Harmonic Things

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Harmonic Things
John D Barrow
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Geometric series
S(n) a ar ar² ar³ ... arn?¹, -1 ltrlt
1 rS(n) 0 ar ar² ar³... arn?¹
arn (1 - r)S(n) a - arn S(n) a(1 - rn)/(1
- r) S(n ? ?) a/(1 - r) So if a ½ and r
½ S(n ? ?) 1 1/2 1/4 1/8 1/16 1/32
........ 1
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Behold..
1/2 1/4 1/8 1/16 1/32 ........ 1
1
1
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The value of your investments can plummet as well
as go down
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VAT in the eternal future

• 17.5 10 5 2.5
• Next step
• 18.75 10 5 2.5 1.25
• Or
• 15 10 5
• And ultimately?
• 10 ? (1 1/2 1/4 1/8 1/16 1/32
  ....) 20
• ie 10 ? 1/(1 ½)

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The Harmonic Series
H 1 1/2 1/3 1/4 1/5 1/6 1/7 1/8
.......
Has an infinite sum
Note that the sum of 1/1p 1/2p 1/3p ..... ?
? if p ? 1 But is finite if p gt 1

Recall that 1?? 1/x dx ? but 1?? 1/xp dx
1/(p-1) for pgt1
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H has an Infinite Sum
H 1 1/2 1/3 1/4 1/5 1/6 1/7 1/8
1/9 1/10 1/11 ....... H 1 (1/2 1/3)
(1/4 1/5 1/6 1/7) (1/8 1/9 1/10
1/11 ..1/15) .. H gt 1/2 (1/4 1/4) (1/8
1/8 1/8 1/8) (1/16 1/16 1/16 1/16
.. 1/16 ) H gt 1/2 1/2 1/2 1/2 . ? ?

Divergent series are the invention of the devil,
and it is a shame to base on them any
demonstration whatsoever Niels Abel
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But H goes to infinity very slowly H(n) 1
1/2 1/3 1/4 1/5 . 1/n

• Then H(1) 1, H(2) 1.5, H(3) 1.833, H(4)
  2.083,
• H(10) 2.93, H(100) 5.19
• The sum grows very slowly as the number of terms
  increases
• H(256) 6.124 but H(1000) 7.49
• H(1,000,000) 14.39.
• When n gets large H(n) only increases as fast as
  the natural logarithm of n and approximately
• H(n) 0.577 logen.

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Rainfall Records

• In year 1 the rainfall must be a record. So the
  number of record years is
• 1
• In year 2, if the rainfall is independent of year
  1, there is a chance of 1/2 of beating the record
  year 1 rainfall and a chance of 1/2 of not
  beating it. So the expected number of record
  years in the first 2 years is
• 1 1/2
• In year 3 there are just two ways in which the 6
  possible rankings
• (123, 132, 321, 213, 312, 231)
• of the rainfall in years 1, 2 and 3 could
  produce a record in year 3 (ie a 1 in 3 chance).
  So the expected number of record years after 3
  years is
• 1 1/2 1/3
• If you keep on going, applying the same reasoning
  to each new year, you will find that after n
  independent years the expected number of record
  years is the sum
• 1 1/2 1/3 1/4 ... 1/n H(n)

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Random Records Are Rare

• Suppose that we were to apply our formula to the
  rainfall records for some place in the UK from
  1748 to 2004 - a period of 256 years.
• Then we predict that we should find only H(256)
  6.124, or about 6 record years. We would have to
  wait for more than a thousand years to have a
  good chance of finding even 8 record years.

I always thought that record would stand
until It was broken Yogi Berra
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H(100) 5.19
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Central England Temperature Record
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Bunched Traffic

• In single-lane traffic, with no overtaking, a
  slow car will be followed by a bunch of cars
  wanting to overtake and go faster. If N cars set
  out, how many bunches will form? That is the same
  as asking how many record low speeds will be
  observed, and we know the answer
• H(N) 1 1/2 1/3 1/N
• Eg H(1000) 7.49
• The bunches are successively slower, so they will
  be more widely spaced. This explains why cars
  near the exit of a long tunnel tend to travel
  faster and in smaller, more widely separated,
  bunches than cars near the entrance of the
  tunnel.

