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Higher Order Derivatives

Module by: Lawrence Baggett. E-mail the author

Summary: A series of definitions of terminology used to describe higher-order derivatives, and a formula for the coefficients of a taylor series function.

Definition 1:

Let SS be a subset of RR (or CC), and Let f:SCf:SC be a function of a real (or complex) variable. We say that ff is continuously differentiable on S0S0 if ff is differentiable at each point xx of S0S0 and the function f'f' is continuous on S0.S0. We say that fC1(S)fC1(S) if ff is continuous on SS and continuously differentiable on S0.S0. We say that ff is 2-times continuously differentiable on S0S0 if the first derivative f'f' is itself continuously differentiable on S0.S0. And, inductively, we say that ff is k-times continuously differentiable on S0S0 if the k-1k-1st derivative of ff is itself continuously differentiable on S0.S0. We write f(k)f(k) for the kkth derivative of f,f, and we write fCk(S)fCk(S) if ff is continuous on SS and is kk times continuously differentiable on S0.S0. Of course, if fCk(S),fCk(S), then all the derivatives f(j)f(j), for jk,jk, exist nd are continuous on S0.S0. (Why?)

For completeness, we define f(0)f(0) to be ff itself, and we say that fC(S)fC(S) if ff is continuous on SS and has infinitely many continuous derivatives on S0;S0; i.e., all of its derivatives exist and are continuous on S0.S0.

As in (Reference), we say that ff is real-analytic (or complex-analytic) on SS if it is expandable in a Taylor series around each point cS0cS0

REMARK Keep in mind that the definition above, as applied to functions whose domain SS is a nontrivial subset of C,C, has to do with functions of a complex variable that are continuously differentiable on the set S0.S0. We have seen that this is quite different from a function having continuous partial derivatives on S0.S0. We will return to partial derivatives at the end of this chapter.

Theorem 1

Let SS be an open subset of RR (or CC).

  1. Suppose WSWS is a subset of R.R. Then, for each k1,k1, there exists a function in Ck(S)Ck(S) that is not in Ck+1(S).Ck+1(S). That is, Ck+1(S)Ck+1(S) is a proper subset of Ck(S).Ck(S).
  2. If ff is real-analytic (or complex-analytic) on S,S, then fC(S).fC(S).
  3. There exists a function in C(R)C(R) that is not real-analytic on R.R. That is, the set of real-analytic functions on RR is a proper subset of the set C(R).C(R).

REMARK Suppose SS is an open subset of C.C. It is a famous result from the Theory of Complex Variables that if ff is in C1(S),C1(S), then ff is necessarily complex analytic on S.S. We will prove this amazing result in (Reference). Part (3) of the theorem shows that the situation is quite different for real-valued functions of a real variable.

Proof

For part (1), see the exercise below. Part (2) is immediate from part (c) of (Reference). Before finishing the proof of part (3), we present the following lemma:

Lemma 1

Let ff be the function defined on all of RR as follows.

f ( x ) = { 0 x 0 p ( x ) e - 1 / x x n x > 0 f ( x ) = { 0 x 0 p ( x ) e - 1 / x x n x > 0
(1)

where p(x)p(x) is a fixed polynomial function and nn is a fixed nonnegative integer. Then ff is continuous at each point xx of R.R.

Proof

The assertion of the lemma is clear if x0.x0. To see that ff is continuous at 0, it will suffice to prove that

lim x 0 + 0 p ( x ) e - 1 / x x n = 0 . lim x 0 + 0 p ( x ) e - 1 / x x n = 0 .
(2)

(Why?) But, for x>0,x>0, we know from part (b) of (Reference) that e1/x>1/(xn+1(n+1)!),e1/x>1/(xn+1(n+1)!), implying that e-1/x<xn+1(n+1)!.e-1/x<xn+1(n+1)!. Hence, for x>0,x>0,

| f ( x ) | = | p ( x ) | e - 1 / x x n < ( n + 1 ) ! x | p ( x ) | , | f ( x ) | = | p ( x ) | e - 1 / x x n < ( n + 1 ) ! x | p ( x ) | ,
(3)

and this tends to 0 as xx approaches 0 from the right, as desired.

Returning to the proof of Theorem 1, we verify part (3) by observing that if ff is as in the preceding lemma then ff is actually differentiable, and its derivative f'f' is a function of the same sort. (Why?) It follows that any such function belongs to C(R).C(R). On the other hand, a nontrivial such ff cannot be expandable in a Taylor series around 0 because of the Identity Theorem. (Take xk=-1/k.xk=-1/k.) This completes the proof.

Exercise 1

  1. Prove part (1) of Theorem 1. Use functions of the form xnsin(1/x).xnsin(1/x).
  2. Prove that any function of the form of the ff in the lemma above is everywhere differentiable on R,R, and its derivative has the same form. Conclude that any such function belongs to C(R).C(R).
  3. For each positive integer n,n, define a function fnfn on the interval (-1,1(-1,1 by fn(x)=|x|1+1/n.fn(x)=|x|1+1/n. Prove that each fnfn is differentiable at every point in (-1,1),(-1,1), including 0.0. Prove also that the sequence {fn}{fn} converges uniformly to the function f(x)=|x|.f(x)=|x|. (See part (h) of (Reference).) Conclude that the uniform limit of differentiable functions of a real variable need not be differentiable. (Again, for functions of a complex variable, the situation is very different. In that case, the uniform limit of differentiable functions is differentiable. See (Reference).)

Exercise 2: A smooth approximation to a step function.

Suppose a<b<c<da<b<c<d are real numbers. Show that there exists a function χχ in C(R)C(R) such that 0χ(x)10χ(x)1 for all x,x,χ(x)1χ(x)1 for x[b,c],x[b,c], and χ(x)0χ(x)0 for x(a,d).x(a,d). (If aa is close to bb and cc is close to d,d, then this function is a CC approximation to the step function that is 1 on the interval [b,c][b,c] and 0 elsewhere.)

  1. Let ff be a function like the one in the lemma. Think about the graphs of the functions f(x-c)f(x-c) and f(b-x).f(b-x). Construct a CC function gg that is 0 between bb and cc and positive everywhere else.
  2. Construct a CC function hh that is positive between aa and dd and 0 everywhere else.
  3. Let gg and hh be as in parts (a) and (b). If j=g+h,j=g+h, show that jj is never 0, and write kk for the CC function k=1/j.k=1/j.
  4. Examine the function hk,hk, and show that it is the desired function χ.χ.

Theorem 2: Formula for the coefficients of a Taylor Series function

Let ff be expandable in a Taylor series around a point c:c:

f ( x ) = a n ( x - c ) n . f ( x ) = a n ( x - c ) n .
(4)

Then for each n,n,an=f(n)(c)/n!.an=f(n)(c)/n!.

Proof

Because each derivative of a Taylor series function is again a Taylor series function, and because the value of a Taylor series function at the point cc is equal to its constant term a0,a0, we have that a1=f'(c).a1=f'(c). Computing the derivative of the derivative, we see that 2a2=f''(c)=f(2)(c).2a2=f''(c)=f(2)(c). Continuing this, i.e., arguing by induction, we find that n!an=f(n)(c),n!an=f(n)(c), which proves the theorem.

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