MathLabs

Problem 2

Let f(x)f(x) and g(x)g(x) be given by f(x)=1x+1x−2+1x−4+⋯+1x−2018f(x)=\dfrac1x+\dfrac1{x-2}+\dfrac1{x-4}+\cdots+\dfrac1{x-2018} and g(x)=1x−1+1x−3+1x−5+⋯+1x−2017g(x)=\dfrac1{x-1}+\dfrac1{x-3}+\dfrac1{x-5}+\cdots+\dfrac1{x-2017}. Prove that ∣f(x)−g(x)∣>2|f(x)-g(x)|>2 for any non-integer real number xx satisfying 0<x<20180<x<2018.
Step 2 of 6: Use the half-turn symmetry
f(2018−x)=f(x),g(2018−x)=g(x)f(2018-x)=f(x),\qquad g(2018-x)=g(x)
Detailed analysis

Replacing xx by 2018−x2018-x permutes the shifts 2k↦2018−2k2k\mapsto 2018-2k and 2k+1↦2017−2k2k+1\mapsto 2017-2k among themselves, so both sums are invariant. Consequently the interval type 2n<x<2n+12n<x<2n+1 maps to the interval type 2m−1<x<2m2m-1<x<2m with m=1009−nm=1009-n, so it suffices to treat only intervals of the form 2n−1<x<2n2n-1<x<2n for 1≤n≤10091\le n\le 1009.