Trigonometric Complex Functions
Algebra
Level
3
Let f:ℝ → ℝ such that 4f(x+y) = (i sin (yx) + cos (yx))^(1/y) (cos(x) – i*sin(x)). Let σ(k)= Σ((x-f(x))^2) from x=1 to k. Find [σ(1000)]+1, where [x] is greatest integer function.
333583313
333583311
333582313
333573312
333583312
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1 solution
By de Moivre's theorem, 4f(x+y) turns out to be (cosx)^2 - i^2 (sinx)^2 = (cosx)^2 + (sinx)^2=1. So, f(k)=1/4 ∀ k∈ ℝ. Therefore, σ(1000)= Σ((x-1/4)^2)= Σx^2 -1/2Σx + 1000 (1/16) = 333833500 - 250250 + 62.5= 333583312.5. Therefore, [σ(1000)] +1 = 333583313.