| z | ||
g(z)= | ||
| 2−z |
| z | z | 1 | z | z | ||||||
g(z)= | = | * ( | )= | * ∑k=0∞ ( | )k=∑k=0∞ | |||||
| 2−z | 2 | 1−z2 | 2 | 2 |
| z | ||
( | )k+1 | |
| 2 |
| 1 | ||
czyli nasz rozkład wynosi P(X=k)= ( | )k+1 dla k=0,1,2,..... | |
| 2 |
| 1 | ||
a ∑k=0∞ ( | )k+1 =1 a powinno wyjść ≠1 | |
| 2 |
?
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |