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Solution to Principles of Mathematical Analysis Chapter 5 Part C


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 F(x)=e1/x2 for x0, F(0)=0, is the classic counterexample of a non-constant, infinitely differentiable real function such that f(n)(0)=0 for all orders n (you can easily find a proof of this, or you can try it yourself). We can use this to define a function F~(x)=F(x) for x>0, F~(x)=0 for x0, which is infinitely differentiable everywhere on R and whose zero set is (,0]. Finally, for a<b let Fa,b(x)=F~(xa)F~(bx), an infinitely differentiable function on R whose zero set is (,a][b,).

The complement of a closed set E of R is an open set consisting of a (possibly infinite) collection of open intervals (an,bn). The function f(x)=Fan,bn(x) is well-defined since at most one of the terms in the sum is non-zero for any given x, infinitely differentiable everywhere on R, and has zero set E.


Exercise 22

(By analambanomenos)

(a) Suppose f(a)=a and f(b)=b for a<b. By Theorem 5.10, there is a point t, a<t<b, such that f(t)=(f(b)f(a))/(ba)=1, contradicting f(t)1 for all real t.

(b) If t=f(t)=t+(1+et)1, then (1+et)1=0, which is impossible. We have f(t)=1et(1+et)2f(t)=et(et1)(1+et)3. Since f(t) has a single zero, at t=0, f(t) decreases from limtf(t)=1 to f(0)=3/4, then increases to limtf(t)=1, so that the range of f is [3/4,1).

(c) Since xn+1=f(xn) and xn=f(xn1), by Theorem 5.10 there is a point t between xn1 and xn such that (xn+1xn)=(xnxn1)f(t). Hence
|xn+1xn|<A|xnxn1|<A2|xn1xn2|<<An1|x2x1|. So for m and n greater than N, we have
|xnxm||xnxn+1|++|xm1xm|(An1++Am2)|x2x1||x2x1|k=NAk|x2x1|1AAN Since A<1, this last term goes to 0 as N, so {xn} is a Cauchy sequence, converging to x. Since f is differentiable, it is continuous, hence
f(x)=limnf(xn)=limnxn+1=x that is, x is a fixed point of f.

(d) This zig-zag path is described in R2 with respect to the graph of f in that space. It starts with the point (x1,x2=f(x1)) on the graph of f, goes horizonatally until it meets the diagonal y=x at (x2,x2) then goes vertically until it hits the graph of f again at (x2,x3=f(x2)), and so forth. It will zig-zag or spiral to the point on the graph of f corresponding to a fixed point of f, where it crosses the diagonal y=x.


Exercise 23

(By analambanomenos) The fixed points of f are the zeros of g(x)=f(x)x, of which there are at most three. Since g(2)=1/3, g(1)=1, g(0)=1/3, g(1)=1/3, and g(2)=1, there are three fixed points of f, lying in the intervals (2,1), (0,1), and (1,2), as asserted.

Note that Δxn=xnxn1=f(xn1)xn1=g(xn1). For x<α, g(x)<0, and g decreases monotonically to as x. Hence if x1<α, then g(xn1)=Δxn<0, so xn<xn1 and Δxn+1=g(xn)<Δxn. Hence xn as n.

Similarly, for x>γ, g(x)>0, and g increases monotonically to + as x+. Hence if x1>γ, then g(xn1)=Δxn>0, so xn>xn1 and Δxn+1=g(xn)>Δxn. Hence xn+ as n.

For x(α,β) we have g(x)>0, and for x(β,γ) we have g(x)<0. I want to compare g with the linear function βx, which also has a zero at x=β and has a slope of 1. Since g(β)=β21>1, we have g(x)<βx for x(α,β) and g(x)>βx for x(β,γ).

Putting this together, we get that if xn(α,β), then 0<Δxn+1<βxn so that xn+1(xn,β). Similarly, if xn(β,γ), then βx<Δxn+1<0 so that xn+1(β,xn).

That is, if x1(α,β), then {xn} is a monotonically increasing sequence with upper bound β, and if x1(β,γ), then {xn} is a monotonically decreasing sequence with lower bound β. In either case they must have a limit β. This must be a zero of g (and so a fixed point of f) since otherwise g(β)0, so for xn close to β, the value of |Δxn| would be larger than |βxn| which would contradict β being a limit point of the monotonic sequence {xn}. Hence β=β.


Exercise 24

(By analambanomenos) The distance between the successive elements of the sequence xn=f(xn1) is given by the function F(x)=f(x)x, while the distance between the successive elements of the sequence yn=g(yn1) is given by the function G(y)=g(y)y. We have
F(x)=12(xαx)F(α)=0,F(α)=1G(y)=αy21+αG(α)=0,G(α)=21+α. By Taylor’s theorem we have near α
F(x)=(xα)+K1(xα)2G(y)=21+α(yα)+K2(yα)2Now αx is the distance from x to α, so we see that as xn approaches α the differences between the successive elements of the first sequence become very close to the distance to α, but not so much in the case of the second sequence, which has the additional factor of 2/(1+α). In other words, the difference between the successive steps of the xn and the distance to α is quadratic in xα, while in the case of the second sequece the difference is only linear in yα.


Exercise 25

(By analambanomenos)

(a) The tangent to the graph of f at the point (xn,f(xn)) has the equation yf(xn)=f(xn)(xxn). Letting y=0 and solving for x, we see that this line intersects the x-axis at the point (xn+1,0).

