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

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 ∈ 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 ∞ 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 P∞ n P∞ 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 ∞ ∞ 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.

∞ X n Theorem 1.1. (Abel) Suppose that r is a real number so that the series anr is con- n=0 ∞ ∞ X n X n vergent. For each |c| < |r|, anc is absolutely convergent. Conversely if anr is n=0 n=0 ∞ X n divergent, then for |c| > |r|, anc is also divergent. n=0

P∞ n n Proof. Since the series anr is convergent, lim anr = 0. Hence there exists N > 0 n=0 n→∞ n −n so that for any n ≥ N, |anr | < 1. Thus |an| < r for n ≥ N. Therefore for any n ≥ N,  c n |a cn| < . n r P∞ n Since |c/r| < 1, the geometric series n=N+1(c/r) is convergent. By the comparison test, P∞ n P∞ n n=N+1 |anc | is convergent and thus n=0 anc is convergent.  Definition 1.1. R > 0 is called the radius convergence of (1.1) if P∞ n (1) for any |c| < R, n=0 anc is absolutely convergent; P∞ n (2) for any |c| > R, n=0 anc is divergent.

pn Theorem 1.2. Suppose that lim |an| = ρ. Then the radius of convergence of (1.1) is n→∞ a R = ρ−1. Similarly, if lim | n+1 | = ρ, then R = ρ−1. n→∞ 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 = ρ|x|. n→∞ an(x) ∞ ∞ X X n We know that if ρ|x| < 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 |x| < 1/ρ. Take R = 1/ρ. If |x| > R, then ρ|x| > 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 ∞ X xn (1) A (x) = (−1)n ; 1 n n=1 ∞ X xn (2) A (x) = ; 2 n2 n=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 P∞ n I is called the interval of convergence if for every c ∈ I, n=0 anc is convergent and P∞ n for any d 6∈ I, n=0 and is divergent.

Corollary 1.1. I can only be [−R,R], or (−R,R], or [−R,R) or (−R,R).

P∞ n If I is the interval of convergence of the formal power series A(x) = n=0 anx , then for P∞ n each c ∈ I, n=0 anc is convergent. For each c ∈ I, define ∞ X n A(c) = anc n=0 Then the formal power series A(x) defines a function on I.

∞ 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→∞(an+1/an) = 0. Thus the radius of convergence is R = ∞. 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 |x| < R. Then A(x) ± B(x), A(x)B(x) are all convergent on |x| < 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 P∞ n Given a power series A(x) = n=0 anx , one might think that if A is differentiable, then ∞ 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 ∞ 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 ∞ ∞ 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 |x| < R and |x0| < R. Then ∞ 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 ∞ 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 |x| < 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 |x| < R, Z A0(x) = B(x), A(x)dx = C(x).

P∞ n We have already see that the radius of convergence of f(x) = n=0 x /n! is R = ∞. Hence the function f(x) is smooth on R. Then by the differentiation theorem for power series, we find that f 0(x) − f(x) = 0, f(0) = 1. One can show that the differential equation has a unique solution and the solution is defined x x to be e . Hence f(x) = e for all x ∈ R, i.e. ∞ X xn ex = . n! n=0 ∞ X n Corollary 1.2. Let A(x) = anx be a convergent power series on |x| < R. Then A is n=0 A(n)(0) k-times differentiable for all k at x = 0 and a = for all n ≥ 0. As a consequence, n n! ∞ X A(n)(0) A(x) = xn, |x| < R. n! n=0 ∞ X n Theorem 1.6. (Taylor’s Theorem) Suppose that A(x) = anx converges on |x| < R. n=0 Let −R < a < R. Then A is k-times differentiable at x = a and for |x − a| < R − |a|, we have ∞ X A(n)(a) A(x) = (x − a)n. n! n=0 By the theorem, we find that A is k-times differentiable on |x| < R, i.e. A is a smooth function on |x| < R. A function f :(a, b) → R is called smooth if f is k-times differentiable at every point of (a, b) for all k ≥ 0. A convergent power series on |x| < R defines a smooth function on |x| < R by the Taylor’s theorem. Conversely, given a smooth function f on |x| < R, can we find a power series which is convergent to f on |x| < R? Definition 1.3. Let f be a C∞-function on |x| < R. The formal power series defined by ∞ X f (n)(0) xn n! n=0 is called the Maclaurin series generated by f at x = 0. In general, the Taylor series generated by f at x = a is the following formal power series ∞ X f (n)(a) (1.2) (x − a)n. n! n=0 6 JIA-MING (FRANK) LIOU

