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For each natural number j, we define the multinomial coefficient  j α  by  j α

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Definition 1.1. An n-dimensional multi-index is an n-tuple of nonnegative integer α = (α1, · · · , αn). The absolute value of α is defined to be |α| = α1+ · · · + αn. For each natural number j, we define the multinomial coefficient  j

α

 by

 j α



= j!

α12! · · · αn!. If x = (x1, · · · , xn) is an n-tuple of real numbers, we set xα= xα11xα22· · · xαnn.

Example 1.1. The triple α = (3, 2, 1) is a three dimensional multi-index. Its absolute value is |α| = 6. The multinomial coefficient  6

α



= 6!

3!2!1!. Furthermore xα = x31x22x3

where x = (x1, x2, x3).

In high school, we have learned the binomial theorem (x + y)m =

m

X

i=0

m i



xiym−i.

Here m is any natural number. If we denote x by x1 and y by x2 and (x1, x2) by x, and α1 = i, α2 = m − i, then

m i



= m!

i!(m − i)! = m!

α12! and xiym−i = xα. Tthe above formula can be rewritten as

(x1+ x2)m = X

|α|=m

m α

 xα. This observation leads to the multimonomial theorem.

Theorem 1.1. (Multinomial Theorem) Let x = (x1, · · · , xn) be a vector in Rn. For any natural number m,

(x1+ x2+ · · · + xn)m= X

|α|=m

m α

 xα.

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 Rn. We set

(Dαf )(x) = ∂|α|f

∂xα11∂xα22· · · ∂xαnn(x), where x = (x1, · · · , xn).

Example 1.2. Let U be an open subset of R3 and f : U → R be a smooth function. Let α = (2, 1, 1). Then |α| = 4 and

Dαf (x) = ∂4f

∂x21∂x2∂x3

(x).

1

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For any h ∈ Rn and any P ∈ U, we define Hi(f )(P )(h) = X

|α|=i

 i α



(Dαf )(P )hα for i ≥ 1 and H0(f )(P )(h) = f (P ).

Theorem 1.2. (Taylor’s Theorem) Let f : U → R be a real-valued function defined on an open subset of Rn. Suppose that f ∈ Ck+1(U ). Let P be a point of U such that B(P, δ) is contained in U for some δ > 0. For any h ∈ Rn with khk < δ, there exists a real number c = cP,h in [0, 1] such that

f (P + h) =

k

X

i=0

1

i!Hi(f )(P )(h)

!

+ 1

(k + 1)!Hk+1(f )(P + ch)(h).

Proof. The proof is the same as that in the case when n = 2. For each h with khk < δ, we define a function Fh: [−1, 1] → R by Fh(t) = F (P + th). Using the one dimensional higher mean value theorem for Fh, we can find c such that

Fh(1) =

k

X

i=0

F(i)(0) i!

!

+F(k+1)(c) (k + 1)! . By induction, we can prove that

F(i)(t) = Hi(f )(P + th)(h), i ≥ 1.

This directly implies the Taylor’s Theorem. 

Definition 1.2. Let f : U → R be a smooth function and P ∈ U ⊆ Rn. We say that f is analytic at P if there exists δ > 0 such that

f (P + h) =

X

i=0

1

i!Hi(f )(P )(h) for any khk < 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 1.3. Let f : B(P, δ) → R be a smooth function. Assume that there exists M > 0 such that

|Dαf (Q)| ≤ M|α| for any Q ∈ B(P, δ) and for any n-dimensional multi-indices α. Then f is analytic at P.

We divide the proof into the following two steps. Suppose f satisfies the assumption given in Theorem 1.3. Let us define

T (f )(P )(h) =

X

i=0

1

i!Hi(f )(P )(h), khk < δ.

In the first step, we prove that TP(f )(h) converges absolutely for any h with khk < δ. In the second step, we prove that

f (P + h) = T (f )(P )(h) for any khk < δ.

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It follows from the assumption and the triangle inequality that for each i ≥ 0,

|Hi(f )(P )(h)| ≤ X

|α|=i

 i α



|Dαf (P )||hα| ≤ Mi X

|α|=i

 i α



|hα|.

The multinomial theorem implies that

(|h1| + · · · + |hn|)i= X

|α|=i

 i α



|hα|.

On the other hand, the Cauchy-Schwarz inequality implies that (|h1| + · · · + |hn|)2 ≤ n(h21+ · · · + h2n) = nkhk2 and hence (|h1| + · · · + |hn|) ≤√

nkhk. We conclude that

|Hi(f )(P )(h)| ≤ Mi(√

nkhk)i = (√

nM khk)i ≤ (√

nM δ)i. For each i ≥ 0,

0 ≤ 1

i!|Hi(f )(P )(h)| ≤ 1 i!(√

nM δ)i. SinceP

i=0(√

nM δ)i/i! = e

nM δis convergent, by the comparison test,P

i=0|Hi(f )(P )(h)|/i!

is convergent for any h with khk < δ. This completes the proof of step 1. Now let us prove the step 2. To prove step 2, let us fix some notation.

Definition 1.3. For any f ∈ Ck+1(U ), we define the m-th Taylor polynomial of f at P ∈ U by

Tm(f )(P )(h) =

m

X

i=0

1

i!Hi(f )(P )(h) for any 0 ≤ m ≤ k.