(b) Note that if xn>ξ then f(xn)>0 so that xn+1=xnf(xn)/f(xn)<xn. By Theorem 5.10, there is a t(ξ,xn) such that f(xn)=f(xn)f(ξ)=f(t)(xnξ)orξ=xnf(xn)f(t)xnf(xn)f(xn)=xn+1 since f0 implies f(t)f(xn). If for some m xm=ξ then xm+1=xm+2==ξ and so the infinite sequence converges to ξ. Otherwise, it is a monotonically decreasing sequence with lower bound ξ and so by Theorem 3.14 converges to a limit yξ.

To see that y=ξ, let g be the continuous function defined in part (e), g(x)=xf(x)/f(x). Assume that y>ξ. Then f(y)>0, so g(y)<y. Let ε>0 be small enough so that g(y)+ε<y. Then there is a δ>0 such that |g(x)g(y)|<ε if |xy|<δ. Let N be large enough so that |xNy|<δ. Then g(xN)<g(y)+ε<y, but this contradicts g(xN)=xN+1y. Hence y=ξ.

(c) Letting α=xn and β=ξ in Theorem 5.15, and using f(xn)=f(xn)(xnxn+1), there is some tn(ξ,xn) such that
f(ξ)=f(xn)+f(xn)(ξxn)+f(tn)2(ξxn)20=f(xn)(xnxn+1)+f(xn)(ξxn)+f(tn)2(ξxn)20=f(xn)(ξxn+1)+f(tn)2(ξxn)2xn+1ξ=f(tn)2f(xn)(xnξ)2(d) Part (c) gives us
(xn+1ξ)A(xnξ)2AA2(xn1ξ)22AA2A22(xn2ξ)23A1+2+22++2n1(x1ξ)2n=A2n1(x1ξ)2n=1A(A(x1ξ))2n.The algorithm described in Exercises 3.16 and 3.18 is Newton’s method applied to the functions x2α and xpα, respectively.

(e) We have g(ξ)=ξ if and only if f(ξ)=0. Since g(x)=1f(x)2f(x)f(x)f(x)2=f(x)f(x)f(x)2f(x)Mδ2, g(x) tends to 0 as x approaches ξ.

(f) For f(x)=x1/3, note that xn+1=x1xn1/3(1/3)xn2/3=2xn so that xn=(2)n1x1. That is, {xn} is an alternating sequence such that |xn| as x.


Exercise 26

(By analambanomenos) I’m going to show this for vector-valued functions f=(f1,,fk) mapping [a,b] into Rk such that f(a)=0 and such that |f(x)|A|f(x)|, since this is needed for Exercise 28.

If A=0, then f=0, so f=f(a)=0 by Theorem 5.11(b). So assume that A>0. Following the hint, let δ>0 be small enough so that δ<1/A and x0=a+δb. For i=1,,k, let Mi,0=sup|fi(x)|,Mi,1=sup|fi(x)| for axx0. By Theorem 5.19, for any x(a,x0) and for every i=1,,k there is a ti(a,x) such that |fi(x)|=|fi(x)fi(a)|=|fi(t)|(xa)Mi,1(xa)AMi,0δ. Since Aδ<1, this can only happen if Mi,0=0, so fi(x)=0 in [a,a+δ] for each i=1,,k, that is, f=0 in [a,a+δ]. Repeating this argument with a+δ replacing a, we get f(x)=0 in [a,a+2δ]. After about (ba)/δ steps, we’ve shown that f(x)=0 in [a,b].


Exercise 27

(By analambanomenos) Following the hint, let f1, f2 be two solutions of the initial value problem and let f=f1f2. Then f is differentiable on [a,b], f(a)=f1(a)f2(a)=0, and
|f(x)|=|f1(x)f2(x)|=|ϕ(x,f1(x))ϕ(x,f2(x))|A|f1(x)f2(x)|=A|f(x)|. Applying Exercise 27, we get f(x)=0 for all x[a,b].


Exercise 28

(By analambanomenos) If there is a constant A such that |ϕ(x,y2)ϕ(x,y1)|A|y2y1|, then y=ϕ(x,y), y(a)=c has at most one solution. For suppose f1 and f2 are two such solutions, and let f=f1f2. Then f is differentiable on [a,b], f(a)=0, and |f(x)|=|ϕ(x,f1(x))ϕ(x,f2(x))|A|f1f2|=A|f|. Hence by the vector-valued version of Exercise 26 shown above, f(x)=0 for all x[a,b].


Exercise 29

(By analambanomenos) Using the notation of Exercise 28, ϕ(x,y)=(ϕ1(x,y),,ϕk(x,y)) where
ϕj(x,y)={yj+1j=1,,k1f(x)+i=1kgi(x)yi=f(x)+g(x)yj=k where g(x)=(g1(x),,gk(x)). Then
|ϕ(x,y2)ϕ(x,y2)|2=j=1k|ϕj(x,y2)ϕj(x,y1)|2=j=1k1|y2,j+1y1,j+1|2+|g(x)(y2y1)|2|y2y1|2+|g(x)|2|y2y1|2=(1+|g(x)|2)|y2y1|2|ϕ(x,y2)ϕ(x,y2)|(1+|g(x)|2)|y2y1|Since the gi(x) are continuous on [a,b], they are bounded, so that there is a constant A such that (1+|g(x)|2)A for all x[a,b]. 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


Linearity

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