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1 . Taylor s Theorem in R n Definition.. An n-dimensional multi-index is an n-tuple of nonnegative integer = (,, n. The absolute value of is defined ( to be = + + n. For each natural j number j, we define the multinomial coefficient by ( j = j!! 2! n!. If x = (x,, x n is an n-tuple of real numbers, we set x = x x 2 2 xn n. Example.. The triple = (3, 2, is a three ( dimensional multi-index. Its absolute 6 value is = 6. The multinomial coefficient = 6! 3!2!!. Furthermore x = x 3 x2 2 x 3 where x = (x, x 2, x 3. In high school, we have learned the binomial theorem m ( m (x + y m = x i y m i. i Here m is any natural number. If we denote x by x and y by x 2 and (x, x 2 by x, and = i, 2 = m i, then ( m m! = i i!(m i! = m!! 2! and x i y m i = x. Tthe above formula can be rewritten as (x + x 2 m = ( m x. =m This observation leads to the multimonomial theorem. Theorem.. (Multinomial Theorem Let x = (x,, x n be a vector in R n. For any natural number m, (x + x x n m = ( m x. =m Proof. The proof will be left to the reader as an exercise. Let f : U R be a real-valued smooth function defined on an open subset of R n. We set where x = (x,, x n. (D f(x = f x x 2 2 xn n Example.2. Let U be an open subset of R 3 and f : U R be a smooth function. Let = (2,,. Then = 4 and D f(x = 4 f x 2 x 2 x 3 (x. (x,

2 2 For any h R n and any P U, we define H i (f(p (h = for i and H 0 (f(p (h = f(p. =i ( i (D f(p h Theorem.2. (Taylor s Theorem Let f : U R be a real-valued function defined on an open subset of R n. Suppose that f C k+ (U. Let P be a point of U such that B(P, δ is contained in U for some δ > 0. For any h R n with h < δ, there exists a real number c = c P,h in [0, ] such that f(p + h = ( k i! H i(f(p (h + (k +! H k+(f(p + ch(h. Proof. The proof is the same as that in the case when n = 2. For each h with h < δ, we define a function F h : [, ] R by F h (t = F (P + th. Using the one dimensional higher mean value theorem for F h, we can find c such that F h ( = By induction, we can prove that ( k This directly implies the Taylor s Theorem. F (i (0 i! + F (k+ (c (k +!. F (i (t = H i (f(p + th(h, i. Definition.2. Let f : U R be a smooth function and P U R n. We say that f is analytic at P if there exists δ > 0 such that f(p + h = i! H i(f(p (h for any h < δ. Let us give you a criterion about the analyticity of a smooth function at a point. Since the analyticity of a function is a local behavior, we can take U = B(P, δ for some δ > 0. Theorem.3. Let f : B(P, δ R be a smooth function. Assume that there exists M > 0 such that D f(q M for any Q B(P, δ and for any n-dimensional multi-indices. Then f is analytic at P. We divide the proof into the following two steps. Suppose f satisfies the assumption given in Theorem.3. Let us define T (f(p (h = i! H i(f(p (h, h < δ. In the first step, we prove that T P (f(h converges absolutely for any h with h < δ. In the second step, we prove that f(p + h = T (f(p (h for any h < δ.

3 It follows from the assumption and the triangle inequality that for each i 0, H i (f(p (h ( i D f(p h M ( i i h. =i The multinomial theorem implies that ( h + + h n i = =i ( i h. On the other hand, the Cauchy-Schwarz inequality implies that =i ( h + + h n 2 n(h h 2 n = n h 2 and hence ( h + + h n n h. We conclude that For each i 0, H i (f(p (h M i ( n h i = ( nm h i ( nmδ i. 0 i! H i(f(p (h i! ( nmδ i. Since ( nmδ i /i! = e nmδ is convergent, by the comparison test, H i(f(p (h /i! is convergent for any h with h < δ. This completes the proof of step. Now let us prove the step 2. To prove step 2, let us fix some notation. Definition.3. For any f C k+ (U, we define the m-th Taylor polynomial of f at P U by m T m (f(p (h = i! H i(f(p (h for any 0 m k. The Taylor s Theorem tells us that f(p + h = T k (f(p (h + R k (f(p (h, where R k (f(p (h = H k+ (f(p + ch(h/(k +! for h < δ. We call R k (f(p the k-th remainder term of f at P for h < δ. The k-th Taylor polynomial of T k (f(p (h of f at P is the k + -th partial sum of the infinite series H i(f(p (h/i! for every h with h < δ. Assume that f satisfies the assumption in Theorem.3. By assumption ( k + R k (f(p (h D f(p + ch h (k +! For any h < δ, =k+ M k+ (k +! ( h + + h n k+ ( nm h k+ (k +! ( nmδ k+. (k +! 0 f(p + h T k (f(p (h ( nmδ k+. (k +! b n In calculus, we have learnt that = 0 for any real number b. By the Sandwich n n! principle, f(p + h T k(f(p (h = 0 k 3