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Testing To Destruction

• Strength of rth component is Br
• Test 1st to destruction so we know B1
• Stress 2nd beam to B1. If OK B2 gt B1. If it
  breaks note B2
• Test 3rd to min of B1 and B2. If it breaks note
  B3 otherwise move to 4th component.
• Expected number of broken components is
• H(n) 1 1/2 1/3 . 1/n
• So with 1000 components you only break about
  H(1000) 7.5 of them to discover the minimum
  breaking stress
• Variance is H(n) ? ?2/6

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Collecting Sets
How many cards should you expect to buy in order
to collect the set of 50 ?
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If All the Cards Exist in Equal Numbers!
1st card I always need the first card. 2nd
card There is a 49/50 chance that I havent
already got it. 3rd card There is a 48/50
chance and so on. After you have got
40 different cards there will be a 10/50 chance
that the next one will be one you havent got.
On the average you will have to buy another
50/10, or 5 more cards, to have a better than
evens chance of getting another one that you
need. Therefore the total number of cards you
will need to buy on average to get the whole set
of 50 will be the sum of 50 terms 50/50
50/49 50/48.50/3 50/2 50/1 Each
successive term tells you how many extra cards
you need to buy to get the 1st, 2nd , 3rd , and
so on, missing members of the set of 50 cards.
The sum is 50 ? (1 1/2 1/3 . 1/50) 50
H(50) ? 225
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Collecting Sets of N Cards
On the average we will have to buy a total
of (N/N) (N/N -1) (N/N-2) . N/2 N/1
cards This is N(1 ½ 1/3 ..1/N)? NH(N)
? N 0.58 ln(N) Its much harder to
complete the second half of the collection than
the first half. The number of cards that you need
to buy in order to collect N/2 cards for half a
set is only N/N N/(N-1) N/(N - 2) . N/(½
N 1) ? N ln(N) 0.58 ln(N/2) -0.58
Nln(2) 0.7N I need on average to buy just 35
cards to get half my set of 50.
The standard deviation is 1.3N so a 66 chance of
needing the average ? 1.3 ? total cards
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Swopping

• Suppose you have F friends and you all pool cards
  in order to build up F1 sets so that you have
  one each. How many cards would you need to do
  this? When the number of cards N is large, and
  you share cards, on average you need
• N ln(N) F ln(lnN) 0.58
• But if you had each collected a set without
  swopping you would have needed about (F1)Nln(N)
  0.58 cards to complete F1 separate sets.
• For N 50 the number of card purchases saved
  would be 156F. Even with F 1 this is a
  considerable economy.

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The Secretary Problem

• N job applicants when N is large
• Interview all of them ?? Takes too much time!
• Pick one at random (1/N chance of the best) !!
• Is there a Goldilocks method between these
  extremes that gives the best chance of getting
  the top candidate quickly?

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An Optimal Strategy

• See the first C of the N candidates
• Keep a note of who is the best candidate seen so
  far
• Then hire them or next one you see who is better
• How should you pick the number C ?

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The Simple Case of Three Candidates

• Imagine we have three candidates 1, 2 and 3,
  where 3 is actually better than 2, who is better
  than 1 the six possible orders that we could
  see them in are
• 123 132 213 231 312 321
• If we always take the FIRST candidate we see
  then we pick the best one (number 3) in only two
  of the six interview patterns
• So we would pick the best person with a
  probability of 2/6, or 1/3.
• If we always let the first candidate go and
  picked the next one we saw who had a higher
  rating, we get the best candidate in the second
  (132), third (213), and the fourth cases (231)
  only, so the chance of getting the best candidate
  is now 3/6, or 1/2.
• If we let the first two candidates go and picked
  the third one we see with a higher rating we get
  the best candidate only in the first (123) and
  third (213) cases.
• The chance of getting the best one is again only
  1/3.
• With 3 candidates the strategy of letting one go
  and picking the next with a better rating gives
  the best chance of getting the best candidate.

