John Von Neumann the World War II era classic mathemician was considered by many to be one of the most brilliant minds of the twentieth century. He reportedly had an IQ of 180. He was a pioneer of Game Theory, which was very important during the nuclear arms race. (Because GT assumes that all players act in their own best enlightened self-interest, GT turned out to be a much better model for evolutionary biology than for human behavior.) He was also one of the two people (Alan Turing being the other) who is credited with being the father of the modern computer.

       Von Neumann the following postulated and would expound the following mathematical puzzle as a problem to solve:

       Two trains are 20 miles apart on the same track heading towards each other at 10 miles per hour, on a collision course. At the same time, a bee takes off from the nose of one train at 20 miles per hour, towards the other train. As soon as the bee reaches the other train, it bangs huwey and heads off at 20 miles per hour back towards the first train. It continues to do this until the trains collide, killing the bee.

     








      The question is, how far does the bee fly (d) before the collision?

      This is a pretty straight-forward problem for anyone who has studied Mathematical Analysis or Pre-Calculus. The initial separation is D = 20 miles (from here on out I will use the variable D, then substitute the value of 20 miles at the end. I think this makes it easier to follow the steps.) The bee is traveling twice as fast as each of the trains, therefore covering twice the distance as the approaching train before the first turn-around. So the distance d1 that the bee travels on the first leg, plus the distance the approaching train travels (one half d1) equals D.
Therefore d1 = 2/3 D.

       As the bee begins leg #2, not only has the turn-around train covered the distance 1/3 D, but so has the train the bee is now heading towards. That means the remaining free distance D' = 1/3 D. By the same argument as above, the bee covers two thirds of this distance on his second leg.
Therefore d2 = 2/3 D' = (2/3)*(1/3)*D.

      For leg #3, by the above argument, the remaining distance D'' = 1/3 D', which the bee covers two thirds of.
Therefore d3 = 2/3 D'' = (2/3)*(1/3)*D' = (2/3)*(1/3)*(1/3)*D.

       I hope you all can see the pattern that's developing. Each trip gets one third shorter. Symbolically speaking, if we move the 2 and the D out to the beginning, we get 2D*(1/3)*(1/3)*... with the number of 1/3's being equal to the leg#. Since this is the same as raising (1/3) to the exponent equal to the leg#, we get
dn = 2D/3^n.

       This means that on the nth leg of its journey, the bee flies 2D times one third raised to the power n (or conversely, 2D divided by three raised to the power n). Now that we know how far the bee travels on each leg, it's time to find out the total distance. The distance the bee travels after n legs (d'n) equals

          d'n = d1 + d2 + ... + dn, which equals
         d'n = 2D/3 + 2D/3^2 + ... + 2D/3^n, which when we factor out the 2D we get
         d'n = 2D*(1/3 + 1/3^2 + ... + 1/3^n), which in mathematical notation is written as
         d'n = 2D * (x=1 --> n)Σ[1/3^x]

         ." The expression inside the brackets is what is being summed. The expression inside the parenthesis tells you over what values you perform the summation. So for the above expression, that means we will be adding up the values of 1/3^x for the values of x=1 through x=n. In other words, 1/3^1 + 1/3^2 + ... + 1/3^n.

        There may be a formula for solving that sum; I may even have been taught that formula. But unfortunately, that's not the kind of knowledge that I tend to retain. Therefore we will solve this using the algorithm I remember and can (kind of) explain. What I will do is sum up the first few terms, look for a pattern, then use induction to prove it. Hopefully my explanation will at least make an inkling of sense to those of you who are completely lost right now. So here we go.

           1/3 =1/3
           1/3 + 1/9 = 4/9
           1/3 + 1/9 + 1/27 = 13/27
           1/3 + 1/9 + 1/27 + 1/81 = 40/81


          The pattern I see here is that the numerator (top part of the fraction) is half of the denominator (bottom part of the fraction) minus one. In other words,

         1/2*(3-1)=1
         1/2*(9-1)=4
         1/2*(27-1)=13
         1/2*(81-1)=40

         Since the denominator is just 3^n, then the numerator is 1/2*(3^n - 1).

        So the formula that I need to test with induction is

            (x=1 --> n)Σ[1/3^x] =? 1/2*(3^n - 1)/3^n

        Before I proceed further, let's do a quick refresher on mathematical induction. The basic idea is that you prove a general expression by first proving that it is consistent, ie. IF it's true for some specific case, then it must necessarily be true for some general case. Then prove that it's true for that specific case, and you're done.

