John von Neumann and the fly

The story was that John von Neumann was asked about the following problem :

  • 2 bikes are headed toward each other at the same speed 60 km/h, they are separated each other by 2 kilometers. A Fly leaves the front of the first bike and travels at 90 km/h (a very fast one) towards the second one then back to the first bike and so on
  • how far does the fly travel ?

i am going to explain the methodical way used by  John von Neumann :

Bikes are traveling at the speed 1 km/minute and the fly 1.5 km/minute

let look at the fist trip when the fly collide with the second bike before going back to the first bike:

we call d(Fly), the distance corresponding to the travel of the fly, and d(Bike 2) the one corresponding to the Bike 2 :

we have  d(Fly) + d(Bike 2) = d(Total) = 2 

at time t, they collide : t *  ( V(Fly) + V(Bike 2) )d(Total)

then td(Total) / ( V(Fly) + V(Bike 2) )  = d(Total) / ( 1 + 1.5 ) =  2/5 * d(Total)

then it comes  d(Fly) = (1.5 * d(Total)) * 2/5  = 6/5  (1)

and d(Bike 2) =  d(Total)) * 2/5 = 4/5 = d(Bike 1)  (2)

Now for the second trip the distance to go back to the first bike leaving the front of second bike is  d(Fly) –  d(Bike 1)   = 2/5

but we don’t have to redo the calculation again, we have to infer 2/5 in d(Total) for equation (1) and (2) which is 1/5 of total distance ( 2 km ),  then  d(Fly) = 6/ 5*5  and so on  :

then we can extract an infinite serie for the fly travel distance :

d(Fly n steps) = 6/5 ( 1 + 1/5 + 1/(5*5 ) + 1/ (5*5*5) + … + 1/(5*5*5 *   n times))

d(Fly) = d(Fly n steps) when n → ∞

d(Fly) = 6/5 * 1/(1 – 1/5) = 3/2 km

 

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