CALCULUS JIA-MING (FRANK) LIOU Abstract. Contents 1. Power Series 1 1.1. Polynomials and Formal Power Series 1 1.2. Radius of Convergence 2 1.3. Derivative and Antiderivative of Power Series 4 1.4. Power Series Expansion defined by Geometric Series 6 1.5. Power Series Defined by Differential Equations 9 1.6. The Convergence of Taylor Series 10 1.7. Definition of ez 12 1.8. More Eamples 12 1. Power Series 1.1. Polynomials and Formal Power Series. Let x be a variable and a0; ··· ; an be real numbers. A polynomial in x with coefficients a0; ··· ; an is an expression n P (x) = a0 + a1x + ··· + anx ; If an 6= 0, n is called the degree of P . The space of polynomials in x with coefficients in R n is denoted by R[x]. For each real number c 2 R, define P (c) = a0 + a1c + ··· + anc . Then P : R ! R defines a function. Hence a polynomial in x can be thought of as a function on (k) R. Evey polynomial function is a smooth function on R which means that P (x) exists for each k ≥ 0. Moreover, P (0) = a0, 0 2 n−1 P (x) = a1 + 2a2x + 3a3x + ··· + nanx 00 2 n−2 P (x) = 1 · 2a2 + 2 · 3a3x + 3 · 4a4x ··· + (n − 1)nanx ··· = ··· (n) P (x) = n!an: 0 00 (n) Hence we have P (0) = a0;P (0) = a1;P (0) = 2!a2; ··· ;P (0) = n!an. Pn i Pm j Suppose that we have two polynomials P (x) = i=0 aix and Q(x) = j=0 bjx . Then n m m+n k ! X X i+j X X k P (x)Q(x) = aibjx = aibk−i x : i=0 j=0 k=0 i=0 Hence the product of two polynomials is again a polynomial. 1 2 JIA-MING (FRANK) LIOU Let (an)n≥0 be a sequence of real numbers. The formal power series in x with coefficients an is an expression 1 2 n X n (1.1) a0 + a1x + a2x + ··· + anx + ··· = anx : n=0 The space of formal power series with coefficients in R is denoted by R[[x]]. Suppose that P1 n P1 n A(x) = n=0 anx and B(x) = n=0 bnx are two formal power series in R[[x]]. Define their sum and their product by 1 1 X n X n A(x) + B(x) = (an + bn)x A(x)B(x) = cnx ; n=0 n=0 Pn where cn = k=0 akbn−k. Hence we know that R[x] is a subset of R[[x]], i.e. every poly- nomial is a formal power series. Here comes a question: When does a formal power series define a function on an interval of R? 1.2. Radius of Convergence. The following theorem gives us an example when a formal power series converges. 1 X n Theorem 1.1. (Abel) Suppose that r is a real number so that the series anr is con- n=0 1 1 X n X n vergent. For each jcj < jrj, anc is absolutely convergent. Conversely if anr is n=0 n=0 1 X n divergent, then for jcj > jrj, anc is also divergent. n=0 P1 n n Proof. Since the series anr is convergent, lim anr = 0. Hence there exists N > 0 n=0 n!1 n −n so that for any n ≥ N, janr j < 1. Thus janj < r for n ≥ N. Therefore for any n ≥ N, c n ja cnj < : n r P1 n Since jc=rj < 1, the geometric series n=N+1(c=r) is convergent. By the comparison test, P1 n P1 n n=N+1 janc j is convergent and thus n=0 anc is convergent. Definition 1.1. R > 0 is called the radius convergence of (1.1) if P1 n (1) for any jcj < R, n=0 anc is absolutely convergent; P1 n (2) for any jcj > R, n=0 anc is divergent. pn Theorem 1.2. Suppose that lim janj = ρ. Then the radius of convergence of (1.1) is n!1 a R = ρ−1. Similarly, if lim j n+1 j = ρ, then R = ρ−1. n!1 an n Proof. Let ρ be the limit of (an+1=an): Let an(x) = anx : Then a (x) a n+1 = n+1 x: an(x) an CALCULUS 3 Hence an+1(x) lim = ρjxj: n!1 an(x) 1 1 X X n We know that if ρjxj < 1; by ratio test, the infinite series an(x) = anx is absolutely n=0 n=0 convergent. In other words, the power series is convergent