Do either (1) or (2):
(1) Prove that ∫1k [x] f '(x) dx = [k] f(k) - ∑1[k] f(n), where k > 1, and [z] denotes the greatest integer ≤ z. Find a similar expression for: ∫1k [x2] f '(x) dx.
(2) A particle moves freely in a straight line except for a resistive force proportional to its speed. Its speed falls from 1,000 ft/s to 900 ft/s over 1,200 ft. Find the time taken to the nearest 0.01 s. [No calculators or log tables allowed!]
[x] is constant over the interval [i, i+1) for i an integer, so we split the range of integration to get ∫1k = ∫[k]k + ∫12 + ∫23 + ... + ∫[k]-1[k]. We can write down each of these integrals, collect terms and get the result.
The same idea works the second integral, except that we divide at √2, √3, √4, ... , √[k2], giving the result: [k2] f(k) - ( f(1) + f(√2) + f(√3) + ... + f(√[k2]) ).
(2) Easy. 1.26 s.
The equation of motion is x'' = - k x'. Integrating: x' = k A e-kt. Integrating again, and putting x(0) = 0, x = A(1 - e-kt). Suppose T is the required time. Then from the speed e-kT = 0.9. x(T) = 1200, so A = 12000. The initial speed is 1000, so k = 1/12 and T = -12 ln 0.9. The only slight snag is that in the exam calculators did not exist and log tables were not allowed. So we have to use the expansion ln(1 + x) = x - x2/2 + x3/3 - ... , or more usefully: - ln(1 - x) = x + x2/2 + x3/3 + ... , giving T = 1.2 + 0.06 + 0.004 + 0.0003 + ... or 1.26 s.
2nd Putnam 1939
© John Scholes
4 Sep 1999