PROBLEM SET #9 PART B OFFICIAL SOLUTIONS TOTAL POINTS: 40

INNA ZAKHAREVICH

(5) B1. We know that

∑_{k=0}^{n} e^{ikx} = \frac{1 − e^{i(n+1)z}}{1 − e^{ix}} .

At the same time, we know that e^{iθ} = cos θ + i sin θ. Thus we see that

∑_{k=0}^{n} cos kx = Re ∑_{k=0}^{n} e^{ikx}  ∑_{k=0}^{n} sin kx = Im ∑_{k=0}^{n} e^{ikx}.

Plugging in that e^{iθ} = cos θ + i sin θ and simplifying we see that

\frac{1 − e^{i(n+1)z}}{1 − e^{ix}} = \frac{1 − cos x − cos(n + 1)x + cos nx}{2 − 2 cos x} + i\frac{sin x − sin(n + 1)x + sin nx}{2 − 2 cos x}.

Thus

∑_{k=0}^{n} cos kx = \frac{1 − cos x − cos(n + 1)x − cos(n + 2)x}{2 − 2 cos x}

∑_{k=0}^{n} sin kx = \frac{sin x + sin(n + 1)x − sin(n + 2)x}{2 − 2 cos x}.

(5) B2. Let P be of degree n. Let P (z_{0}) = re^{iθ}, and write

P (z) = re^{iθ} + b_{1}(z − z_{0}) + ⋯ + b_{n}(z − z_{0})^{n}.

(We can do this because P is a polynomial.) Let j be the minimum index such that b_{j} ≠ 0. For 0 < δ, let

z = z_{0} + \frac{δ}{b_{j}^{1/j}} e^{θi/j}.

Then

P (z) = re^{iθ} + b_{j} \frac{δ^{j}}{b_{j}} e^{iθ} + O(δ^{j+1})

which will have bigger absolute value than P (z_{0}). Thus for all ϵ > 0 we can find a point z such that |z − z_{0}| < ϵ but |P (z)| > |P (z_{0})| so P has no local maximum.

(15) B3.

(5) (1) Suppose that P is split with n distinct roots α_{1}, …, α_{n} (α_{1} < ⋯ < α_{n}) with multiplicities μ_{1}, …, μ_{n}. Note that, by the mean value theorem, P′ will have a zero between α_{i} and α_{i+1}. Also, since P (x) = (x − α_{i})^{μ_{i}}Q(x) we know that P′(x) = μ_{i}(x − α_{i})^{μ_{i}−1}Q(x) + (x − α_{i})^{μ_{i}}Q′(x), so P′(x) will have a 0 of multiplicity μ_{i} − 1 at each α_{i} (where a zero of multiplicity 0 is not a zero). Thus we have found

∑_{j=1}^{n}(μ_{i} − 1) + (n − 1) = d − 1

zeroes of P′. Since deg P′ = d − 1 we have found all zeroes, so all zeroes are real. Therefore P′ is split.

(5) (2) Let n, α_{i}, μ_{i} be as in the previous part. Also, note that P′ + λP will have a zero of multiplicity μ_{i} − 1 at each α_{i}. Notice that P (x)e^{λx} will have zeroes at each α_{i}; therefore its derivative must have a zero between α_{i} and α_{i+1}. Since e^{λx} ≠ 0 for all x ∈ R, and

(P (x)e^{λx})′ = (λP (x) + P′(x))e^{λx}

we see that λP + P′ must have a zero between α_{i} and α_{i+1}. Thus we have found d − 1 zeroes for this polynomial. Note, however, that as x → −∞ P (x)e^{λx} → 0. Since it is not uniformly 0 below α_{1} there must be some 0 of the derivative less than α_{1}. Therefore we have found d real zeroes for a polynomial of degree d, so we have found them all.

(5) (3) We will prove this by induction on d. In the previous part we showed the base case. Now suppose that the desired conclusion holds for d − 1. Then let Q = b_{0} + b_{1}x + ⋯ + b_{d}x^{d} be split. Then we know that Q = (x − α)(c_{0} + c_{1}x + ⋯ + c_{d−1}x^{d−1}) for some polynomial of degree d − 1. Let

R = c_{0}P (x) + c_{1}P′(x) + ⋯ + c_{d−1}P^{(d−1)}(x).

Note that

R − αR′ = b_{0}P (x) + b_{1}P′(x) + ⋯ + b′_{d}P^{(d)}(x).

We know that R is split, by the inductive hypothesis. Therefore R − αR′ is split, so the desired conclusion holds.

