∑ω=1∞(1/ω)sin(ωt) = Im∑ω=1∞(1/ω)ejωt
= Im[- ln(1 - ejt)*] =
Im{- ln[1 - cos(t) - jsin(t)]**} = - atg{- sin(t) / [1 -
cos(t)]} =
atg{sin(t) / [1 - cos(t)]} = atg[ctg(t/2)***] =
atg[tg(π/2 - t/2)] = (π - t) / 2
----------
* ln(1 - x) = - ∑n=1∞(1/n)xn; - 1
≤ x < 1
- 1 ≤ ejt
< 1 → 0 ≤ ejt + 1 < 2 → 0 < t ≤ π → π/2 > ∑ω=1∞(1/ω)sin(ωt)
≥ 0
** ln(rejφ = a + jb) = ln(r) + lnejφ
= ln√(a² + b²) + jφ = (1/2)ln(a² + b²) + jatg(b/a)
*** tg(φ/2) = [1 - cos(φ)] / sin(φ); tg(π/2 - φ) = ctg(φ)
0 ≤ ∑ω=1∞(1/ω)sin[ω(π - t)] = t/2
< π/2; 0 ≤ t < π ↔* ∫0∞(1/ω)sin(ωt)dω = π/2; t >
0
----------
* f(x1) =
f(x2); x2 > x1 → (x0)
"π" (x0+); x2 = x1 + Δx
Допълнението към статията ми "Дяволите да ни вземат", pdf: link
