The Heine-Borel Theorem

Can't seem to make sense of the definition in my Calc book.  Time to lay it out.

Consider a finite number of points c_k on an interval I:
     
If I is an open interval (a,b), then  a < c₁,   and c_n < b.
If I is a closed interval [a,b], we may choose   c₁ = a    and/or    c₂ = b, but we do not have to.

Widder says:

6.1 THE HEINE-BOREL THEOREM
Let there correspond to each point c of the closed interval a ≤ x ≤ b a  number δ_c and an interval I_c of length 2δ_c with c the center point,

The Heine-Borel theorem states that a finite number of the intervals (1) can be chosen which will cover the whole interval a ≤ x ≤ b.  That is, every point of a ≤ x ≤ b will be in at least one of the above mentioned finite number of intervals (1).  In order to emphasize the need for proving this result, let us give an example to show it false if the interval (a, b) were open instead of closed.
    Let a = 0, b = 1, and define I_c, for 0 < c < 1, as

That is, δ_c = c/2.  No finite set of intervals (2) will cover the interval 0 < x < 1.  For, consider such a set, I_c1, I_c2, . . . , I_cn, where 0 < c₁ < c₂ . . . < c_n.  Of these, the interval I_c1  reaches farthest to the left.  Hence, no point to the left of c₁/2 is covered by the set.
                    − David Widder, Advanced Calculus; Dover Press,1989; pg 170

OBJECTIONS

1) But with δ_c= c/2, the closed interval [0, 1] cannot be covered by a finite number of intervals, either.  For consider a set   I_c1, I_c2, . . . , I_cn, where 0 = c₁ < c₂ . . . < c_n=1.  We cannot have c₁ = 0:
     
Which leaves us with:
    
And we cannot cover the range    0  ≤ x  <  c₁/2.

Allowing equalities at endpoints  does not ease the bind.  Let c₁ = 0.  Then we have
     
Now consider I_c2.  Its leftmost point is c₂/2.  We still cannot cover  0 < x <  c₂/2.

2) "Covered" remains undefined.

3) According to (1), each of the I_c has a nonzero length, 2δ_c.  If we allow the endpoints to be among the c_i, the "cover" can extend beyond the edges of the interval.  But if the cover may exceed the interval, we can cover an open interval as well: every point of (a, b) is covered by the same set that covers [a, b].
We may do this at any time by choosing a fixed width for δ_c.  Perhaps that is a trivial approach.  Perhaps Widder should specify what he expects of δ_c.

Note that, although we can approach 0 from above in such a way that we never arrive, division is bounded from below, and the limit exists:
       lim _{{n→ ∞}} 1/n  = 0.
 Enumeration is not bounded from above:
       lim_{{n → ∞}} n  = ∞.
{Oops, did I give away the mystery?}

If we do not allow ourselves to hold the limit point 0, and then ask how large we have to make n  to obtain zero, then of course there is no answer.

4) We can always use circular logic to prove our assumptions back to ourselves.
For suppose we cannot.
But this is contrary to our assumption, and so we can always use circular logic to prove our assumptions back to ourselves.

Widder measures differently when he has a closed, versus open, interval.  He uses offset (finite number of subintervals) when he has the limit point, and proceeds by infinite division when the limit cannot be used.
If we break his procedure up into distinct steps, he neglects to mention the steps in red for the closed interval:
    - subdivide any finite distance c − a
    - No finite number of divisions (intervals/inequalities) can obtain a.
    - By hypothesis, the limit exists (c − a is a finite quantity), so
    - Obtain the limit, a, by  infinite subdivision.
    - Is the limit a part of the interval?  
                           (yes)
    - agree to use it, instead of the limit sequence.
    - go back to ordinary measure and observe that we may lay off a finite number of subintervals between c and a.

The question in bold is where the two cases diverge.   In the middle of the argument, he asks if he can skip the proof altogether.  The answer is yes for the closed interval, no for the open.  Otherwise, his procedure is equally incapable of obtaining a closed interval in a finite number of steps.  The proof and definitions are irrelevant to the conclusion he obtains (see 5).

The problem remains: how ought they relate?

5) Suppose we choose our endpoint a arbitrarily small, but nonzero.  Leave δ_c,=c/2 as in the book example (2).  We get a different result.  The whole, open  interval a < x < b can be convered by a finite number of I_ck.  And likewise, a closed interval.
If the interval is open, let    a < c₁ < 3a/2.
  If the interval is closed,  we may use this same range for c₁.  In addition, we can let c₁=a, and the interval c₁/2 < x < 3c₁/2  is nonzero.