If the Taylor series (1.2) converges to f, we say that the Taylor series (1.2) is the Taylor expansion of f. The Taylor series generated by f may not be convergent to the original function. For example,

Example 1.3. Let f(x) be the following function on R:  e−1/x2 , for x 6= 0; f(x) = 0, x = 0. Then its Taylor series is 0 but f(x) is not a zero function. Remark. Note that if the function f is defined by a convergent power series, the Taylor series generated by f is convergent to f by Taylor’s theorem. 1.3.1. Uniqueness of Power Series. Theorem 1.7. If |x| < R, the power series ∞ X n A(x) = anx = 0. n=0

Then an = 0 for all n ≥ 0. (k) (k) Proof. Since A(x) on 0 on |x| < R, then A (x) = 0 on |x| < R. Hence an = A (0)/n! = 0 for all n.  ∞ X n Corollary 1.3. (Uniqueness of Power Series) Suppose that A(x) = anx and B(x) = n=0 ∞ X n bnx are both convergent on |x| < R. Then A(x) = B(x) if and only if an = bn for all n=0 n ≥ 0. P∞ n P∞ n Proof. For each |x| < R, n=0 anx and n=0 bnx are both convergent. Hence ∞ ∞ ∞ X n X n X n (an − bn)x = anx − bnx = 0. n=0 n=0 n=0 By the previous theorem, an − bn = 0 for all n ≥ 0, i.e. an = bn for all n ≥ 0.  Now, we have already shown that the convergence power series on |x| < R defines a function on |x| < R. We might ask ourself the following questions. To which function does the power series converge? 1.4. Power Series Expansion defined by Geometric Series. Example 1.4. Find a function f(x) so that ∞ X (1) f(x) = xn. n=0 ∞ X (2) g(x) = nxn−1. n=1 ∞ X xn+1 (3) h(x) = . n + 1 n=1 CALCULUS 7

Sol. The radius of convergence of these three functions are 1. These three functions are all convergent on |x| < 1. The power series f is a geometric series and f(x) = 1/(1 − x). One also finds that g(x) = f 0(x) = 1/(1 − x)2 for |x| < 1. Moreover, h0(x) = f(x) and h(0) = 0. Hence h(x) = − ln(1 − x) for |x| < 1. Example 1.5. Find a function f(x) so that ∞ X (1) f(x) = nxn. n=1 ∞ X xn (2) f(x) = . n n=1 ∞ X xn (3) f(x) = . n2 n=1 P∞ n−1 P∞ n 0 Sol. For x 6= 0, f(x)/x = n=1 nx . Let g(x) = n=0 x . Then g (x) = f(x)/x. We know that g(x) = 1/(1 − x). Hence g0(x) = 1/(1 − x)2. Thus f(x) = x/(1 − x)2 on |x| < 1. 1.4.1. Taylor Expansion of ln(1 + x). Let f(x) = ln(1 + x), x > 0. Then f 0(x) = 1/(1 + x). For |x| < 1, we know that ∞ 1 X (1.3) = 1 − x + x2 − x3 + ··· + (−1)nxn + ··· = (−1)nxn. 1 + x n=0 By the fundamental theorem of calculus, for any x > −1, Z x dt ln(1 + x) = . 0 1 + t For |x| < 1, we can integrate (1.3) term by term and obtain ∞ ∞ X Z x X xn+1 ln(1 + x) = (−1)n tndt = (−1)n . n + 1 n=0 0 n=0 We can compute the following series by the Taylor expansion of f(x) = ln x. ∞ X (−1)n 1 1 1 (−1)n (1.4) = 1 − + − + ··· + + ··· . n + 1 2 3 4 n + 1 n=0 By (1.3), for |x| < 1, n 1 X (−1)nxn+1 = (−1)nxn + . 1 + x 1 + x k=0 Hence for |x| < 1, n X xn+1 Z x tn+1 ln(1 + x) = (−1)n + (−1)n dt. n + 1 1 + t k=0 0 For each n ∈ N, the partial sum of (1.4) is n 1 1 1 (−1)n X (−1)k s = 1 − + − + ··· + = . n 2 3 4 n k + 1 k=0 Then we obtain the following identity: Z 1 n+1 n t ln 2 = sn + (−1) dt. 0 1 + t 8 JIA-MING (FRANK) LIOU

