Chapter 5. Differentiation
5.5. DERIVATIVES OF HIGHER ORDER. (p.110)
Relevant exercises in Rudin:
5:R11. Computing second derivatives with just one limit-operation. (d : 3)
Rudin’s Hint at the end refers to the result you are to prove; so it should come before the sentence asking for an example. Hints for finding the example asked for: If there is such an example, then by subtracting a constant, you can get one where f (x) = 0. The easiest way to get the limit to exist is to make the numerator everywhere zero.
5:R12. | x3| has some derivatives, but not many. (d : 1) 5:R13(e-g ). Behavior of xasin ( | x |– c) (continued). (d : 1)
See comments under section 5.1 on what you may assume in doing this exercise.
5:R14. Convexity and f′′. (d : 2)
The term ‘‘convex function’’ used here was defined in 4:R23, p.101.
5:R21. Smooth functions with arbitrary zero-sets. (d : 4, 5)
The rating d : 4 applies to everything up to the last phrase, about ‘‘derivatives of all orders’’. To the student who wishes to attempt that part, I suggest first doing 5.6:1 below.
5:R25(a, b, f ). Newton’s method: the basics. (d : 3)
At the end of the first sentence, where Rudin writes ‘‘f′(x) ≥ δ > 0 [...] for all x∈[a, b]’’, he means
‘‘there exists δ > 0 such that for all x∈[a, b], f′(x)≥δ’’; similarly, the condition on f′′ means that there exists an M such that for all x, the stated inequalities hold. In part (a), after the first sentence add the instruction ‘‘Show inductively that if xn is defined and lies in (ξ, b), then the same is true of xn+1’’. The last sentence of part (a) asks you for a geometric interpretation; this interpretation is not hard to find, but if you don’t see it, it is possible for you to do the remaining parts without it.
I discuss parts (c) through (e) in the next section.
Exercises not in Rudin:
5.5:1. Derivatives and higher derivatives of bell-shaped curves. (d : 3, 2, 3, 4-5. > 5.2:4)
(a) Let f : R→ R be infinitely differentiable (meaning that f(n) exists for all n ≥ 0), and suppose that for all n ≥ 0, limx→+∞ f(n)(x) = limx→–∞ f(n)(x) = 0. (In the case n = 0, we understand f(0) to mean f .) Show that for each n ≥ 0 there exist at least n distinct real numbers x such that f(n)(x) = 0.
(b) Show that the functions 1 ⁄ (x2+ 1), 1 ⁄ (x4+ 1) and e–x
2
all satisfy the assumptions of (a). (You may assume familiar properties of the exponential function ex.) Thus by the conclusion of (a), the nth derivative of each of these functions is zero at at least n points of the line.
(c) Show that of the functions named in (b), the first and third have the property that for every n their ...
Answers to True/False question 5.3:0. (a)T. (b)F.
nth derivative is zero at exactly n points, but the second does not. (To get a feel for the problem, you might begin by sketching for yourself the functions and their first two or three derivatives. You will be able to prove that exact equality does not hold for one of the functions by finding > n zeroes of f(n)(x) for a particular value of n. In proving that it does hold for the other two, you may use the fact that a polynomial equation of degree d, adxd+ ... + a1x + a0 = 0, has at most d solutions.)
(d) If one looks for other functions f which satisfy the conditions of (a) and whose nth derivative is zero at only n points for every n, one easily sees that this is true of the trivial variants of the above two examples, A ⁄ ((B x + C )2+ 1) and A e– (B x + C )2, for all real numbers A , B , C with A and B nonzero. In each case, this is a 3-parameter family of functions. Can you find any functions with these properties that does not belong to one of these families? Any families of such functions with more than three parameters? If so, how large can you make the number of parameters? (By an n-parameter family let us understand a set of functions fA
1, A2, ... , An(x) depending on real numbers A1, A2, ... , An such that if (A1, A2, ... , An) ≠ (A1′, A2′, ... , An′), then the corresponding functions are distinct.)
5.5:2. The relation between boundedness of f′ and uniform continuity of f. (d : 1, 2, 3, 4) Let f : R→ R be a differentiable function.
(a) Show that if f′ is bounded, then f is uniformly continuous.
