How do you solve the following question: Show that set of real numbers xx, which satisfy the inequality sum_(k=1)^70 k/(x-k) ge 5/470∑k=1kx−k≥54 is a union of disjoint intervals, the sum of whose lengths is 1988?

1 Answer
Cesareo R.
Sep 13, 2016

See below.

Explanation:

f(x) = sum_(k=1)^70 k/(x-k)f(x)=70∑k=1kx−k is a function with 7070 vertical assymptotes located at 1,2,3, cdots, 701,2,3,⋯,70

p(x) = 5Pi_(k=1)^70(x-k) -4(sum_(k=1)^70 k Pi_(j ne k)^70(x-j)) is an associated polynomial to f(x) with 70 roots. Due continuity considerations, it's roots are interleaved with the assymptotes so the claimed sum of interval is

Delta = sum_(k=1)^70 r_k - sum_(k=1)^70 k where r_k are the roots of p(x) = 0

but sum_(k=1)^70 r_k = a_(69) the coefficient of x^69 in the p(x) polynomial. Computing a_69 gives

5a_69=-5sum_(k=1)^70 k-4 sum_(k=1)^70k = -9sum_(k=1)^70k

or

a_69 = -9/5 (70 xx71)/2 = -9 xx 7 xx 71

Finally, the sum of intervals gives

Delta =abs(- 63 xx 71 + 35 xx 71) = 28 xx 71 = 1988

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