100 Essential Things You Didn't Know You Didn't Know

by John D. Barrow

  100

  The Global Village

  Imagine all the people

  Sharing all the world.

  John Lennon

  Sometimes you can’t see the wood for the trees. Large numbers swamp our understanding. It’s hard to imagine a million, let alone a billion. Cutting things down to size helps us to make things more concrete and immediate. In 1990 a famous illustration of the state of the World was suggested,fn1 which asked us to imagine that the World’s population was scaled down to create a single village of 100 people, with all their attributes scaled down accordingly. What would that village be like?

  It would contain 57 Asians and 21 Europeans, 14 members from the Western hemisphere, both north and south, and just 8 Africans. There would be 70 villagers who were non-white and 30 who were white. Only 6 people would possess 59% of the entire village’s wealth and all 6 would be from the United States. Of the village’s 100 members, 80 would live in substandard housing, 70 would be unable to read, 50 would suffer from malnutrition, only one would own a computer, and only one would have had a college education. A strange village, isn’t it?

  fn1 This picture of a global village was first proposed in 1990 by Donatella Meadows and was based on a population of 1,000. After hearing Meadows talking on the radio, the environmental activist David Copeland tracked down Meadows and revised her statistics to reflect a village population of 100. He added it to a poster prepared for distribution to 50,000 people at the 1992 Earth Summit in Rio. Meadows’s original example of the State of the Village report was published in 1990 as Who Lives in the Global Village? A myth grew up that the originator of this example was a Stanford University professor, Philip Harter, but in fact he only forwarded an email of the Meadows and Copeland facts over the internet. See http://www.odtmaps.com/behind_the_maps/population_map/state-of-villagestats.asp

  Notes

  ‘Immersing myself in a book or a lengthy article used to be easy. My mind would get caught up in the narrative or the turns of the argument, and I’d spend hours strolling through long stretches of prose. That’s rarely the case any more. Now my concentration often starts to drift after two or three pages. I get fidgety, lose the thread, begin to look for something else to do. I feel as if I’m always dragging my wayward brain back to the text. The deep reading that used to come naturally has become a struggle.’

  Nicholas Carr

  1. The chain is assumed to be of uniform mass per unit length, is perfectly flexible and has zero thickness. The mathematical function cosh(. . .) is defined in terms of the exponential function by cosh(x) = (ex + e-x)/2.

  2. Assume the maximum acceleration you can exert is +A, and hence the maximum deceleration is – A, and you have to push the car from position x = 0 to x = 2D. You start at zero speed and finish at zero speed. If you apply +A for the distance from x = 0 to x = D, then it will take a time √(2D/A) to get to x = D and you will arrive there with speed √(2DA). If you immediately apply deceleration A to the car then it will return to zero speed at x = 2D after a period of deceleration lasting a time √(2D/A). As expected by the symmetry of the situation, this is exactly equal to the time taken to reach halfway. The total time to get the car in the garage is therefore 2√(2D/A).

  3. The random walk is a diffusion process governed by the diffusion equation. For the spread of y in time, t, and one dimension of distance, x, this is ∂y/∂t = K∂2y/∂x2, where K is a constant measure of how easy it is to diffuse through the medium. Hence, by inspecting the dimensions we have that y/t should be proportional to y/x2. Hence, t is proportional to x2 and we will need S2 steps to go the straight-line distance x = S.

  4. This length is known as the Planck length to physicists, after Max Planck, one of the pioneers of quantum theory, who first determined it. It is the only quantity with the dimensions of a length that can be formed from the three great constants of nature, c the speed of light, h, the quantum constant of Planck, and G, the constant of gravitation. It is equal to (Gh/c5)½ and is a unique reflection of the relativistic, quantum and gravitational character of the universe. It is a unit of length that is not picked out by human convenience and this is why it seems so small in terms of our everyday units.