(b) Show by example that the converse is not true. (Suggestion: Find a function that is uniformly continuous by 4.7:1, but whose derivative is unbounded because the function wiggles rapidly for large values of x.)
(c) Show that if an example such as you are asked for in (b) is twice differentiable, then its second derivative must also be unbounded. Equivalently (in view of (a)), show that a twice differentiable function R → R whose second derivative is bounded is uniformly continuous if and only if its first derivative is bounded.
(d) Can one strengthen (c) to say for every integer n > 1 that if f : R → R is an n times differentiable function which is uniformly continuous, but such that f′ is unbounded, then f′′, f(3), ... , f (n) must all be unbounded?
5.5:3. Lagrange interpolation. (d : 3)
(a) Let f be a function on [a, b] which is n times differentiable, i.e., such that f(n) exists, and suppose there are at least n + 1 points x on [a, b] such that f (x) = 0. Show that there is at least one point f on [a, b] such that f(n)(x) = 0.
(Suggestion if you have trouble getting started: Draw a sketch of such an f for n = 3, and see how many points x in your picture satisfy f′(x) = 0. Can you prove there must be that many, or come up with a picture in which there are fewer? Once you can prove something about the number of zeroes of f′, see whether you can get from this a conclusion about the number of zeroes of f′′, and so on.)
(b) Let f be any function on [a, b] and let x0, ... , xn be distinct points of [a, b]. Show that there exists a polynomial p(x) of degree ≤ n such that p(xi) = f (xi) for i = 0, ... , n.
(Hint: If you have found a polynomial that agrees with f at x0, ... , xi –1, show that by adding a scalar multiple of (x – x0) ... (x – xi –1) you can get a polynomial that agrees with f at x0, ... , xi.)
(c) Let f and p be as in part (b). Because p is a polynomial of degree ≤n, its nth derivative p(n) is a constant c. Show with the help of (a) above that if f is n times differentiable, then for some x∈[a, b], f(n)(x) = c.
(d) Let f be as in part (b), and let q be the polynomial of degree ≤ n –1 which agrees with f at the n points x0, ... , xn –1. Then the number E = f (xn) – q(xn) represents the error resulting when we use this polynomial to approximate f at the n+1st point xn; also, this number E determines what multiple of (x – x0) ... (x – xn –1) must be added to q as in the hint to part (b) to make equality at xn hold. Write ...
Answers to True/False question 5.4:0. (a)T. (b)F.
down the formula that describes the polynomial of degree ≤n agreeing with f at all n + 1 points in terms of q and E, take its nth derivative, and apply the result of (c). Conclude that if we know a bound on f(n) on [a, b], then we can bound the error arising when we use q to approximate f at xn.
(Remark: The polynomial q is called a ‘‘Lagrange interpolation polynomial’’ for f, and the result of (d) is the ‘‘remainder formula for Lagrange interpolation’’.)
5.6. TAYLOR’S THEOREM. (pp.110-111)
Relevant exercises in Rudin:
5:R15. Bounds relating f, f′ and f′′. (d : 3)
To motivate this result, consider any twice-differentiable function f on an infinite interval (a, +
∞
). Iff′′ is everywhere zero, then the graph of f is a straight line, so that the only way f can be bounded is if its derivative is everywhere zero. Similarly, if f′′ is everywhere very small, then the curve bends very slowly, so if f ever achieves even a moderate nonzero slope, | f | must become large before it can return to a small value. Thus, if the values of | f | and of | f′′| are both everywhere small, then that of | f′|
must also be. Thus, it is natural to look for an explicit bound for | f′| in terms of | f | and | f′′|. That is what you obtain here.
Incidentally, here and in the next exercise, when Rudin writes (a,
∞
), he should, strictly, write (a, +∞
).5:R16. Bounded second derivative prevents wriggling near +
∞
. (d : 2. > 5:R15)5:R17. Specified values at three points lead to a lower bound on the third derivative. (d : 2) The assumption f (0) = 0 is not really needed.
The words ‘‘Note that equality holds for 1⁄2(x3+ x2)’’ simply point out that that function is an example showing that the conclusion ‘‘≥3’’ cannot be strengthened to ‘‘≥c’’ for any c > 3. It does not require you to do anything, unless your instructor tells you to verify that computation.
