CALCULUS JIA-MING (FRANK) LIOU

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1 CALCULUS JIA-MING (FRANK) LIOU Abstract. Contents. Power Series.. Polynomials and Formal Power Series.2. Radius of Convergence 2.3. Derivative and Antiderivative of Power Series 4.4. Power Series Expansion defined by Geometric Series 6.5. Power Series Defined by Differential Equations 9.6. The Convergence of Taylor Series 0.7. Definition of e z 2.8. More Eamples 2. Power Series.. Polynomials and Formal Power Series. Let x be a variable and a 0,, a n be real numbers. A polynomial in x with coefficients a 0,, a n is an expression P (x) = a 0 + a x + + a n x n, If a n 0, n is called the degree of P. The space of polynomials in x with coefficients in R is denoted by R[x]. For each real number c R, define P (c) = a 0 + a c + + a n c n. Then P : R R defines a function. Hence a polynomial in x can be thought of as a function on R. Evey polynomial function is a smooth function on R which means that P (k) (x) exists for each k 0. Moreover, P (0) = a 0, P (x) = a + 2a 2 x + 3a 3 x na n x n P (x) = 2a a 3 x + 3 4a 4 x 2 + (n )na n x n 2 = P (n) (x) = a n. Hence we have P (0) = a 0, P (0) = a, P (0) = 2!a 2,, P (n) (0) = a n. Suppose that we have two polynomials P (x) = n i=0 a ix i and Q(x) = m j=0 b jx j. Then ( m m+n k ) P (x)q(x) = a i b j x i+j = a i b k i x k. i=0 j=0 Hence the product of two polynomials is again a polynomial. i=0

2 2 JIA-MING (FRANK) LIOU Let (a n ) n 0 be a sequence of real numbers. The formal power series in x with coefficients a n is an expression (.) a 0 + a x + a 2 x a n x n + = a n x n. The space of formal power series with coefficients in R is denoted by R[[x]]. Suppose that A(x) = a nx n and B(x) = b nx n are two formal power series in R[[x]]. Define their sum and their product by A(x) + B(x) = (a n + b n )x n A(x)B(x) = c n x n, where c n = n a kb n k. Hence we know that R[x] is a subset of R[[x]], i.e. every polynomial is a formal power series. Here comes a question: When does a formal power series define a function on an interval of R?.2. Radius of Convergence. The following theorem gives us an example when a formal power series converges. Theorem.. (Abel) Suppose that r is a real number so that the series a n r n is con- vergent. For each c < r, a n c n is absolutely convergent. divergent, then for c > r, a n c n is also divergent. Conversely if a n r n is Proof. Since the series a nr n is convergent, lim a nr n = 0. Hence there exists N > 0 n so that for any n N, a n r n <. Thus a n < r n for n N. Therefore for any n N, ( c ) n a n c n <. r Since c/r <, the geometric series n=n+ (c/r)n is convergent. By the comparison test, n=n+ a nc n is convergent and thus a nc n is convergent. Definition.. R > 0 is called the radius convergence of (.) if () for any c < R, a nc n is absolutely convergent; (2) for any c > R, a nc n is divergent. Theorem.2. Suppose that lim n R = ρ. Similarly, if lim n a n+ a n = ρ, then R = ρ. n an = ρ. Then the radius of convergence of (.) is Proof. Let ρ be the limit of (a n+ /a n ). Let a n (x) = a n x n. Then a n+ (x) a n (x) = a n+ a n x.

3 CALCULUS 3 Hence lim a n+ (x) n a n (x) = ρ x. We know that if ρ x <, by ratio test, the infinite series a n (x) = a n x n is absolutely convergent. In other words, the power series is convergent if x < /ρ. Take R = /ρ. If x > R, then ρ x >. By ratio test, the power series is divergent. Hence we prove that R = /ρ is the radius of convergence of the power series. Let us take a look at some examples. Example.. Find the radius of convergence of the following three formal power series () A (x) = ( ) n xn n ; n= x n (2) A 2 (x) = n 2 ; n= (3) A 3 (x) = nx n. n= Sol. By ratio test, the radii of convergence of the above three formal power series are. Note that n= ( )n /n is convergent by the Leibnitz test. A(x) is also convergent at x = but when x =, the series n= /n is divergent by the p-test. Hence A (x) is convergent on < x or I = (, ]. Similarly, A 2 (x) is convergent on both x = ±. Therefore A 2 (x) is convergent on I 2 = [, ]. We also find that A 3 is convergent on I 3 = (, ) and divergent at x = ±. I, I 2, I 3 are called the interval of convergence of A, A 2, A 3 respectively. Definition.2. An interval I on R is of the form [a, b] or [a, b) or (a, b] or (a, b). An interval I is called the interval of convergence if for every c I, a nc n is convergent and for any d I, a nd n is divergent. Corollary.. I can only be [ R, R], or ( R, R], or [ R, R) or ( R, R). If I is the interval of convergence of the formal power series A(x) = a nx n, then for each c I, a nc n is convergent. For each c I, define A(c) = a n c n Then the formal power series A(x) defines a function on I. Example.2. The radius of convergence of x n.