  5. A simple example is the monkey puzzle where you have, say, 25 square pieces that each have 4 sides. On each side is half of a monkey (top or bottom). There are four colours of monkey and you have to put the 25 pieces together into a larger 5 × 5 square so that each border completes a monkey’s body by joining a top to a bottom of the same colour. How many alternative ways of putting the cards down would you have to search through in order to find the correct ‘answer’ to the puzzle in which all the half monkeys are joined to other pieces of the same colour that match the other part of their body? There are 25 ways to put down the first piece, followed by 24 ways for the next one, 23 ways for the next, and so on. Hence, there are 25 × 24 × 23 × 22 . . . × 3 × 2 × 1 = 25! ways to match up the 25 cards in this way. But there are 4 possible orientations for each card and this creates another 425 possibilities. The total number of configurations to be tried in order to find the correct pattern is therefore 25! × 425. This number is stupendously large. If we tried to write it out in full it would not fit on the page. If a computer were to search through this huge number of moves at a rate of one million per second it would still take more than 5,333 trillion trillion years to examine them all in order to find the right answer. For comparison, our Universe has been expanding for only 13.7 billion years.

  6. 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 occur only 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/(r+5) + . . . + r/(N-1)] ≈ 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 ] = 1, 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.

  7. More precisely, if we skip the first N/e applicants then our probability of finding the best one will approach 1/e as N becomes large, where e = 2.7182 . . . = 1/0.37, is the mathematical constant that defines the base of the natural logarithms.

  8. The chance that one person will not have your birthday is 364/365 and so, if there are G guests and their birthdays are independent, the probability that none of them will share your birthday is P = (364/365)G. Therefore the chance that there will be one of the guests who shares your birthday will be 1 - P. As G gets larger and larger P approaches zero and the chance of sharing your birth date with another guest approaches 1, and we can check that 1 - P exceeds 0.5 when G exceeds log(0.5)/ log(364/365) which is approximately 253.

  9. Again, it is easiest to start by working out the probability that they don’t share birthdays. If there are N people then this probability equals P = 365/365 × 364/365 × 363/365 × . . . × [365 – (N-1)]/365. The first term is from allowing the first person to choose their birthday freely (any one of the 365 days is allowed). The second term is then the fraction allowed for the second person if they are not to have the same birthday as the first – so they can only have 364 of the 365 possibilities. Then the third person can be on only 363 of the 365 days to avoid having the same birthday as either of the first two, and so on, down to the Nth person who can only choose 365 minus (N-1) of the 365 days to avoid having the same birthday as any of the other N-1 before him. So the chance that two of them do share a birthday is just 1 – P = 1 – N!/{365N(365 – N)!} which is bigger than 0.5 when N exceeds 22.

  10. If you alter the confidence probability from 95% to some other number, say P%, then the three intervals in the diagram have lengths ½×(1-P/100), P/100 and ½×(1-P/100), respectively
(check the previous answer is obtained when you put P = 95). Following the same reasoning as before, we see that at A the future is [P/100 + ½×(1-P/100)]/½×(1-P/100) = [100 + P]/[100 – P] times longer than the past. So, with P% probability something will last at least [100 – P]/[100 + P] times its present age, but at most [100 + P]/[100 – P] times that age. As P gets close to 100% the predictions get weaker and weaker because they have to be so much more certain. If we want 99% probability then it will last more than 1/199 times the present age but less than 199 times that age. On the other hand, if we let the certainty of being right drop to, say, 50% then the likely interval is 1/3 to 3 times the present age: a very narrow interval but the chance of it being correct is too low to worry about.

  11. A Maclaurin series is a series approximation to a function f of one variable x by a series of polynomial terms. Tamm would write f(x) = f(0) + xf’(0) + . . . xnf(n)(0)/n! + Rn where R n is the error, or remainder, after approximating f(x) by n of the preceding terms, where n = 1,2,3, . . . as required. The remainder term that Tamm was asked to deduce can be calculated to be Rn = ∫oxn f(n+1)(t)/n! dt. If the bandit leader was particularly awkward he might have required the further simplification of this expression that is possible by using the mean value theorem to obtain R n = xn+1 f(n+1)(y)/(n+1)!, where y is some value between 0 and x. Colin Maclaurin was a Scottish contemporary of Isaac Newton.

  12. This approximation is best for very small values of the interest rate r. If we want to make it more accurate, then we can approximate loge(1+r) by r–r2/2 and obtain the estimate that n = 0.7/r(1-½r). For the case of 5% interest rates, with r = 0.05, this gives the number of years for the investment to double as 14.36 instead of 14 years.

