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Homework 11, Advanced Calculus 1
1. Rudin Chapter 8 Exercise 1
Solution: By induction on n, we can readily compute that for each n,
f(n)(x) = Pn(x) x3n e−1x2, where Pn is a polynomial for all x 6= 0 (do it!).
Then we apply induction again to show that f(n)(0) = 0 for all n, where each step uses the following L’Hospital computation with k ∈ N:
lim
x→0
e−1x2 xk = lim
x→0
x−k e−1x2
=k 2 lim
x→0
x−k+2 e−1x2
= · · · = 0
For n = 1, we have
f0(0) = lim
x→0
f (x) − f (0)
x = lim
x→0
f (x) − f (0)
x = lim
x→0
e−1x2 x = 0.
Suppose that f(n)(0) = 0, then
lim
x→0
f(n)(x) − f(n)(0)
x = Pn(0) lim
x→0
e−1x2 x3n+1 = 0 and therefore f(n+1)(0) = 0 and the induction is completed.
2. Rudin Chapter 8 Exercise 9a
Solution: Let aN = sN − log N . We show that it is a bounded and monotonic sequence and therefore convergent.
Since log N =RN 1
1
x dx, or the area under the curve f (x) = x1 from 1 to N . The number sN can be seen as the sum of areas of rectangles of base 1 and height 1n, 1 ≤ n ≤ N . In fact, it is U (f, PN) on [1, N ], where PN is the partition of N -equally subdivisions. One may see that for the same partition, L(f, PN) = sN − 1. Both of these follow from the fact that f is a decreasing function and so the supremum (infimum) of f occurs at left (right) endpoint.Therefore, log N ≤ sN and log N ≥ sN− 1 and we have
0 ≤ sN − log N ≤ 1.
That is, aN is bounded. For monotonicity, we have, for all N ,
aN +1− aN = 1 N + 1 −
Z N +1 N
1
x dx ≤ 1
N + 1− 1 N + 1 = 0 since again 1x is decreasing.
3. Rudin Chapter 8 Exercise 23
Solution: The curve γ(t) in this problem may be written in the polar form
γ(t) = r(t)eiθ(t)
where r(t) > 0 for all t ∈ [a, b] and θ(t) ∈ R. Since γ(a) = γ(b), it follows that r(a) = r(b) and θ(b) = θ(a) + 2kπ for some k ∈ Z. We will show that Ind(γ) = k.
With γ(t) 6= 0 ∀t, we have γ0(t)
γ(t) = r0(t)
r(t) − iθ0(t) = d
dt(log(r(t)) − iθ(t)).
Then, by fundamental theorem of calculus (for complex valued function) and discussion above, we have
1 2πi
Z b a
γ0(t)
γ(t) dt = θ(b) − θ(a) = k := Ind(γ) ∈ Z.
For [a, b] = [0, 2π], we have θ(t) = t and it is clear that θ(2π) − θ(0) = 2π, or Ind(γ) = 1.
The quantity is called winding number since it counts how many circles around origin (or a loop containing origin that can be continuously deformed into a circle) the curve γ has travelled during the time [a, b] with signs (+1 for counterclockwise loop and −1 for clockwise).
4. Rudin Chapter 8 Exercise 24
Solution: Intuitively, if γ does not intersect negative real axis, it can never travel a complete loop and therefore the winding number has to be 0. Precisely, for c ∈ [0, ∞), the closed curve
γc(t) := γ(t) + c
is never zero (otherwise γ(t) = −c is on the negative real axis). Also, since γc0 = γ, we have
Ind(γc) = 1 2πi
Z b a
γ(t) γ(t) + c dt
which tends to 0 as c → ∞. However, since the expression above is continuous in c (integral of a continuous function) and integer valued, it must be constant. It then follows that Ind(γ) = Ind(γc) = 0 for all c ≥ 0.
5. Rudin Chapter 8 Exercise 25
Solution: Following hint, let γ =γγ2
1, then
|1 − γ(t)| = |γ1(t) − γ2(t) γ1(t) | < 1
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since |γ1(t) − γ2(t)| < |γ1(t)| for all t. Therefore, γ(t) never lies on negative real axia and by previous problem, Ind(γ) = 0.
On the other hand, straightforward computations show that γ0
γ = γ20 γ2
−γ10 γ1
and therefore Ind(γ) = Ind(γ2) − Ind(γ1) = 0. The result then follows.
6. Rudin Chapter 8 Exercise 26
Solution: Take trigonometric polynomials P1, P2 as in the hint. By triangle inequality, we have
|γ(t)| − |P1(t)| ≤ |γ(t) − P1(t)| ≤ δ4, or |P1(t)| ≥ |γ(t)| −δ4 ≥ 3δ4.
Also, we have |P1(t) − P2(t)| ≤ |P1(t) − γ(t)| + |γ(t) − P2(t)| ≤ δ4+δ4 =δ2. The two inequalities we have so far imply
|P1(t) − P2(t)| < |P1(t)| ∀t
and therefore by Problem 5, Ind(P1) = Ind(P2). We call the common value Ind(γ).
Since both P1 and P2 have norms ≥ 3δ4 > 0, they will never be 0. Moreover, P1(a) = P1(b) = P2(a) = P2(b) by Theorem 8.15 (with [a, b] rescaled and translated to [0, 2π]). Problems 4 and 5 apply to P1and P2. We prove the two problems for γ not necessarily differentiable.
For Problem 4, if γ does not intersect negative real axis, by picking δ small enough, Re(γ(t)) > 2δ > 0 for all t. Therefore, if |γ(t) − P1(t)| < δ, P1does not intersect negative real axis either and we have Ind(P1) = 0. Therefore, Ind(γ) = 0.
For Problem 5, suppose γ1 and γ2 satisfy |γ1(t) − γ2(t)| < |γ1(t)|. Take trigonometric polynomials P1, P2 uniformly converge to γ1 and γ2 respectively. We then have
|1 −P2
P1
| ≤ |1 −γ2
γ1
| + |γ2
γ1
−P2
P1
| < 1 + |γ2
γ1
−P2
P1
|.
But since P1, P2 converge uniformly to γ1, γ2, there exist P1, P2 so that |1 −PP2(t)
1(t)| < 1 for all t.
We may then repeat Problem 5 for P1 and P2 to conclude that Ind(P1) = Ind(P2) and therefore Ind(γ1) = Ind(γ2)
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