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The Situation with N Candidates

• Note that if the best candidate is in the (r1)st
  position and we are skipping the first r
  candidates then we will choose the best candidate
  for sure, but this situation will only occur with
  a chance 1/N.
• If the best candidate is in the (r2)st position
  we will pick them with a chance 1/N ? (r/r1).
  Carrying on for the higher positions we see that
  the overall probability of success is just the
  sum of all these quantities, which is
• P(N, r) 1/N ?1 r/(r1) r/(r2) r/(r3)
  r/(r4) . r/(N-1)
• P(N, r) ? 1/N1 r ln(N-1)/r.
• This last quantity, which the series converges
  towards as N gets large has its maximum value
  when the logarithm ln(N-1)/r e, so e (N-1)/r
  ? N/r when N is large.
• Hence the maximum value of P(N, r) occurring
  there is
• P ? r/N ? ln(N/r) ? 1/e ? 0.37.

In life we tend to stop searching too soon !
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The Magic Recipe

• Reject the first C N/2.7 0.37N applicants then
  pick the first one that is better than all of
  those rejected
• We will find the best one with probability 1/e
  0.37
• Consider the case where we have 100 candidates.
  The optimal strategy is to see 37 of them and
  then pick the next one that we see who is better
  than any of them and then see no one else. This
  will result in us picking the best candidate for
  the job with a probability of about 37 -- quite
  good compared with the 1 chance if we had picked
  a candidate at random

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The Exploration Problem

• Lots of jeeps and fuel
• How do you go as far as you like using minimum
  fuel?
• I jeep goes 1 unit
• 2 jeeps use 1/3 then move 1/3 tank from jeep 2 to
  1,
• jeep 2 goes back and jeep one goes
• 11/3 units
• 3 jeeps stop after using 1/5. Put 1/5 from jeep 3
  into jeeps 1 and 2 which go on as before with 2
  coming back empty to join 3. They return home but
  jeep 1 goes
• 11/31/5 units
• 4 jeeps allow jeep 1 to go
• 11/31/51/7 units

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N Jeeps and You Can Go Forever

• Using N jeeps you can organise refuelling so
  that jeep 1 goes a distance
• 1 1/3 1/5 1/7 1/(2N-1)
• By making N large this grows as
• ½ ln(N)
• and gets as large as you like..

An unlimited supply chain!
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Max overhang is
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House of Cards
Max overhang in book lengths is
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52 card trick
When considering the stacking of a deck of 52
cards so that maximum overhang occurs, the total
amount of overhang achieved after sliding over 51
cards leaving the bottom one fixed is
In order to find the number of stacked books
required to obtain    book-lengths of overhang,
solve the      equation for   , and take the
ceiling function. For             , 2, ...
book-lengths of overhang, 4, 31, 227, 1674,
12367, 91380, 675214, 4989191, 36865412,
272400600, ... (Sloane's A014537) books are
needed.
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Gabriels Horn or Torricellis trumpet Infinite
surface but finite volume
To understand this for sense it is not required
that a man should be a geometrician or a
logician, but that he should be mad. Thomas
Hobbes
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The Painters Paradox

• Infinite amount of paint to cover the surface
• Finite amount to fill the interior volume
• You are not comparing like with like area?volume

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Picture of the y 1/x2
The area below the graph is much bigger than the
thin line of the graph above it so how can you
need less ink to colour the area that draw the
line
In practice, no pen could draw a line that had
zero thickness and so as you drew the curve off
to the right with a real pen you would eventually
find your line actually running into the
horizontal axis of the graph. It would therefore
only be finite in extent
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365-day rainfall, Dublin, Ireland
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Harmonic Things

• John D Barrow