       The normal route is to assume that an expression is true for some value x=n, then attempt to prove that it must necessarily be true for x=n+1. Which of course means it's also true for n+2 (substitute n+1 for n), n+3 and all x greater than n. Now if you prove it's true for x=n=1, then it must be true for all n. Since we've already proven our formula true for n=1 (and 2 and 3 and4), we only need to prove that if it's true for x=n, then it must also be true for x=n+1.

We have

         1. (x=1 --> n)Σ[1/3^x] = 1/2*(3^n - 1)/3^n

which we are assuming to be true for x=n, and trying to see if

          2. (x=1 --> n+1)Σ[1/3^x] =? 1/2*(3^{n+1} - 1)/3^{n+1}

    is also true.   If we recall that adding 1 to an exponent is the same as multiplying by the base

          3. 3^{n+1}=3*3^n, we get

          4. (x=1 --> n+1)Σ[1/3^x] =? 1/2*(3*3^n - 1)/3*3^n

      Going back to the meaning of Σ as summation, summing up to x=n+1 is the same as summing up to n, then adding the next term where x=n+1. In other words,

           5. (x=1 --> n+1)Σ[1/3^x] = (x=1 --> n)Σ[1/3^x] + 1/3^{n+1}

     Substituting statements 1. and 3. into 5. we get

           6. (x=1 --> n+1)Σ[1/3^x] = 1/2*(3^n - 1)/3^n + 1/3*3^n
= 3*(3^n - 1)/2*3*3^n + 2/2*3*3^n
=(3*3^n - 3 +2)/2*3*3^n = (3*3^n - 1)/2*3*3^n = 1/2*(3*3^n - 1)/3*3^n
which is the expression we were trying to prove in statement 4.

       Since we've already proven it for n=1, then expression 1. has now been proven true for all n.

           1. (x=1 --> n)Σ[1/3^x] = 1/2*(3^n - 1)/3^n

       Back to our friend the bee. We now have an expression for how far the bee flies after n legs

          7. d'n = 2D * 1/2*(3^n - 1)/3^n = D*(3^n - 1)/3^n

and we need to solve it for how far the bee flies before dying. In actuality, the bee will stop flying (okay, "in actuality" this would never happen) when the distance between the trains is less than the body length of the bee. However, since the summation quickly converges on the solution, we can assume that the bee is a point, ignore the famous paradox, and do the summation up to n=infinity.

          7. d'n = D*(3^n - 1)/3^n = D*(3^n/3^n - 1/3^n) = D*(1 - 1/3^n)

Quick refresher on infinite limits: as we let n get infinitely large, 3^n approaches infinity, and 1/3^n approaches zero. Therefore the limit as n approaches infinity is

          d = limit(n-->infinity)[D*(1 - 1/3^n)] = D*(1-0) = D = 20 miles!

       We have just solved this problem by the infinite series method. Infinite series are very important in Mathematical Analysis and Pre-Calculus as they form the basis for derivatives and integration and everything which is Calculus and Differential Equations. That's why all students of Math, Physics and Engineering get to do a ton of infinite series problems before they graduate. The reason I call this problem the "Mathematician's Trap," is because virtually all mathematicians who see this problem will try to solve it the way we just did.

       However, if you were to give the problem to someone who's only had basic Algebra, they might solve it differently. The trains crash at the midway point which is at 10 miles. Since each train is going 10 mph, this takes one hour. During that same hour, the bee is flying at 20 mph, therefore the bee flies 20 miles! Wow, that was a lot easier!

       (note: Some of you may have noticed that for the infinite series solution, I didn't use the respective speeds of the bee and the trains, but only their ratio. This would also work for the algebraic solution, but it would make the math a little more complicated, which rather defeats the purpose of what I was trying to prove. In fact, as a general case, as long as the speeds of the two trains adds up to the speed of the bee, the distance traveled by the bee will always be the initial separation of the trains, ie. d=D. You can see this in the following thought experiment. Since the speeds of the trains added together equals the speed of the bee, at any given moment in time, the distance covered by the two trains together is equal to the distance covered by the bee. That means that at any given moment in time, the available flying space left is equal to what would be the remaining distance if the bee were flying unimpeded from point A to point B. Therefore d=D. Think about it.)

       The moral of the story is that being smarter or better educated can often times put you at a disadvantage. When someone is trained at doing something a certain way, that action is virtually automatic. It takes great insight to be able to "step outside of the box" and ask if there's an easier way to do it. (I'm currently working on another post on just this subject which should hopefully be up soon.) Even brilliant mathematicians will fall into the trap, which brings us back to John Von Neumann.

      When posed with the above problem (or some variation of it), JVN took all of five to ten seconds to come up with the correct solution. This floored the questioner who said "I'm impressed that you didn't fall for the Mathematician's Trap." After getting a perplexed look from our genius, he asked "How did you solve the problem?"

     "By infinite series, of course!"