5:R18. Taylor’s Theorem with a different error estimate. (d : 3)
I haven’t thought through what is involved in this proof, so my difficulty-estimate is even more of a guess than usual.
5:R25(c, d, e). Newton’s method: convergence estimates. (d : 3)
Parts (a), (b) and ( f ) are discussed in the preceding section. In part (c), think carefully about which of xn, xn + 1, ξ, tn should correspond to which of α, β, x in the statement of Taylor’s Theorem; only one set of choices will give the result you have to prove. In the display in part (d ), note that the exponent is 2n (not 2 n ). The last sentence of that part, ‘‘(Compare ...)’’, is for your interest only, not something to write up and hand in. To part ( f ) add at the end of the first sentence the words ‘‘starting with x1 = 1’’, and add after the last sentence the additional question ‘‘Why does this not contradict what you proved in parts (a) and (b)?’’
Exercises not in Rudin:
5.6:0. Say whether each of the following statements is true or false.
(a) If f : R → R is a function such that f(n) exists for all positive integers n, then f (x) = Σ∞
n = 0( f(n)(0) xn) ⁄ n ! for all real numbers x such that this series converges.
(b) If f : R → R is a function such that f(n)(x) exists and is ≤1,000 for all positive integers n and all x∈R, then f (x) = Σ∞
n = 0( f(n)(0) xn) ⁄ n ! for all x∈R.
5.6:1. A function whose Taylor polynomials converge to the wrong result. (d : 2, 1, 1, 2, 1)
Here is a well-known example concerning Taylor series which Rudin doesn’t give till Chapter 8 (where it is Exercise 8:R1 ), because it uses properties of the exponential function, which he develops there. Since we don’t reach that chapter in this course, let us assume for this exercise a few basic properties of that function: That ex is everywhere nonzero, that e– x = 1 ⁄ ex, that ex is differentiable, with derivative equal to itself, and that ex→ +
∞
as x →+∞
. We begin by deducing from these a few more facts.(a) Show that for every polynomial p(x), as x→ +
∞
we have p(x) e–x → 0. (Hint: Use L’Hospital’s rule and induction on the degree of p.)(b) Sketch an argument showing from this that for every polynomial p(x), limx→0 p(x–1) e–x–2= 0.
(I say ‘‘sketch’’ because, although Rudin proves in Theorem 4.7 that for continuous functions f and g with appropriate domains and codomains, the composite function g f is also continuous, he does not develop general results on limits of composite functions; in particular limits at infinity; and although it is not too difficult to prove such results, it would be time-consuming to have you do so here. So simply sketch how these functions behave, assuming that in this example, limits of composites behaves as one would expect. Note that ‘‘x→ 0’’ involves values both above and below 0.)
We can now give the example. Let f : R→ R be defined by f (x) = e–x–2 if x≠ 0, and f (0) = 0.
(c) Show that on R – {0}, f has derivatives f(n) for all nonnegative integers n, and that for each n, f(n)(x) has the form pn(x–1) e– x– 2 for some polynomial pn(x).
(d) Show that for every nonnegative integer n, f(n)(0) is also defined, and equals 0. Now compute the Taylor polynomials shown in display (23) on p.110 for α = 0. Show that these converge as n→
∞
, butthat the limit is not the original function f (t).
(e) As a variant of the above, suppose we define a function g by g(x) = e–x–2 if x > 0, and g(0) = 0 if x ≤ 0. Show that g is also infinitely differentiable, and is given by its Taylor series for all negative x, but not for any positive x.
5.6:2. Lagrange interpolation with multiplicities. (d : 4. > 5.5:3)
Suppose f is a function on [a, b], and for some x∈[a, b] and n≥0, f is m –1 times differentiable at x, and f (x) = f′(x) = ... = f(m –1)(x) = 0. Then one says that f ‘‘has a zero of multiplicity at least m’’ at x. (For instance, this is the behavior of a polynomial p(t) that is divisible by (x – t)m.) If there are points x1, ... , xr in [a, b] and positive integers m1, ... , mr such that f has a zero of multiplicity at least m1 at x1, a zero of multiplicity at least m2 at x2, etc., then writing n = m1+ ... + mr, we say that ‘‘f has at least n zeroes on [a, b], counting multiplicities’’.
