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?

1 Answer
Cesareo R.
Sep 13, 2016

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#

Related questions
See all questions in Summation Notation
Impact of this question
2354 views around the world
You can reuse this answer
Creative Commons License