if jxj < 1/ρ. Take R = 1/ρ. If jxj > R; then ρjxj > 1: By ratio test, the power series is divergent. Hence we prove that R = 1/ρ is the radius of convergence of the power series. Let us take a look at some examples. Example 1.1. Find the radius of convergence of the following three formal power series 1 X xn (1) A (x) = (−1)n ; 1 n n=1 1 X xn (2) A (x) = ; 2 n2 n=1 1 X n (3) A3(x) = nx . n=1 Sol. By ratio test, the radii of convergence of the above three formal power series are 1. Note that P (−1)n=n is convergent by the Leibnitz test. A(x) is also convergent at n=1 P x = 1 but when x = −1, the series n=1 1=n is divergent by the p-test. Hence A1(x) is convergent on −1 < x ≤ 1 or I1 = (−1; 1]. Similarly, A2(x) is convergent on both x = ±1. Therefore A2(x) is convergent on I2 = [−1; 1]. We also find that A3 is convergent on I3 = (−1; 1) and divergent at x = ±1. I1;I2;I3 are called the interval of convergence of A1;A2;A3 respectively. Definition 1.2. An interval I on R is of the form [a; b] or [a; b) or (a; b] or (a; b). An interval P1 n I is called the interval of convergence if for every c 2 I, n=0 anc is convergent and P1 n for any d 62 I, n=0 and is divergent. Corollary 1.1. I can only be [−R; R], or (−R; R], or [−R; R) or (−R; R). P1 n If I is the interval of convergence of the formal power series A(x) = n=0 anx , then for P1 n each c 2 I, n=0 anc is convergent. For each c 2 I, define 1 X n A(c) = anc n=0 Then the formal power series A(x) defines a function on I. 1 X xn Example 1.2. The radius of convergence of . n! n=0 4 JIA-MING (FRANK) LIOU 1 Sol. Since an = n! , an+1=an = 1=(n + 1) for all n. Hence limn!1(an+1=an) = 0. Thus the radius of convergence is R = 1. Remark. Later we will show that this power series converges to ex. Theorem 1.3. (Multiplication Theorem) Suppose that two power series A(x), B(x) are convergent on jxj < R. Then A(x) ± B(x), A(x)B(x) are all convergent on jxj < R. 1.3. Derivative and Antiderivative of Power Series. Suppose that p(x) = a0 + a1x + n Pn k ··· + anx = k=0 akx be a polynomial. Then n 0 n−1 X k−1 p (x) = a1 + 2a2x + ··· + nanx = kakx k=0 P1 n Given a power series A(x) = n=0 anx , one might think that if A is differentiable, then 1 0 X n−1 A (x) = nanx : n=0 On the other hand, an antiderivative of p is given by Z n a1 an X ak p(x)dx = C + a x + x2 + ··· + an+1 = C = xk+1: 0 2 n + 1 k + 1 k=0 Hence one might guess Z 1 X an A(x)dx = c + xn+1: n + 1 n=0 Theorem 1.4. Suppose that the radius of convergence of (1.1) is R. Then the power series of 1 1 X X an B(x) = na xn−1;C(x) = c + xn+1 n n + 1 n=0 n=0 are also R. Suppose the radius of convergence of (1.1) is R. Suppose that jxj < R and jx0j < R. Then 1 X n n A(x) − A(x0) = an(x − x0 ): n=0 In order to calculate the difference quotient of A, we need: Lemma 1.1. Let x; y be real numbers. Then xn − yn = (x − y)(xn−1 + xn−2y + ··· + xyn−2 + yn): Hence the difference quotient of A at x0 is 1 A(x) − A(x0) X = a (xn−1 + xn−2x + ··· + xxn−2 + xn−1): (x − x ) n 0 0 0 0 n=0 CALCULUS 5 We know that n−1 n−2 n−2 n−1 n−1 lim (x + x x0 + ··· + xx0 + x0 ) = nx0 : x!x0 0 0 Hence one finds that (not so rigorous) A (x0) = B(x0). Thus A (x) = B(x) for any jxj < R. One could also see that C0(x) = A(x). Hence C(x) = c + R A(x)dx. Theorem 1.5. Suppose that the radius of convergence of (1.1) is R. Then for all jxj < R, Z A0(x) = B(x); A(x)dx = C(x): P1 n We have already see that the radius of convergence of f(x) = n=0 x =n! is R = 1. Hence the function f(x) is smooth on R.