ln(1+x)	=	x - x2 2 + x3 3 - x4 4 + ¼
   	=	¥ å k = 0  (-1)k xk+1 k+1

   which converges only for -1 < x £ 1.

The Taylor Series in (x-a) is the unique power series in (x-a) converging to f(x) on an interval containing a. For this reason,

• By Example 1,

  e-2x = 1 - 2x + 2x2 -

  4 3

  x3+¼

  where we have substituted -2x for x.

• By Example 2, since d/dx[ln(1+x)] = 1/(1+x), we can differentiate the Taylor series for ln(1+x) to obtain

  1 1+x

  = 1-x+x2-x3+¼.

  Substituting -x for x,

  1 1-x

  = 1+x+x2+x3+¼.

In the Exploration, compare the graphs of various functions with their first through sixth degree Taylor polynomials about x = 0.

Taylor's Theorem

Suppose f has n+1 continuous derivatives on an open interval containing a. Then for each x in the interval,