4 4 JIA-MING (FRANK) LIOU Sol. Since a n =, a n+/a n = /(n + ) for all n. Hence lim n (a n+ /a n ) = 0. Thus the radius of convergence is R =. Remark. Later we will show that this power series converges to e x. Theorem.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..3. Derivative and Antiderivative of Power Series. Suppose that p(x) = a 0 + a x + + a n x n = n a kx k be a polynomial. Then p (x) = a + 2a 2 x + + na n x n = ka k x k Given a power series A(x) = a nx n, one might think that if A is differentiable, then A (x) = na n x n. On the other hand, an antiderivative of p is given by p(x)dx = C + a 0 x + a 2 x2 + + a n n + an+ = C = Hence one might guess A(x)dx = c + a n n + xn+. a k k + xk+. Theorem.4. Suppose that the radius of convergence of (.) is R. Then the power series of B(x) = na n x n a n, C(x) = c + n + xn+ are also R. Suppose the radius of convergence of (.) is R. Suppose that x < R and x 0 < R. Then A(x) A(x 0 ) = a n (x n x n 0 ). In order to calculate the difference quotient of A, we need: Lemma.. Let x, y be real numbers. Then x n y n = (x y)(x n + x n 2 y + + xy n 2 + y n ). Hence the difference quotient of A at x 0 is A(x) A(x 0 ) = a n (x n + x n 2 x xx0 n 2 + x0 n ). (x x 0 )

5 CALCULUS 5 We know that lim (x n + x n 2 x xx n 2 x x 0 + x n 0 ) = nx n 0. 0 Hence one finds that (not so rigorous) A (x 0 ) = B(x 0 ). Thus A (x) = B(x) for any x < R. One could also see that C (x) = A(x). Hence C(x) = c + A(x)dx. Theorem.5. Suppose that the radius of convergence of (.) is R. Then for all x < R, A (x) = B(x), A(x)dx = C(x). We have already see that the radius of convergence of f(x) = xn / is R =. Hence the function f(x) is smooth on R. Then by the differentiation theorem for power series, we find that f (x) f(x) = 0, f(0) =. One can show that the differential equation has a unique solution and the solution is defined to be e x. Hence f(x) = e x for all x R, i.e. Corollary.2. Let A(x) = e x = x n. a n x n be a convergent power series on x < R. Then A is k-times differentiable for all k at x = 0 and a n = A(n) (0) for all n 0. As a consequence, A (n) (0) A(x) = x n, x < R. Theorem.6. (Taylor s Theorem) Suppose that A(x) = a n x n converges on x < R. Let R < a < R. Then A is k-times differentiable at x = a and for x a < R a, we have A (n) (a) A(x) = (x a) n. 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.3. Let f be a C -function on x < R. The formal power series defined by f (n) (0) x n 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 f (n) (a) (.2) (x a) n.

6 6 JIA-MING (FRANK) LIOU If the Taylor series (.2) converges to f, we say that the Taylor series (.2) is the Taylor expansion of f. The Taylor series generated by f may not be convergent to the original function. For example, Example.3. Let f(x) be the following function on R: { e /x 2, for x 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..3.. Uniqueness of Power Series. Theorem.7. If x < R, the power series A(x) = a n x n = 0. Then a n = 0 for all n 0. Proof. Since A(x) on 0 on x < R, then A (k) (x) = 0 on x < R. Hence a n = A (k) (0)/ = 0 for all n. Corollary.3. (Uniqueness of Power Series) Suppose that A(x) = a n x n and B(x) = b n x n are both convergent on x < R. Then A(x) = B(x) if and only if a n = b n for all n 0. Proof. For each x < R, a nx n and b nx n are both convergent. Hence (a n b n )x n = a n x n b n x n = 0. By the previous theorem, a n b n = 0 for all n 0, i.e. a n = b n 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?.4. Power Series Expansion defined by Geometric Series. Example.4. Find a function f(x) so that () f(x) = x n. (2) g(x) = (3) h(x) = nx n. n= n= x n+ n +.

