Section 1.5. Taylor Series Expansions In the previous section, we learned that any power series represents a function and that it is very easy to di¤erentiate or integrate a power series function. In this section, we are going to use power series to represent and then to approximate general functions. Let us start with the formula 1 1 = xn, for x < 1. (1) 1 x j j n=0 ¡ X We call the power series the power series representation (or expansion) for the function 1 f (x) = about x = 0. 1 x ¡ 1 It is very important to recognize that though the function f (x) = (1 x)¡ is de…ned for all x = 1, the representation holds only for x < 1. In g¡eneral, if a function f (x) c6 an be represented by a power series asj j 1 n f (x) = cn (x a) n=0 ¡ X then we call this power series power series representation (or expansion) of f (x) about x = a. We often refer to the power series as Taylor series expansion of f (x) about x = a. Note that for the same function f (x) , its Taylor series expansion about x = b, 1 f (x) = d (x b)n n ¡ n=0 X if a = b, is completely di¤erent from the Taylor series expansion about x = a. Gen6erally speaking, the interval of convergence for the representing Taylor series may be di¤erent from the domain of the function. Example 5.1. Find Taylor series expansion at given point x = a : 1 (a) f (x) = , a = 0; 1 + x2 1 x (b) g (x) = , a = 0; x + 2 1 (c) h (x) = , a = 1. 2x + 3 Solution: (a) We shall use (1) by …rst rewriting the function as follows: 1 1 y= x2 1 1 = =¡ = yn, for y < 1. 1 + x2 1 ( x2) 1 y n=0 j j ¡ ¡ ¡ X Formula (1) leads to 1 1 1 n 1 = yn = x2 = ( 1)n x2n, for y < 1. 1 + x2 n=0 n=0 ¡ n=0 ¡ j j X X ¡ ¢ X Note that, since y = x2,and ¡ y < 1 x2 < 1 x < 1 , j j () ¡ () j j we know conclude ¯ ¯ ¯ ¯ 1 1 = ( 1)n x2n, for x < 1. 1 + x2 n=0 ¡ j j X (b) Write x 1 1 g (x) = = x = x (2) x + 2 2 + x 2 (1 + x/2) µ ¶ µ ¶ x 1 x 1 = = . 2 ¢ 1 + x/2 2 ¢ 1 ( x/2) ¡ ¡ We now use (1) to derive 1 y= x/2 1 1 =¡ = yn 1 ( x/2) 1 y ¡ ¡ ¡ n=0 X n 1 x n 1 ( 1) = = ¡ xn, for y < 1. 2 2n n=0 ¡ n=0 j j X ³ ´ X Substituting this into (2), we obtain n x x 1 x 1 ( 1) = = ¡ xn x + 2 2 ¢ 1 ( x/2) 2 2n ¡ ¡ n=0 n X 1 ( 1) = ¡ xn+1, for y < 1. 2n+1 n=0 j j X 2 Now since x y = < 1 x < 2, j j 2 () j j we conclude ¯ ¯ ¯ ¯ n x ¯1¯ ( 1) = ¡ xn+1, for x < 2. x + 2 2n+1 j j n=0 X (c) For 1 h (x) = , a = 1, 2x + 3 we need to rewrite the denominator in terms of (x 1)as follows: ¡ 1 1 1 = = 2x + 3 2 [(x 1) + 1] + 3 2 (x 1) + 5 ¡ 1 1 ¡ 1 = = 5 [1 + 2 (x 1) /5] 5 1 + 2 (x 1) /5 ¡ ¡ y= 2(x 1)/5 1 1 1 1 ¡ =¡ = yn (for y < 1). 5 1 y 5 n=0 j j ¡ X 2 (x 1) We then substitute y = ¡ back to obtain ¡ 5 1 1 1 h (x) = = yn 2x + 3 5 n=0 X n 1 1 2 (x 1) 2 (x 1) = ¡ (for y = ¡ < 1) 5 ¡ 5 j j 5 n=0 µ ¶ ¯ ¯ X n ¯ ¯ 1 1 2 ¯ ¯ = ( 1)n (x 1)n ¯ ¯ 5 ¡ 5 ¡ n=0 µ ¶ X n 1 ( 1) 2n 5 = ¡ (x 1)n , for x 1 < . 5n+1 ¡ j ¡ j 2 n=0 X Example 5.2. Find Taylor series about a = 0 for 1 (a) f (x) = ; (1 x)2 (b) g (x) = ln (¡1 x) ; (c) h (x) = arcta¡n x. 3 Solution: (a) Di¤erentiate 1 1 = xn, for x < 1, 1 x n=0 j j ¡ X we obtain 1 1 0 1 n 1 n 1 f = = = (x )0 = nx ¡ . 2 1 x (1 x) n=0 n=1 ¡ µ ¡ ¶ X X (b) Take anti-derivative on both sides of 1 1 = xn, for x < 1, 1 x n=0 j j ¡ X we obtain 1 1 1 xn+1 dx = xndx = + C. 1 x n + 1 n=0 n=0 Z ¡ X µZ ¶ X So 1 1 xn+1 ln (1 x) = dx = C. 1 x n + 1 ¡ ¡ ¡ n=0 ¡ Z ¡ X To determine the constant, we insert x = 0 into both sides: 1 xn+1 0 = ln (1 0) = C = C. n + 1 ¡ ¡ "n=0 # ¡ ¡ X x=0 We have to choose C = 0 and 1 xn+1 1 xn ln (1 x) = = . (Memorize it) n + 1 n ¡ ¡ n=0 ¡ n=1 X X (c) Note d 1 h = arctan x = . 