The procedure offered in (2) fails for any interval containing zero, where the width of the intervals I_ck  is some proper fraction of c.  Perhaps this is the result he wants.  He is about to use the Heine-Borel theorem to "prove" that a function which never escapes to infinity, over a finite range of x, has a *cough* upper limit.  
 But if, with a finite number of intervals of width 2δ_c,  where 2δ_c is a proper fraction of c, we can cover the entire range 0 ≤ x ≤ 1, then a finite number of intervals centered at c'_k = 1/c_k  and of width  δ'_c = 1/δ_c will cover the range 1 > x > ∞.  

6)  No interval containing 0 is compatible with both the definition (1) and the assumption of δ_c = c/2.  So there can be no subinterval centered at 0.  But in the following Theorem (see part 2), he requires that every point of the closed interval a ≤ x ≤ b  be the center of such a subinterval.  

7) We can obtain contrary results by choosing different functions for δ_c.  We have a situation where
Given ε there is a corresponding δ such that there are a finite number of ...
Given ε there is a corresponding δ such that there are not a finite number of...

Which is not a contradiction; δ_c is not unique or prescribed: we can measure in whatever fashion we wish. We may divide up a length by crossing half the distance between ourselves and the other side, and never arrive at the other side, as Xeno does, AND we can cross the whole distance in a single step.

Onward:                                                                                                   Part II: →

The Fourier Transform

HOMEWORK

The Fourier Transform and its Applications, Brad Osgood

Free Online Course, Stanford University {course home}

 Problem Set 1

The complete course materials, including all problem sets and answers can be found on the course home page.  I don't look at the answers.  Comprehension.  Comprehension.  Comprehension.  I'm also using the assignments to integrate SAGE into my math studies, to build worksheets and functioning code as I move forward.

QUESTION 1(a)

Find the sum of a geometric series for arbitrary start and end points:

          

where w ≠ 1 is real or complex, p and q integers.

Consider the cases p = −∞ or q = ∞.  What about (p = −∞ and q = +∞)?

A: Let S denote the sum.
       
  Then
       
  And hence
       
  for real or complex w.

Limits at infinity:
  i) w real
I have included unbounded limits.
  p → −∞
     • w > 1:   Then w^p → 0.    S = w^q+1/(w-1)
     • 0 < w < 1:  Then w^p → +∞.  S = ∞  (S is unbounded.  The series does not converge)  
     • −1< w < 0: Then w^p → ± ∞.  S =  ± ∞.
  (Consecutive terms have opposite signs. The alternating sequences w^−2k,  w^−(2k+1)  approach +∞ and −∞ respectively.)
      • w < −1 :  Then w^p → 0.   S = w^q+1/(w-1)
  (The alternating sequences of even and odd terms approach 0⁺ and 0⁻, respectively.)

  q → +∞
     • w > 1:   Then w^q+1 → +∞.    S = +∞
     • 0 < w < 1:  Then w^q → 0⁺.  S = w^p/(1−w)
     • −1< w < 0: Then w^q → 0^±.  S = w^p/(1−w)
  (Alternating sequences of terms w^2k,  w^(2k+1)  approach 0⁺ and 0⁻ respectively.)
      • w < −1 :  Then w^q → ± ∞.    S = ± ∞
  (The alternating sequences w^2k,  w^(2k+1)  approach +∞ and −∞, respectively)

  p → −∞    and   q → +∞
     • w > 1:           Then     w^p → 0,  w^q+1 → +∞.      S = +∞
     • 0 < w < 1:  Then   w^p → +∞,    w^q → 0⁺.       S = +∞
     • −1< w < 0: Then  w^p → ± ∞,    w^q → 0^±.      S = ± ∞
     • w < −1 :        Then    w^p → 0,     w^q → ± ∞.     S = ± ∞

 ii) w complex:  w = a + bi,   b ≠ 0
Now that is a good question.
What are the conditions of convergence?
 Magnitude: 
   • Case 1:  |w| = √(a² + b²) > 1.   Then  w^q → 0 as  q → ∞,
   • Case 2:  |w| < 1.                        Then     w^p → 0  as  p → −∞
In both cases, whenever wⁿ does not go to zero, (I believe)  both a and b → ± ∞.  This is easy to see if we begin with w purely imaginary.  For any four consecutive integral powers, the resulting single term will cycle through the four states of iⁿ:  {+,  +i,  −,  −i}.  Both the sequence and series approach the limits ∞ −∞, ∞i, −∞i   in four alternating sequences.  I believe the series approaches the same four limits simultaneously for all |w| > 1, a≠0, b ≠0, by interleaved combinations of ±a and ±b.  It should be possible to determine, for any particular w, the arrangement of those terms for sequential powers by examining the argument of w:
       Arg(w) = θ,
      where   a = cos θ,     b = sin θ
It seems obvious that the series cannot converge to a particular angle measure, as even division of the circle length cannot be given by a rational expression.  However, I have left off there; my impression is that this work is beyond the scope of the question.
For complex w there is also a
     • Case 3:  |w| = 1.   Then there is no limit for a or b.  Arg(w) should likewise have no limit, as successive powers are polynomial expressions.  However, |w^p|, and |w^q+1| = 1 for all values of p and q.  The resultant vectors are rotations around the unit circle; each successive power should produce an angle different from all preceding angles.
I have not verified this, except for double-checking that it is impossible for any angle but zero to be preserved under squaring.  Hard to know how much detail other people think is normal.

