Question

How do you solve the following question: Show that set of real numbers `x`, which satisfy the inequality `sum_(k=1)^70 k/(x-k) ge 5/4` is a union of disjoint intervals, the sum of whose lengths is 1988?

Answer 1

See below.

Explanation:

`f(x) = sum_(k=1)^70 k/(x-k)` is a function with `70` vertical assymptotes located at `1,2,3, cdots, 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`