7 CALCULUS 7 Sol. The radius of convergence of these three functions are. These three functions are all convergent on x <. The power series f is a geometric series and f(x) = /( x). One also finds that g(x) = f (x) = /( x) 2 for x <. Moreover, h (x) = f(x) and h(0) = 0. Hence h(x) = ln( x) for x <. Example.5. Find a function f(x) so that () f(x) = nx n. (2) f(x) = (3) f(x) = n= n= n= x n n. x n n 2. Sol. For x 0, f(x)/x = n= nxn. Let g(x) = xn. Then g (x) = f(x)/x. We know that g(x) = /( x). Hence g (x) = /( x) 2. Thus f(x) = x/( x) 2 on x <..4.. Taylor Expansion of ln( + x). Let f(x) = ln( + x), x > 0. Then f (x) = /( + x). For x <, we know that (.3) + x = x + x2 x ( ) n x n + = ( ) n x n. By the fundamental theorem of calculus, for any x >, x dt ln( + x) = + t. For x <, we can integrate (.3) term by term and obtain x ln( + x) = ( ) n t n dt = 0 0 ( ) n xn+ n +. We can compute the following series by the Taylor expansion of f(x) = ln x. ( ) n (.4) n + = ( )n n + +. By (.3), for x <, Hence for x <, ln( + x) = + x = ( ) n x n + ( )n x n+. + x For each n N, the partial sum of (.4) is ( ) n xn+ n + + x s n = ( )n n Then we obtain the following identity: ln 2 = s n ( ) n tn+ + t dt. = ( ) n tn+ + t dt. ( ) k k +.

8 8 JIA-MING (FRANK) LIOU This shows that We conclude that lim n s n = ln 2. s n ln 2 0 t n+ dt = n Taylor Expansion of tan x. Let f(x) = tan x, x R. Then f (x) = /( + x 2 ) for all x R. For x <, f (x) has the following series expansion (.5) Integrating (.5), we obtain Similarly, one can show that + x 2 = ( ) n x 2n. tan x = ( ) n x2n+ 2n +. π 4 = ( )n 2n + + = ( ) n 2n +. We leave it to the reader as an exercise. Theorem.8. (Abel Theorem) Suppose that know that f(x) = a n is convergent. Then we have already a n x n converges uniformly on x <. Then lim f(x) = a n. x Proof. Let s n be the n-th partial sum of (a n ). Then a n = s n s n. Hence n (s k s k )x k = ( x) s k x k + s n x n. Hence for x <, we obtain f(x) = ( x) s n x n. Denote s = n s n. Then for ɛ > 0, there exists N > 0 so that for n N, s n s < ɛ/2. By xn = /( x) for x <, we find N f(x) s ( x) s n s x n + ɛ 2. Since (s n ) is convergent, so is (s n s). Thus there exists M > 0 so that s n s M for all n. Therefore f(x) s ( x)nm + ɛ 2. For the same ɛ > 0, choose δ = ɛ/(2mn) > 0. Then whenever x < δ, we have x MN < ɛ/2. In this case we find that for x < δ, f(x) s < ɛ. Thus we prove that lim x f(x) = s

9 .5. Power Series Defined by Differential Equations. CALCULUS Taylor Expansion of sin x and cos x. Let us consider the following second order differential equation (.6) y + y = 0. Suppose that y = a nx n is a solution. Then we find the following recursive relation a n+2 = (n + )(n + 2) a n, n 0. Hence one finds that a 2n = ( )n (2n)! a 0, a 2n+ = ( )n (2n + )! a. Then y is formally given by ( ) n ( ) n y = a 0 (2n)! x2n + a (2n + )! x2n+. Let C(x) and S(x) be the following formal power series ( ) n ( ) n C(x) = (2n)! x2n, S(x) = (2n + )! x2n+. Then y = a 0 C(x) + a S(x). Proposition.. The radii of convergence of C and S are both and they are both solution to (.6). Moreover, C(0) =, C (0) = 0 and S(0) = 0 and S (0) =. One can also check that c(x) = cos x and s(x) are both solutions to (.6); c(0) =, c (0) = 0 and s(0) = 0, s (0) =. 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 g (0) = 0 and g also satisfies (.6). Define h(x) = (g(x)) 2 + (g (x)) 2. Then for all x, h (x) = 2g(x)g (x) + 2g (x)g (x) = 2g (x) [ g(x) + g (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 Taylor Expansion of ( + x) α. Let f(x) = ( + x) α. Then (.7) ( + x)f (x) = αf(x), f(0) =. Assume that A(x) = a nx n is a solution to (.7). By (.7), one can show that a n+ = α n n + a n, n and a = αa 0. One can check that α(α ) α(α )(α 2) α(α ) (α n + ) a 2 = a 0, a 3 = a 0, a n = a ! Since f(0) =, a 0 =. One can show that the radius of convergence of the formal power series is. Hence the power series is convergent on x <. We also denote ( ) α α(α ) (α n + ) =. n

