jukaukor

keskiarvojono.jl

Apr 26th, 2023
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  1. function keskiarvojono(n,k)
  2. # aluksi a= 1,2,..k
  3. # sitten seuraava on k:n edellisen keskiarvo jne
  4. # a[k+1] = 2,3,...k,ak jne
  5. a = rand(Float64,n) # arvotaan satunnaisjono, n lukua
  6. # alkuu lukujono 1,2,..,k
  7. for i = 1:k
  8. a[i] = i
  9. end
  10. # tulostetaan jonon alun k termiΓ€
  11. for i= 1:k
  12. println(a[i])
  13. end
  14. for i = k+1 : n
  15. a[i] = sum(a[i-k:i-1])/k # edellisten k termin keskiarvo
  16. println(a[i]) # keskiarvotermi
  17. end
  18. println("jonon teoreettinen raja-arvo = (2k+1)/3 = ",(2*k+1)/3)
  19. end
  20.  
  21.  
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