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John D Barrow

Harmonic Things. John D Barrow. Geometric series. S(n) = a + ar + ar² + ar³ +...+ arⁿ⁻¹, -1 <r< 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

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John D Barrow

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  1. Harmonic Things John D Barrow

  2. Geometric series S(n) = a + ar + ar² + ar³ +...+ arⁿ⁻¹, -1 <r< 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

  3. Behold….. 1/2 + 1/4 + 1/8 + 1/16 + 1/32 +........ = 1 1 1

  4. The value of your investments can plummet as well as go down

  5. 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)

  6. 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 > 1 Recall that 1 1/x dx =  but 1 1/xp dx = 1/(p-1) for p>1

  7. 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 > 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 > 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

  8. But H goes to infinity very slowlyH(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.

  9. 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)

  10. 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

  11. H(100) = 5.19

  12. Central England Temperature Record

  13. 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.

  14. 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 > 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

  15. Collecting Sets How many cards should you expect to buy in order to collect the set of 50 ?

  16. 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 haven’t 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 haven’t 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

  17. 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)] It’s 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

  18. Swopping • Suppose you have F friends and you all pool cards in order to build up F+1 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 (F+1)N[ln(N) + 0.58] cards to complete F+1 separate sets. • For N = 50 the number of card purchases saved would be 156F. Even with F = 1 this is a considerable economy.

  19. “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?

  20. 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 ?

  21. 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.

  22. The Situation with N Candidates • Note that if the best candidate is in the (r+1)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 (r+2)st position we will pick them with a chance 1/N  (r/r+1). 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/(r+1) + r/(r+2) + r/(r+3) + r/(r+4) + ….+ r/(N-1)] • P(N, r)  1/N[1 + 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 !

  23. 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

  24. 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 1+1/3 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 1+1/3+1/5 units • 4 jeeps allow jeep 1 to go 1+1/3+1/5+1/7 units

  25. 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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