Prove the analog 5.5:3(a) with the condition that f be zero at at least n +1 distinct points replaced by the condition that it have at least n +1 zeroes counting multiplicities, and use this to get results similarly generalizing parts (b), (c) and (d) of that exercise. Show that Taylor’s Theorem (Theorem 5.15, p.110) is a case of the analog of part (d) of that exercise.
5.7. DIFFERENTIATION OF VECTOR-VALUED FUNCTIONS. (pp.111-113) Relevant exercises in Rudin:
5:R8 (last sentence). ‘‘Uniform differentiability’’ for vector-valued functions. (d : 3) The first part of this exercise was discussed under section 5.2.
5:R10. A case of L’Hospital’s rule valid for complex-valued functions. (d : 3) 5:R20. Taylor’s Theorem for vector-valued functions. (d : 2)
The formula Rudin asks you to get should look like that of Taylor’s Theorem, but instead of giving a precise expression for the remainder in terms of the value of the f(n) at some point, it should give an upper bound in terms of an upper bound on f(n). You can get this by obtaining such a result for real-valued functions from Taylor’s Theorem, and applying it to the components of a vector-real-valued function.
5:R28. A uniqueness theorem for systems of differential equations. (d : 3. > 5:R26, 5:R27) 5:R29. A uniqueness theorem for systems of linear differential equations. (d : 3. > 5:R28)
...
Answers to True/False question 5.6:0. (a)F. (b)T.
Exercises not in Rudin:
5.7:0. Say whether each of the following statements is true or false.
(a) If f , g : R → Rk are differentiable functions such that for all x∈R, f (x) · g (x) = 1, then for all x∈R, f′(x) · g (x) + f (x) · g′(x) = 0.
(b) If f : R → R2 is a differentiable function such that f (0) = (0, 1) and f (1) = (1, 0), then there exists c with 0 < c < 1 such that f′(c) = (1, –1).
5.7:1. Derivative of a product of a scalar- and a vector-valued function. (d : 1)
Rudin mentions on p.112, four lines below display (31), that Theorem 5.3(b) is true for vector-valued functions if ‘‘f g’’ is replaced by the inner product ‘‘f · g ’’. Prove a version of Theorem 5.3(b) in which f g is replaced by the product f g , where f is a scalar function and g a vector-valued function. (Here f g is defined by ( f g )(x) = f (x) g (x).)
5.7:2. Another version of L’Hospital’s Rule for complex-valued functions. (d : 2, 4)
(a) Show that if in Theorem 5.13, we replace f (x) by an Rk-valued function f (x), and A by a vector a∈Rk, but keep g(x) a real-valued function, then the statement remains true.
(Here a fraction such as f (x) ⁄ g(x) is understood to mean g(x)–1f (x). Note that the variable x remains real-valued, and a , b continue to be extended reals. We could generalize this exercise by taking a to be a ‘‘possibly infinite vector’’ in the sense of either of 3.4:5 or 3.4:6 above – in fact, this is what led me to put together those exercises! – but I finally decided not to bring that added complication into this one.)
(b) Deduce that Theorem 5.13 becomes true for valued functions (i.e., with f and g complex-valued functions of a real variable x, and A a complex number) if we add the hypothesis that either limx→a Im (g′(x)) ⁄ Re (g′(x)) or limx→a Re (g′(x)) ⁄ Im (g′(x)) exists. (Clearly this hypothesis does not hold in Rudin’s Example 5.18.)
5.7:3. Disconnected derivative-loci. (d : 4, 1)
Theorem 5.12 is equivalent to the statement that if f is a differentiable function on an interval, then the set of values of f′ is connected.
However –
(a) Show by example that there exists a differentiable function f from an interval [a , b] to R2, such that { f′(x) | x∈[a , b] } is not connected.
On the other hand –
(b) Show that the Corollary to Theorem 5.12 is true for vector-valued functions; i.e., that if f : [a , b] → Rk is differentiable, its derivative f′ has no simple discontinuities. (Suggestion: Apply that Corollary to the components of f .)