QUESTION 1(b)

Find the sum
          
and explain your answer geometrically.
A: Let S denote the sum.
     
  The ratio of consecutive terms is
     
  Then
     
  And hence
     
Geometrically, the situation is straightforward.  By construction, the N terms are evenly spaced points along the arc [0, 2π) of the imaginary unit circle.  The angle measure at each point is is the same as the arc length from zero; i.e.
  • the cosine of each angle = the real component of each term = the projection of the point on the real axis
   •the sine of each angle = the imaginary component = the projection onto the imaginary axis
Each of these sums (real and imaginary) is zero for all whole numbers N > 1.

QUESTION 1(c)

   (c) Derive the formula

         

A: Let S denote the sum
     
  The ratio of consecutive terms is
     
  Following the same procedure,
     
 This could get ugly.  Preliminaries:
  
  Basic formulas:
     cos(−θ) =  cos(θ),       sin(−θ)  =  −sin(θ)
     sin(2θ)  =   2sinθ cosθ,
     cos(2θ) =  2cos²θ −1  =  1− 2sin²θ
     the addition and subtraction formulas for sine and cosine
  Collecting some terms,

     
  Applying the double-angle formulas, and cleaning up a bit:
     
  Go away, Red guys. Move the leading −i into the square brackets:
     
  Which has the conspicuous form of an angle sum:
     
Tits.  Maybe I missed an easier way.  I'll look at the answers after I have posted mine.

QUESTION 2

Represent each of the figures (a) - (d) as the sum of two isosceles triangles,  in the form,
     
  where

When the midpoints of the triangles do not coincide, call the leftmost triangle Λ_a1, and the rightmost, Λ_a2.

I have also written each figure as an adjacency list (a vector). Reading left-to-right, connecting adjacent points with straight lines gives the figure.

(a)   [(-2, 0), (0, 1), (2, 1), (4, 0)]   = Λ₂(t) + Λ₂(t −2)

(b) [(-2, 0), (0, 2), (1, 1), (2, 1), (3, 2), (5, 0)]   = 2Λ₂(t) + 2Λ₂(t −3)

For both (c) and (d), let's invent a petit theorem:
Theorem: The apex of one (or both) of the triangles occurs at the maximum value of the envelope.
{Note that the maximum value is not restricted to occur only at the apex positions}.

PROOF: The maximum value of each triangle is the apex, at c_i.  Consider their sum.
Case 1: Let the peaks of the two triangles coincide.  The highest point of the sum is of course at this common point.  In math,
     Let c₁ = c₂
     Then max(Λ_a1) + max(Λ_a2) = max(Λ_a1 + Λ_a2),   at    t = c₁ = c₂.
Case 2: Separate the peaks by a small amount, δ, so that c₁ ≠ c₂. For all t < c₁, and t > c₂, the direction of increasing Λ_a1 and Λ_a2 is into the interval c₁ ≥ t ≥ c₂.  The maximum value must then lie in c₁ ≥ t ≥ c₂.
Let the slopes of the two triangles be m₁ = (ΔΛ_a1 / Δt), and  m₂ = (ΔΛ_a2 / Δt).
Now consider m₁ + m₂, only over the interval  c₁ ≥ t ≥ c₂.   The slopes have opposite sign: there is some α > 0 such that m₁ = −α m₂.
(I consider only the case where the triangles are sufficiently close that neither m₁ nor m₂ falls to zero in the interval.  If the triangles are moved farther apart... oh, hell... the proof is even easier that way, but I will finish it this way, and it should be clear the proof always holds as the triangles are moved farther apart.)
Suppose α=1, then the maximum value of  m₁+ m₂ = 0.  The sum is a horizontal line, with breakpoints at c₁ and c₂.  Then max(Λ_a1 + Λ_a2) occurs at both c₁ and c₂ (and everywhere in between).
Suppose α ≠ 1.  Then one of the two, |m₁| or |m₂| is greater.  Suppose |m₁| > |m₂|.  Now, the minimum value of Λ_a1 on the interval is at the rightmost point, c₂.  Moving to the left, m₁ is positive, m₂ is negative, and |m₁| > |m₂| and therefore the sum m₁+m₂ is positive over the entire interval, to the maximum value of Λ_a1, at c₁.  
If |m₂| > |m₁|, the procedure is the same, but moving from right to left.
Hence, the maximum value of  Λ_a1+ Λ_a2  invariably coincides with the apex of one (or both) of the triangles.