0 dx 1 + x2 So 1 h (x) = arctan x = dx. 1 + x2 Z 4 From Example 5.1 (a), we know 1 1 = ( 1)n x2n. 1 + x2 ¡ n=0 X Thus n 1 1 1 ( 1) arctan x = dx = ( 1)n x2ndx = ¡ x2n+1 + C. 1 + x2 2n + 1 n=0 ¡ n=0 Z X Z X By setting x = 0 above, we …nd C = 0. So n 1 ( 1) arctan x = ¡ x2n+1. 2n + 1 n=0 X Taylor Series for General Functions. Consider power series expansion 1 n 2 3 f (x) = cn (x a) = c0 + c1 (x a) + c2 (x a) + c3 (x a) + ... (3) n=0 ¡ ¡ ¡ ¡ X for general function f (x) about x = a. Setting x = a, we obtain f (a) = c0. Next, we take derivative on (3) so that 1 n 1 2 3 f 0 (x) = cnn (x a) ¡ = c1+c2 2 (x a)+c3 3 (x a) +c4 4 (x a) +... n=1 ¡ ¢ ¡ ¢ ¡ ¢ ¡ X (4) Setting x = a, we have f 0 (a) = c1. We repeat the same process again and again: take derivative again on (4) 1 n 2 2 f 00 (x) = c n (n 1) (x a) ¡ = c 2 1+c 3 2 (x a)+c 4 3 (x a) +... n ¡ ¡ 2¢ ¢ 3¢ ¢ ¡ 4¢ ¢ ¡ n=2 X (5) 5 and set x = a to obtain f 00 (a) f 00 (a) = c 2 1 = c = ; 2 ¢ ¢ ) 2 2! take derivative again on (5) 1 (3) n 3 2 f (x) = cnn (n 1) (n 2) (x a) ¡ = c33 2 1+c44 3 2 (x a)+c55 4 3 (x a) +... n=3 ¡ ¡ ¡ ¢ ¢ ¢ ¢ ¡ ¢ ¢ ¡ X and insert x = a to obtain (3) (3) f (a) f (a) = c33 2 1 = c3 = . ¢ ¢ ) 3! In general, we have f (n) (a) c = , n = 0, 1, 2, ... n n! here we adopt the convention that 0! = 1. All above process can be carried on as long as any number of order of derivative at x = a exists, i.e., f (x) must be a smooth function near a. Then, we have Taylor series expansion formula 1 f (n) (a) f (x) = (x a)n . (Taylor Series) n! n=0 ¡ X When a = 0, it becomes 1 f (n) (0) f (x) = xn, (Maclaurin Series) n! n=0 X we call it Maclaurin Series of f (x) . Example 5.3. Find Maclaurin series for (a) f (x) = ex; (b) g (x) = bx (b > 0) Solution: (a) For f = ex, we know x x (n) x f 0 = e , f 00 = e , ..., f = e . Thus f (n) (0) 1 c = = , n n! n! 6 1 f (n) (0) 1 1 ex = xn = xn. (Maclaurin Series For ex) n! n! n=0 n=0 X X This is one of the most useful Taylor series, and must be memorized. (b) We o¤er two methods to solve this problem. First is the direct method by using formula for Maclaurin Series. To this end, we compute derivatives x g0 = b ln b x x x 2 g00 = (b )0 ln b = (b ln b) ln b = b (ln b) , ... g(n) = bx (ln b)n . So n 1 g(n) (0) 1 (ln b) bx = xn = xn. n! n! n=0 n=0 X X Another method is to use Taylor series for ex above. Write n x y=x ln b 1 1 1 1 1 (ln b) bx = eln(b ) = e(x ln b) = ey = yn = (x ln b)n = xn. n! n! n! n=0 n=0 n=0 X X X Example 5.4. Find Maclaurin series for (a) f (x) = sin x; (b) g (x) = cos x. Solution: (a) We observe that f = sin x = f (0) = 0, ) f 0 = cos x = f 0 (0) = 1, ) f" = sin x = f" (0) = 0, ¡ ) f (3) = cos x = f (3) (0) = 1 ¡ ) ¡ f (4) = sin x = f (4) (0) = 0, ) and that this cyclic pattern repeats every 4 times di¤erentiations. In partic- ular, we see that when n is even, i.e., n = 2m, f (n) (0) = 0. When n is odd, 7 i.e., n = 2m + 1, f (n) (0) equals 1 and 1 alternating. Thus, ¡ 1 f (n) (0) sin x = xn n! n=0 X 1 f (n) (0) = xn n! n=0 nX=odd 1 f (2m+1) (0) = x2m+1 (2m + 1)! m=0 X m 1 ( 1) = ¡ x2m+1 (Maclaurin Series for cos x) (2m + 1)! m=0 X x3 x5 x7 = x + + ...