  13. This is a beautiful application of what is known as the ‘method of dimensions’ in physics. Taylor wants to know the radius, R, of a spherical explosion at a time t after it occurs (call that detonation time t=0). It is assumed to depend on the energy, E, released by the bomb and the initial density, , of the surrounding air. If there exists a formula R = kEaρbtc, where k, a, b and c are numbers to be determined, then, because the dimensions of energy are ML2T-2, of density are ML-3, where M is mass, L is length and T is time, we must have a = 1/5, b = -1/5 and c = 2/5. The formula is therefore R = kE1/5ρ-1/5t2/5. Assuming that the constant k is fairly close to 1, we see that the unknown energy is given approximately by E = ρR5/t2. By comparing several pictures you can determine k as well.

  14. It’s an elephant!

  15. Any 3-digit number ABC can be written out as 100A + 10B + C. The first step is to take away its reverse, that is 100C + 10B +A. The result is 99|A-C|, where the straight brackets mean take the magnitude with a positive sign. Now the magnitude of A-C must lie between 2 and 9 and so when you multiply it by 99 you get one of the multiples of 99 that has three digits. There are only 8 of these that are possible answers: the numbers 198, 297, 396, 495, 594, 693, 792 and 891. Notice that the middle digit is always a 9 and the other two digits always add up to 9 in each of these numbers. Therefore, when you add any one of them to its reverse you always end up with the answer 1089.

  16. If the Leader spoke the truth with probability p then the probability that his statement was indeed true is Q = p2/[p2 + (1-p)2]. In our case, where p = ¼, this gives a probability of Q = 1/10. You can check that Q is less than p whenever p < ½ and Q exceeds p whenever p > ½. Notice that when p = ½ that Q = ½ also.

  17. For a more detailed discussion see J. Haigh, Taking Chances, Oxford UP, Oxford (1999), chap. 2. The essential quantity required to work out any result is that the odds for matching r of the drawn numbers is 1 in 49C6/[6Cr43C6-r]where nCr=n!/(n-r)!r! is the number of different combinations of r different numbers that can be chosen from n numbers.

  18. If you create the Fibonacci sequence of numbers (in which each number from the third onwards is the sum of the two before it), 1,1,2,3,5,8,13,21,34,55,89 and so on forever, with the nth number in the sequence labelled as Fn, then you can generalise the example here by rearranging an Fn × Fn square into an Fn-1 × Fn+1 rectangle. The example we have drawn is the n = 6 case as F6 = 8. The area difference between the square and the rectangle is given by Cassini’s identity, (Fn × Fn) − (Fn-1 × Fn+1) = (–1)n+1. So when n is an even number the right-hand side equals -1 and the rectangle is bigger than the square; but when n is an odd number we ‘lose’ an area of 1 when we go from the square to the rectangle. Remarkably the difference is always +1 or -1 no matter what the value of n. This puzzle was invented by Lewis Carroll and became popular in Victorian England. See S. D. Collingwood, The Lewis Carroll Picture Book, Fisher Unwin, London (1899) pp. 316–7.

  19. For more information see Donald Saari, ‘Suppose You Want to Vote Strategically’, Maths Horizons, November 2000, p. 9.

  20. Tolkowsky showed that for total internal reflection of incoming light to occur at the first face it strikes, the inclined facets must make an angle of more than 48º 52’ with the horizontal. After the first internal reflection the light will strike the second inclined face and will be completely reflected if the face makes an angle of less than 43º 43’ with the horizontal. In order for the light to leave along a line that is close to vertical (rather than close to the horizontal plane of the diamond face) and for the best possible dispersion in the colours of the outgoing light, the best value was found to be 40º 45’. Modern cutting can deviate from these values by a small amount in order to match properties of individual stones and variations in style.

  21. That is for any two quantities x and y it is always true that ½ (x+y (xy)½, which is a consequence of the fact that the square (x½–y½)2≥0. This formula also displays an attractive feature that the geometric mean possesses that the arithmetic mean does not. If the quantities x and y were measured in different units ($ per ounce and £ per kilogram, for instance) then you could not compare the value of ½ (x+y) at different times meaningfully. To do that, you need to make sure that x and y are both expressed in the same units. However, the geometric mean quantity (xy)½ can be compared in value at different times even if different units are being used to measure x and y.

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