(c)  [(1, 0), (3, 6), (7, 0)] =   6Λ₂(t−3) + 3Λ₂(t −5)

(d) The question appears to be, "under what conditions is the sum a triangle with unequal sides?"

Working this out in Sage, I have a couple of clunky cases, with proofs.  The answer is simple, and but piecewise functions make the algebraic arguments a bit messy.  The main point is this: Each triangle has 3 corners where the slope changes:  The function is zero; it goes up; it goes down. No mystery here: it's a triangle.  The sum then has at most six corners.  What are the conditions for a sum with only three?  First, there are three corners if the x-coordinates of the three corners are identical for both triangles.  That is,

Case 0: The sum is a triangle with equal sides.  The two triangles have the same base, and the same center.  a₁=a₂, c₁=c₂.  That is, the peaks and base coincide.  But we're really interested in the next two cases where,

The sum is a triangle with unequal sides.

Suppose only two of the three x-coordinates are shared between the triangles.  

Consider the following diagram/solution:

Case 1:
The apex of each triangle coincides with an endpoint of the other:  a₁=a₂.
Then c₂ divides the longer side into two equal segments.  For the two sections to form an unbroken line, over the same change in t we must have:

rise(Λ₁) − rise(Λ₂) = rise(Λ₂)
rise(Λ₁)  =   2 rise(Λ₂)
Hence b₁ = 2b₂.  Equivalently, the slope of Λ₁ is twice that of Λ₂.
In terms of Lambda,  the sum is
          
where the first is a left-shouldered triangle, the second is right-shouldered.
case 2:

Here, the triangles share a common endpoint

Again, in terms of Lambda, we have

          
What is common to both cases? The conditions are:
1) One of the triangles must have a slope twice that of the other.
2) the c_i satisfy c₂ = c₁ + a, independently of how we choose to construct the 2:1 slope ratio.
Synthetically, in the expression
          
we can place the numeral 2 in one of four possible positions: before a or b in either of the bΛ_a.
{remaining to do: double check that it is not possible have a triangle sum with only one shared corner.}

There are a number of ways to assemble the same conclusion.  Note, for example, that, by the little Theorem one of the triangles must have a peak at x₂.  Its base can extend no further than x₁, and therefore the second triangle must have a base which extends to x₃.  This insures that the peak of the second triangle must occur to the right of x₂.  The furthest to the left c₂ may be is the midpoint of the base.  There are two cases.  Suppose c₂ lies on the midpoint.  Then for the left endpoint we have c₁−a₁ = c₂−a₂.
Suppose c₂ lies to the right of the midpoint... etc.
{Document getting long.  To be continued}

Blizzard, June

To Our Brown-eyed Cousin,

Selky daughter, split-tongued; who wears the river, skirt-and-trouser;

I have always preferred to mock the donning of garlands, and laugh at each pronouncement of the phases of the moon.  Today the stories are no longer a game.  Your friends have been murdered and a white sill advances into the chest of the living.  None of this is alright, and I was foolish to say otherwise.

We know the Queen abducts the body.  She took us as presents; bound in furs and sleep.  This is why we practiced.   The chill curtain is our common enemy, and we are not without recourse: we are at ease with winter; the ice moon is ours.

There are three months between us and the boyish Harvest-moon, and then the Blood.  You intend to our shore by September. Then, come.  The vote has been called, and counted.  We request your presence among us, by the Harvest moon.

This date passing without sign of your arrival, I am prepared to hunt you, bring you across the river. You never have to go back.

The objections have been heard, and dismissed.  We summarize:

That we are allied with the Snow Queen.
We stole the bodies of boys and unknotted her violet dresses.  It is true.  I wear the bolt through the chest, still.  The allegiance is renounced.

On the division Seele - Unseele
We declare it null.  Let them say fortune, misfortune.  Favor and disfavor.  Nonsense.  We have not found our hearts so divined, nor the navigable world so divided.  The storm is everywhere.  Dreams come, regardless.  Terror creeps under the windowsill; I shake in the arms of lovers.  Neither pole withholds itself.
The borderlands are undivided and we believe this fact is plain to you.

That our mouths are dangerous
Well.  We are toothed.  So are you.

We love our neighbor, crooked, and hope that she is crooked in turn.
Let me walk alongside you in the dark.

It is therefore decided to send for you , to live the winter among us.  All hands assent.
Promulgated this second rose day of June.


     We remain your most humble & c.

     The wolves.