This shows that Z 1 n+1 1 |sn − ln 2| ≤ t dt = . 0 n + 2 We conclude that lim sn = ln 2. n→∞ −1 −1 0 2 1.4.2. Taylor Expansion of tan x. Let f(x) = tan x, x ∈ R. Then f (x) = 1/(1 + x ) 0 for all x ∈ R. For |x| < 1, f (x) has the following series expansion ∞ 1 X (1.5) = (−1)nx2n. 1 + x2 n=0 Integrating (1.5), we obtain ∞ X x2n+1 tan−1 x = (−1)n . 2n + 1 n=0 Similarly, one can show that ∞ π 1 1 1 (−1)n X 1 = 1 − + − + ··· + + ··· = (−1)n . 4 3 5 7 2n + 1 2n + 1 n=0 We leave it to the reader as an exercise. ∞ X Theorem 1.8. (Abel Theorem) Suppose that an is convergent. Then we have already n=0 ∞ X n know that f(x) = anx converges uniformly on |x| < 1. Then n=0 ∞ X lim f(x) = an. x→1− n=0

Proof. Let sn be the n-th partial sum of (an). Then an = sn − sn−1. Hence n n−1 X k X k n (sk − sk−1)x = (1 − x) skx + snx . k=0 k=0 Hence for |x| < 1, we obtain X n f(x) = (1 − x) snx . n=0 P Denote s = n→∞ sn. Then for  > 0, there exists N > 0 so that for n ≥ N, |sn − s| < /2. P∞ n By n=0 x = 1/(1 − x) for |x| < 1, we find N X  |f(x) − s| ≤ (1 − x) |s − s||x|n + . n 2 n=0

Since (sn) is convergent, so is (sn − s). Thus there exists M > 0 so that |sn − s| ≤ M for all n. Therefore  |f(x) − s| ≤ (1 − x)NM + . 2 For the same  > 0, choose δ = /(2MN) > 0. Then whenever |x − 1| < δ, we have |1 − x|MN < /2. In this case we find that for |1 − x| < δ, |f(x) − s| < . Thus we prove that limx→1− f(x) = s  CALCULUS 9

1.5. Power Series Defined by Differential Equations. 1.5.1. Taylor Expansion of sin x and cos x. Let us consider the following second order dif- ferential equation (1.6) y00 + y = 0. P∞ n Suppose that y = n=0 anx is a solution. Then we find the following recursive relation 1 a = − a , n ≥ 0. n+2 (n + 1)(n + 2) n Hence one finds that (−1)n (−1)n a = a , a = a . 2n (2n)! 0 2n+1 (2n + 1)! 1 Then y is formally given by ∞ ∞ X (−1)n X (−1)n y = a x2n + a x2n+1. 0 (2n)! 1 (2n + 1)! n=0 n=0 Let C(x) and S(x) be the following formal power series ∞ ∞ X (−1)n X (−1)n C(x) = x2n,S(x) = x2n+1. (2n)! (2n + 1)! n=0 n=0