4 4 and hence T (f(p (h = T k(f(p (h = f(p + h k for any h < δ. We complete the proof of step 2. Now let us observe the property of the k-th remainder term of a(ny function f C k+ (U at a point P U. Here we do not assume that f satisfies the assumption in Theorem.3. Lemma.. Let f C k+ (U and P U and R k (f(p (h be the k-th remainder term of f at P. Then R k (f(p (h Proof. Choose δ > 0 so that the open ball B(p, δ is contained in U. Let K be the closure of the open ball B = B(P, δ/2. Then K is closed and bounded; hence it is compact. Since f C k+ (U, D f is continuous on U for any with k +. By compactness of K and the Weierstrass extreme value Theorem, we can find M > 0 so that D f(x M on K for any k +. Let M = max{m : k + }. Since { : k + } is a finite set, M > 0. This shows that D f(q M for any Q B for any with k +. For h < δ/2, For h < δ/2, Since h = prove our assertion. R k (f(p (h (k +! =k+ ( k + M (k +! ( h + + h n k+ M (k +! ( n h k+. 0 R k(f(p (h h k Mn k+ 2 (k +! h. 0 = 0, the Sandwich principle implies D f(p + ch h R k (f(p (h h k = 0; we The above property allows us to show that the k-th Taylor polynomial of a function f at a point P is unique if f C k+ (U. More precisely, the above property characterizes the k-th Taylor polynomial of a function at a point P. We have the following result: Theorem.4. Suppose f C k+ (B(P, δ and there exists a polynomial of Q of degree k and a function E defined on B(0, δ such that ( f(p + h = Q(h + E(h for h < δ, E(h h k = 0. (2 Then Q(h = T k (f(p (h. To prove this theorem, we need one more lemma. Lemma.2. If P (h is a real polynomial of degree at most k such that then P is the zero polynomial. P (h h k = 0,

5 Proof. This technical lemma will be proved later. Let us prove Theorem.4. Write f(p + h = T k (f(p (h + R k (f(p (h for h < δ. Since f(p + h = Q(h + E(h, we find T k (f(p (h Q(h = E(h R k (f(p (h. Let P (h = T k (f(p (h Q(h for any h. Then P (h is a real polynomial of degree k. Furthermore, by assumption and the property of R k, we have E(h R k (f(p (h E(h h k = h k + R k (f(p (h This implies that P (h h k = E(h R k (f(p (h By Lemma.2, P is the zero polynomial, i.e. Q(h = T k (f(p (h for any h R n. Now let us go back to the proof of Lemma.2. Denote P (h = k a h and each 0 j k, P j (h = =j a h. For each 0 j k, either P j is the zero polynomial or P j is a homogenous polynomial of degree j. Furthermore, P (h = P 0 (h + P (h + + P k (h. If all P j are zero polynomial, we are done. Suppose not. Let j < k be the smallest natural number so that P j 0. Then P = P j + P j+ + + P k. Therefore Then t j = t j P j (h + t j+ P j+ (h + + t k P k (h. = t k h k tk j h k = th k tk j h k = 0. On the other hand, t j = P j (h + tr(t, h where R(t, h is a polynomial in t, h. This shows that P j (h is the zero polynomial which leads to the contradiction to our assumption. Department of Mathematics, National Cheng Kung University, Taiwan, fjmliou@mail.ncku.edu.tw NCTS, Mathematics 5

e x = 1 + x + x2 2! + x3 If the function f(x) can be written as a power series on an interval I, then the power series is of the form

e x = 1 + x + x2 2! + x3 If the function f(x) can be written as a power series on an interval I, then the power series is of the form Taylor Series Given a function f(x), we would like to be able to find a power series that represents the function. For example, in the last section we noted that we can represent e x by the power series