Chapter 5 Differentiation
Exercise 21
(By analambanomenos) I’m going to show this for the infinitely differentiable case (and even that won’t be complete). Cauchy’s function for , , is the classic counterexample of a non-constant, infinitely differentiable real function such that for all orders (you can easily find a proof of this, or you can try it yourself). We can use this to define a function for , for , which is infinitely differentiable everywhere on and whose zero set is . Finally, for let an infinitely differentiable function on whose zero set is .
The complement of a closed set of is an open set consisting of a (possibly infinite) collection of open intervals . The function is well-defined since at most one of the terms in the sum is non-zero for any given , infinitely differentiable everywhere on , and has zero set .
Exercise 22
(By analambanomenos)
(a) Suppose and for . By Theorem 5.10, there is a point , , such that , contradicting for all real .
(b) If , then , which is impossible. We have Since has a single zero, at , decreases from to , then increases to , so that the range of is .
(c) Since and , by Theorem 5.10 there is a point between and such that Hence
So for and greater than , we have
Since , this last term goes to 0 as , so is a Cauchy sequence, converging to . Since is differentiable, it is continuous, hence
that is, is a fixed point of .
(d) This zig-zag path is described in with respect to the graph of in that space. It starts with the point on the graph of , goes horizonatally until it meets the diagonal at then goes vertically until it hits the graph of again at , and so forth. It will zig-zag or spiral to the point on the graph of corresponding to a fixed point of , where it crosses the diagonal .
Exercise 23
(By analambanomenos) The fixed points of are the zeros of , of which there are at most three. Since , , , , and , there are three fixed points of , lying in the intervals , , and , as asserted.
Note that . For , , and decreases monotonically to as . Hence if , then , so and . Hence as .
Similarly, for , , and increases monotonically to as . Hence if , then , so and . Hence as .
For we have , and for we have . I want to compare with the linear function , which also has a zero at and has a slope of . Since , we have for and for .
Putting this together, we get that if , then so that . Similarly, if , then so that .
That is, if , then is a monotonically increasing sequence with upper bound , and if , then is a monotonically decreasing sequence with lower bound . In either case they must have a limit . This must be a zero of (and so a fixed point of ) since otherwise , so for close to , the value of would be larger than which would contradict being a limit point of the monotonic sequence . Hence .
Exercise 24
(By analambanomenos) The distance between the successive elements of the sequence is given by the function , while the distance between the successive elements of the sequence is given by the function . We have
By Taylor’s theorem we have near
Now is the distance from to , so we see that as approaches the differences between the successive elements of the first sequence become very close to the distance to , but not so much in the case of the second sequence, which has the additional factor of . In other words, the difference between the successive steps of the and the distance to is quadratic in , while in the case of the second sequece the difference is only linear in .
Exercise 25
(By analambanomenos)
(a) The tangent to the graph of at the point has the equation Letting and solving for , we see that this line intersects the -axis at the point .
(b) Note that if then so that . By Theorem 5.10, there is a such that since implies . If for some then and so the infinite sequence converges to . Otherwise, it is a monotonically decreasing sequence with lower bound and so by Theorem 3.14 converges to a limit .
To see that , let be the continuous function defined in part (e), . Assume that . Then , so . Let be small enough so that . Then there is a such that if . Let be large enough so that . Then , but this contradicts . Hence .
(c) Letting and in Theorem 5.15, and using , there is some such that
(d) Part (c) gives us
The algorithm described in Exercises 3.16 and 3.18 is Newton’s method applied to the functions and , respectively.
(e) We have if and only if . Since tends to 0 as approaches .
(f) For , note that so that . That is, is an alternating sequence such that as .
Exercise 26
(By analambanomenos) I’m going to show this for vector-valued functions mapping into such that and such that , since this is needed for Exercise 28.
If , then , so by Theorem 5.11(b). So assume that . Following the hint, let be small enough so that and . For , let for . By Theorem 5.19, for any and for every there is a such that Since , this can only happen if , so in for each , that is, in . Repeating this argument with replacing , we get in . After about steps, we’ve shown that in .
Exercise 27
(By analambanomenos) Following the hint, let , be two solutions of the initial value problem and let . Then is differentiable on , , and
Applying Exercise 27, we get for all .
Exercise 28
(By analambanomenos) If there is a constant such that then , has at most one solution. For suppose and are two such solutions, and let . Then is differentiable on , , and Hence by the vector-valued version of Exercise 26 shown above, for all .
Exercise 29
(By analambanomenos) Using the notation of Exercise 28, where
where . Then
Since the are continuous on , they are bounded, so that there is a constant such that for all . Hence by Exercise 28, the equation has at most one solution.
Baby Rudin 数学分析原理完整第五章习题解答