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Harmonic Things

John D Barrow

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

Behold..

1/2 1/4 1/8 1/16 1/32 ........ 1

1

1

The value of your investments can plummet as well

as go down

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)

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

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

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.

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)

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

H(100) 5.19

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Central England Temperature Record

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.

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

Collecting Sets

How many cards should you expect to buy in order

to collect the set of 50 ?

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

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

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.

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?

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 ?

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.

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 !

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

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

House of Cards

Max overhang in book lengths is

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.

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

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

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

365-day rainfall, Dublin, Ireland

Harmonic Things

- John D Barrow