Then y = a0C(x) + a1S(x). Proposition 1.1. The radii of convergence of C and S are both ∞ and they are both solution to (1.6). Moreover, C(0) = 1, C0(0) = 0 and S(0) = 0 and S0(0) = 1. One can also check that c(x) = cos x and s(x) are both solutions to (1.6); c(0) = 1, c0(0) = 0 and s(0) = 0, s0(0) = 1. Now, we can show that C(x) = c(x) and S(x) = s(x). To show that C(x) = c(x) is equivalent to show that C(x) − c(x) = 0. Let g(x) = C(x) − c(x). Then g(0) = 0 and g0(0) = 0 and g also satisfies (1.6). Define h(x) = (g(x))2 + (g0(x))2. Then for all x, h0(x) = 2g(x)g0(x) + 2g0(x)g00(x) = 2g0(x) g(x) + g00(x) = 0. Hence h is a constant function by the mean value theorem. Thus h(x) = 0 for all x ∈ R which shows that g(x) = 0 for all x ∈ R. Therefore C(x) = c(x) for all x ∈ R. 1.5.2. Taylor Expansion of (1 + x)α. Let f(x) = (1 + x)α. Then (1.7) (1 + x)f 0(x) = αf(x), P∞ n f(0) = 1. Assume that A(x) = n=0 anx is a solution to (1.7). By (1.7), one can show that α − n a = a , n ≥ 1 n+1 n + 1 n and a1 = αa0. One can check that α(α − 1) α(α − 1)(α − 2) α(α − 1) ··· (α − n + 1) a = a , a = a , ··· a = a ··· . 2 2 0 3 3! 0 n n! 0 Since f(0) = 1, a0 = 1. One can show that the radius of convergence of the formal power series is 1. Hence the power series is convergent on |x| < 1. We also denote α α(α − 1) ··· (α − n + 1) = . n n! 10 JIA-MING (FRANK) LIOU

Since the power series is convergent on |x| < 1, the power series solve (1.7) on |x| < 1. Hence by the differentiation theorem for power series, ∞ X α A(x) = xn n n=0 satisfies (1.7). Now, we want to show that f(x) = A(x) for |x| < 1. Define h(x) = f(x) − A(x). Then h(0) = 0 and h also satisfies (1.7). Multiplying (1.7) by (1 + x)−α−1, we find d (1 + x)−αh0(x) − α(1 + x)−α−1h(x) = (1 + x)−αh(x) = 0. dx −α α Hence (1 + x) h(x) = C for all x ∈ R. Thus h(x) = C(1 + x) , x ∈ R. Since h(0) = 0, C = 0. We find that h(x) = 0 for |x| < 1. We conclude that ∞ X α f(x) = xn. n n=0 We can also compute the Taylor series generated by f at x = 0 by direct computing the derivatives: f (n)(x) = α(α − 1) ··· (α − n + 1)(1 + x)α−n, n ≥ 0. Hence f (n)(0) = α(α − 1) ··· (α − n + 1). Thus the Taylor series generated by f at x = 0 is given by ∞ X α(α − 1) ··· (α − n + 1) xn n! n=0 and its radius of convergence is 1. How do we know that the Taylor series generated by f at x = 0 is convergent to f? By using differential equations, we can find a power series which converges to f and thus the power series is its Taylor series by the Taylor’s theorem. Although the computation of the Taylor series generated by f is simpler, we didn’t know if the Taylor series is convergent to f. Therefore, we need to study the convergence of Taylor series.

1.6. The Convergence of Taylor Series. Suppose that f is a differentiable function on an interval I. Then for each a < x, the slope of the secant line is given by f(x) − f(a) s = . x − a The mean value theorem states that we can find a point c lies between x and y so that the slope of the tangent line to x = c is s, i.e. f(x) − f(a) = f 0(c). x − a Thus f(x) − f(a) = f 0(c)(x − a), i.e. we have f(x) = f(a) + f 0(c)(x − a). Suppose that f is n + 1-times differentiable. Similarly, for a < x, we can always find a < c < x so that f 0(a) f (n)(a) f (n+1)(c) (1.8) f(x) = f(a) + (x − a) + ··· + (x − a)n + (x − a)n+1. 1! n! (n + 1)! CALCULUS 11

Definition 1.4. The polynomial Pn(x) defined by n X f (k)(a) P (x) = (x − a)k n k! k=0 is called the n-th Taylor polynomial of f at x = a. The last term in (1.8) is called the remainder of f at x = a and denoted by Rn(x, a). ∞ Theorem 1.9. (Taylor’s Theorem) Let f be a C -function on R. Let a < x be points in R. Then there exists a < c < x Then (1.8) holds. Proof. By the fundamental theorem of calculus, we find that Z x f(x) = f(a) + f 0(t)dt. a Using integration by parts, Z x f(x) = f(a) + f 0(a)(x − a) − (t − x)f 00(t)dt. a Inductively, n X f (k)(a) Z x (x − t)n f(x) = (x − a)k + f (n+1)(t)dt. k! n! k=0 a By the mean value theorem, show that there exists a ≤ cx,a ≤ x so that

n (k) (n+1) X f (a) f (cx,a) f(x) = (x − a)k + (x − a)n+1. k! (n + 1)! k=0

Therefore f(x) = Pn(x) + Rn(x, a), where f (n+1)(c ) R (x, a) = x,a (x − a)n+1 n (n + 1)!  (n+1 Theorem 1.10. Suppose that |f (t)| ≤ M for all a ≤ t ≤ x for all n ∈ N. Then the Taylor series generated by f is convergent to f for |x − a| < R. for R > 0.