10 0 JIA-MING (FRANK) LIOU Since the power series is convergent on x <, the power series solve (.7) on x <. Hence by the differentiation theorem for power series, ( ) α A(x) = x n n satisfies (.7). Now, we want to show that f(x) = A(x) for x <. Define h(x) = f(x) A(x). Then h(0) = 0 and h also satisfies (.7). Multiplying (.7) by ( + x) α, we find ( + x) α h (x) α( + x) α h(x) = d dx ( + x) α h(x) = 0. Hence ( + x) α h(x) = C for all x R. Thus h(x) = C( + x) α, x R. Since h(0) = 0, C = 0. We find that h(x) = 0 for x <. We conclude that ( ) α f(x) = x n. n We can also compute the Taylor series generated by f at x = 0 by direct computing the derivatives: f (n) (x) = α(α ) (α n + )( + x) α n, n 0. Hence f (n) (0) = α(α ) (α n + ). Thus the Taylor series generated by f at x = 0 is given by α(α ) (α n + ) x n and its radius of convergence is. 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..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 s = f(x) f(a). 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) x a Thus f(x) f(a) = f (c)(x a), i.e. we have = f (c). f(x) = f(a) + f (c)(x a). Suppose that f is n + -times differentiable. a < c < x so that (.8) f(x) = f(a) + f (a)! Similarly, for a < x, we can always find (x a) + + f (n) (a) (x a) n + f (n+) (c) (n + )! (x a)n+.

11 Definition.4. The polynomial P n (x) defined by f (k) (a) P n (x) = (x a) k k! CALCULUS is called the n-th Taylor polynomial of f at x = a. The last term in (.8) is called the remainder of f at x = a and denoted by R n (x, a). Theorem.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 (.8) holds. Proof. By the fundamental theorem of calculus, we find that Using integration by parts, Inductively, f(x) = f(a) + x f(x) = f(a) + f (a)(x a) f(x) = f (k) (a) x (x a) k + k! a a f (t)dt. x a (t x)f (t)dt. (x t) n f (n+) (t)dt. By the mean value theorem, show that there exists a c x,a x so that f (k) (a) f(x) = (x a) k + f (n+) (c x,a ) (x a) n+. k! (n + )! Therefore f(x) = P n (x) + R n (x, a), where R n (x, a) = f (n+) (c x,a ) (x a) n+ (n + )! Theorem.0. Suppose that f (n+ (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 P n (x) be the n-th Taylor polynomial of f at x = a. By the Taylor s theorem, P n (x) f(x) = R n (x, a) MRn+ (n + )!. Since lim n MR n+ /(n + )! = 0, by the Sandwich principle, Hence the series lim P n(x) = f(x). n f (n) (a) (x a) n is convergent to f(x) for x a < R. Example.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 2 JIA-MING (FRANK) LIOU.7. Definition of e z. For a complex number z C, define e z z n =. Theorem.. Given a real number θ, let z = iθ. Then e iθ = cos θ + i sin θ, θ R. Proof. For n = 2k, i n = ( ) k and for n = 2k +, i n = ( ) k i. Hence e iθ i n θ n ( ) k ( ) k = = (2k)! θ2k + i (2k + )! θ2k+ = cos θ + i sin θ. Theorem.2. For real numbers θ, ϕ, we have () e i(θ+ϕ) = e iθ e iϕ ; (2) for n Z, e inθ = (e iθ ) n..8. More Eamples. Example.7. Find the Taylor expansion of the following functions () f (x) = x at x = and x = 2. (2) f 2 (x) = at x =. x2 (3) f 3 (x) = at x =. x3 (4) f 4 (x) = x at x = 0. x Sol. Since x = x +, f (x) = + (x ) = ( ) n (x ) n = We also find that f (n) (0) = ( ) n. Since x = x 2 + 2, f (x) = 2 + (x 2) = 2 + (x 2)/2 = ( ) x 2 n ( ) n 2 2 ( ) n = 2 n+ (x 2)n, x 2 < 2. ( ) n (x ) n, x <. We also find that f (n) (2) = ( ) n /2 n+. Note that f (x) = f 2(x) and f 2 (x) = 2f 3(x). As long as you find f 2 by taking the derivative of f, you can find f 2 and thus f 3. For the last problem, consider x = x n, x <.

13 Then by the multiplication theorem for power series, on x <, CALCULUS 3 f 4 (x) = x x = x n+. Example.8. Find the Taylor expansion of the following functions: () f (x) = cos 2x at x = 0. (2) f 2 (x) = x 2 sin x at x = 0. (3) f 3 (x) = x 2 cos x 2 at x = 0. (4) f 4 (x) = e x at x =. (5) f 5 (x) = xe x at x = 0. Sol. Since cos t = ( ) n t2n (2n)!, cos 2x = ( ) n 2n x2n 2. Similarly, one can find the (2n)! Taylor expansion of f 2, f 3, f 5 at x = 0. Since e x = ee x, we know that f 4 (x) = e x (x ) n e = e = (x )n. Department of Mathematics, University of California, Davis, CA 9566, U.S.A. address: frankliou@math.ucdavis.edu