The Taylor’s Theorem tells us that

f (P + h) = Tk(f )(P )(h) + Rk(f )(P )(h),

where Rk(f )(P )(h) = Hk+1(f )(P + ch)(h)/(k + 1)! for khk < δ. We call Rk(f )(P ) the k-th remainder term of f at P for khk < δ. The k-th Taylor polynomial of Tk(f )(P )(h) of f at P is the k + 1-th partial sum of the infinite series P

i=0Hi(f )(P )(h)/i! for every h with khk < δ. Assume that f satisfies the assumption in Theorem 1.3. By assumption

|Rk(f )(P )(h)| ≤ 1 (k + 1)!

X

|α|=k+1

k + 1 α



|Dαf (P + ch)|hα|

≤ Mk+1

(k + 1)!(|h1| + · · · + |hn|)k+1

≤ (√

nM khk)k+1 (k + 1)! ≤ (√

nM δ)k+1 (k + 1)! . For any khk < δ,

0 ≤ |f (P + h) − Tk(f )(P )(h)| ≤ (√

nM δ)k+1 (k + 1)! . In calculus, we have learnt that lim

n→∞

bn

n! = 0 for any real number b. By the Sandwich principle,

k→∞lim |f (P + h) − Tk(f )(P )(h)| = 0

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and hence

T (f )(P )(h) = lim

k→∞Tk(f )(P )(h) = f (P + h) for any khk < 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 ∈ Ck+1(U ) at a point P ∈ U. Here we do not assume that f satisfies the assumption in Theorem 1.3.

Lemma 1.1. Let f ∈ Ck+1(U ) and P ∈ U and Rk(f )(P )(h) be the k-th remainder term of f at P. Then

khk→0lim

Rk(f )(P )(h) khkk = 0.

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 ∈ Ck+1(U ), Dαf is continuous on U for any α with |α| ≤ k + 1. By compactness of K and the Weierstrass extreme value Theorem, we can find Mα > 0 so that |Dαf (x)| ≤ Mα

on K for any |α| ≤ k + 1. Let M = max{Mα : |α| ≤ k + 1}. Since {α : |α| ≤ k + 1} is a finite set, M > 0. This shows that |Dαf (Q)| ≤ M for any Q ∈ B for any α with |α| ≤ k + 1.

For khk < δ/2,

|Rk(f )(P )(h)| ≤ 1 (k + 1)!

X

|α|=k+1

k + 1 α



|Dαf (P + ch)|hα|

≤ M

(k + 1)!(|h1| + · · · + |hn|)k+1

≤ M

(k + 1)!(√

nkhk)k+1. For khk < δ/2,

0 ≤ |Rk(f )(P )(h)|

khkk ≤ M nk+12 (k + 1)!khk.

Since lim

khk→0khk = lim

khk→00 = 0, the Sandwich principle implies lim

khk→0

Rk(f )(P )(h)

khkk = 0; we

prove our assertion. 

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 ∈ Ck+1(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 1.4. Suppose f ∈ Ck+1(B(P, δ)) and there exists a polynomial of Q of degree

≤ k and a function E defined on B(0, δ) such that (1) f (P + h) = Q(h) + E(h) for khk < δ, (2) lim

khk→0

E(h) khkk = 0.

Then Q(h) = Tk(f )(P )(h).

To prove this theorem, we need one more lemma.

Lemma 1.2. If P (h) is a real polynomial of degree at most k such that

khk→0lim P (h) khkk = 0, then P is the zero polynomial.

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Proof. This technical lemma will be proved later.  Let us prove Theorem 1.4. Write f (P + h) = Tk(f )(P )(h) + Rk(f )(P )(h) for khk < δ.

Since f (P + h) = Q(h) + E(h), we find

Tk(f )(P )(h) − Q(h) = E(h) − Rk(f )(P )(h).

Let P (h) = Tk(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 Rk, we have

khk→0lim

E(h) − Rk(f )(P )(h)

khkk = lim

khk→0

E(h)

khkk + lim

khk→0

Rk(f )(P )(h) khkk = 0.

This implies that

khk→0lim P (h)

khkk = lim

khk→0

E(h) − Rk(f )(P )(h)

khkk = 0.

By Lemma 1.2, P is the zero polynomial, i.e. Q(h) = Tk(f )(P )(h) for any h ∈ Rn. Now let us go back to the proof of Lemma 1.2. Denote P (h) = P

|α|≤kaαhα and each 0 ≤ j ≤ k, Pj(h) = P

|α|=jaαhα. For each 0 ≤ j ≤ k, either Pj is the zero polynomial or Pj is a homogenous polynomial of degree j. Furthermore,

P (h) = P0(h) + P1(h) + · · · + Pk(h).

If all Pj are zero polynomial, we are done. Suppose not. Let j < k be the smallest natural number so that Pj 6= 0. Then P = Pj+ Pj+1+ · · · + Pk. Therefore

P (th) = tjPj(h) + tj+1Pj+1(h) + · · · + tkPk(h).

Then

t→0lim P (th)

tj = lim

t→0

P (th)

tkkhkk · tk−jkhkk= lim

t→0

P (th) kthkk · lim

t→0tk−jkhkk= 0.

On the other hand,

P (th)

tj = Pj(h) + tR(t, h)

where R(t, h) is a polynomial in t, h. This shows that Pj(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

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