Proof. Let Pn(x) be the n-th Taylor polynomial of f at x = a. By the Taylor’s theorem, MRn+1 |P (x) − f(x)| = |R (x, a)| ≤ . n n (n + 1)! n+1 Since limn→∞ MR /(n + 1)! = 0, by the Sandwich principle,

lim Pn(x) = f(x). n→∞ ∞ X f (n)(a) Hence the series (x − a)n is convergent to f(x) for |x − a| < R. n!  n=0 Example 1.6. Calculate the Taylor series of C(x) = cos x and S(x) = sin x at x = 0 directly from the definition and show that their Taylor series converge to C and S. Sol. We leave it to the reader as an exercise. 12 JIA-MING (FRANK) LIOU

z 1.7. Definition of e . For a complex number z ∈ C, define ∞ X zn ez = . n! n=0 Theorem 1.11. Given a real number θ, let z = iθ. Then iθ e = cos θ + i sin θ, θ ∈ R. Proof. For n = 2k, in = (−1)k and for n = 2k + 1, in = (−1)ki. Hence ∞ ∞ ∞ X inθn X (−1)k X (−1)k eiθ = = θ2k + i θ2k+1 n! (2k)! (2k + 1)! n=0 k=0 k=0 = cos θ + i sin θ.

 Theorem 1.12. For real numbers θ, ϕ, we have (1) ei(θ+ϕ) = eiθeiϕ; inθ iθ n (2) for n ∈ Z, e = (e ) . 1.8. More Eamples. Example 1.7. Find the Taylor expansion of the following functions 1 (1) f (x) = at x = 1 and x = 2. 1 x 1 (2) f (x) = at x = 1. 2 x2 1 (3) f (x) = at x = 1. 3 x3 x (4) f (x) = at x = 0. 4 1 − x Sol. Since x = x − 1 + 1, ∞ ∞ 1 X X (−1)nn! f (x) = = (−1)n(x − 1)n = (x − 1)n, |x − 1| < 1. 1 1 + (x − 1) n! n=0 n=0 We also find that f (n)(0) = (−1)nn!. Since x = x − 2 + 2, 1 1 1 f (x) = = 1 2 + (x − 2) 2 1 + (x − 2)/2 ∞ n 1 X x − 2 = (−1)n 2 2 n=0 ∞ X (−1)n 1 = (x − 2)n, |x − 2| < . 2n+1 2 n=0 (n) n n+1 0 0 We also find that f (2) = (−1) n!/2 . Note that f1(x) = −f2(x) and f2(x) = −2f3(x). As long as you find f2 by taking the derivative of f1, you can find f2 and thus f3. For the last problem, consider ∞ 1 X = xn, |x| < 1. 1 − x n=0 CALCULUS 13

Then by the multiplication theorem for power series, on |x| < 1, ∞ 1 X f (x) = x · = xn+1. 4 1 − x n=0 Example 1.8. Find the Taylor expansion of the following functions:

(1) f1(x) = cos 2x at x = 0. 2 (2) f2(x) = x sin x at x = 0. 2 2 (3) f3(x) = x cos x at x = 0. x (4) f4(x) = e at x = 1. x (5) f5(x) = xe at x = 0. ∞ ∞ X t2n X x2n Sol. Since cos t = (−1)n , cos 2x = (−1)n22n . Similarly, one can find the (2n)! (2n)! n=0 n=0 x x−1 Taylor expansion of f2, f3, f5 at x = 0. Since e = ee , we know that ∞ ∞ X (x − 1)n X e f (x) = ex = e · = (x − 1)n. 4 n! n! n=0 n=0

Department of Mathematics, University of California, Davis, CA 95616, U.S.A. E-mail address: [email protected]