(15) B4.

(7) (1) We will first prove, by induction, that the convex hull of n points is the smallest convex polygon that contains all of the points. For two points, a and b we have that the convex hull is the set of λa + (1 − λb) for λ ∈ [0, 1]. However, note that this is b + λ(a − b) which is clearly the segment between a and b. Now suppose that the statement holds for n − 1 points. Consider the case with n points, a_{1}, …, a_{n}. Fix a nonnegative λ_{n}. Note that for any nonnegative λ_{1}, …, λ_{n−1} such that their sum is less than or equal to 1,

λ_{1}a_{1} + ⋯ + λ_{n−1}a_{n−1} = λ(t_{1}a_{1} + ⋯ + t_{n−1}a_{n−1})

where

λ = λ_{1} + ⋯ + λ_{n−1} and t_{i} = λ_{i}/λ.

Note that t_{1}a_{1} + ⋯ + t_{n−1}a_{n−1} is a point in the convex hull of the first n − 1 points. Then we know that λ_{n} = 1 − λ. So the point obtained by these lambdas will be on the segment between t_{1}a_{1} + ⋯ + t_{n−1}a_{n−1} and a_{n}. Therefore it will be contained in sone triangle with one vertex on an a_{n} and the other two among the a_{i}. Therefore the convex hull will be contained in the smallest convex polygon containing all a_{i}. Now consider the smallest convex polygon containing all a_{i}. Clearly the sides of this polygon will be contained in the convex hull of the a_{i} since we know that a segment between two points is contained in the convex hull. Now consider any other point inside the polygon. Draw a line through the point and any vertex (WLOG a_{1}) on the sides of the polygon. This line will intersect the polygon in one more point, x. We know, by a similar argument to before, that the segment a_{1}x is contained in the convex hull. Thus all points inside the polygon are contained in the convex hull. Thus the convex hull of n points is the smallest convex polygon containing the points. Notice that this can be written as the union of the interiors of all triangles with the a_{i} as vertices. We know that the roots of x^{n} − 1 are a regular n-gon centered at the origin with one vertex on 1. Thus the convex hull of these roots is the n-gon including its interior. Thus for n = 2 it is the segment from 1 to −1, for n = 3 it is the equilateral triangle, for n = 4 it is the square, and for n = 5 it is the regular pentagon. All of these are centered at the origin and have one vertex at 1.

(3) (2) Let P = (z − z_{1}) ⋯ (z − z_{d}). Then

P′ = ∑_{j=1}^{d}(z − z_{1}) ⋯ (z − z_{j−1})(z − z_{j+1}) ⋯ (z − z_{d}) = ∑_{j=1}^{d} \frac{P (z)}{z − z_{i}} .

Thus

\frac{P′(z)}{P (z)} = ∑_{j=1}^{d} \frac{1}{z − z_{i}} .

(5) (3) Solution 1: Consider a point z_{0} that is outside of the convex hull of the z_{i}. Then we know that there is some line L through z_{0} such that all of the z_{i} are on the same side of the line. Then the vectors z_{i} − z_{0} will all point to one side of this line (and not along it). Therefore the vectors 1/(z_{i} − z_{0}) will all point to one side of \overline{L}, the reflection of L over the real axis. However, this is a contradiction since these vectors cannot add to 0, since they all contain a positive component along the line perpendicular to \overline{L}. Therefore z_{0} is not a 0 of ∑ 1/(z − z_{i}) so it is not a zero of P′(z). Thus all zeroes of P′(z) lie inside the convex hull. Solution 2: Suppose that w is a zero of P′. Let l_{i} = 1/|w − z_{i}|^{2}, and let l = ∑_{j=1}^{d} l_{j}. Then

∑_{j=1}^{d} \frac{1}{w − z_{j}} = 0 ⟹ ∑_{j=1}^{d} \frac{\overline{w}}{|w − z_{j}|^{2}} = ∑_{j=1}^{d} \frac{\overline{z}_{j}}{|w − z_{j}|^{2}} .

Thus we have that

\overline{w}l = ∑_{j=1}^{d} z_{j}l_{i} ⟹ \overline{w} = ∑_{j=1}^{d} \frac{l_{j}}{l} \overline{z}_{j} ⟹ w = ∑_{j=1}^{d} \frac{l_{j}}{l} z_{j}.

Clearly, each l_{i} is nonnegative. In addition, ∑ l_{i}/l = 1. Thus this point will be in the convex hull of the z_{i}.