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dadiw96

Zdanie 2 czerwone

Jun 5th, 2017
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Octave 0.32 KB | None | 0 0
  1. clear
  2. clc
  3. format long
  4. n=0;
  5. nmax=0;
  6. s=0;
  7. while 2^n/factorial(n)>=0.01
  8.     if nmax<2^n/factorial(n)
  9.         nmax=2^n/factorial(n);
  10.     end
  11.     n=n+1;
  12. end
  13.  
  14. for i=0:n
  15.     s=s+2^i/factorial(i);
  16. end
  17.  
  18. disp('najwieksza wartosc')
  19. disp(2^n/factorial(n))
  20. disp('indeks najwiekszej wartosci')
  21. disp(n)
  22. disp('suma')
  23. disp(s)
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